chapter 1 to 11 final
Deviation from the mean is the conventional form of variation. It measures the deviation from a common central tendency. It can tell us how an ________________ as well us how accurate this set of scores is to a population. When there is a small vulnerability then it is a good representation because it means that more scores are contained in an area. (10.02.2014, stats lec 4 variability)
individual stands in relation to the other scores
What are the differences between interval and ratio scales?
interval and ratio scales are categorized from a series of intervals and go up in size. Interval scale has a non zero starting point and is most often used for stuff like C and F temperature while ratio starts at 0.
The estimated standard error (sm)
is used as an estimate of the real standard error, QM, when the value of a is unknown. It is computed from the sample variance or sample standard deviation and provides an estimate of the standard distance between a sample mean, M, and the population mean, u.
The t statistic
is used to test hypotheses about an unknown population mean, u, when the value of a is unknown. The formula for the t statistic has the same structure as the z-score formula, except that the t statistic uses the estimated standard error in the denominator
A repeated-measures design, or a within-subject design, is one in which the dependent variable is
measured two or more times for each individual in a single sample. The same group of subjects is used in all of the treatment conditions.
The second is the Interquartile range of 25th to 75th percentile. This is a better representation of the outliers, but may not give the entire picture of the variables. To achieve this you look for the the second quarter of the value. The general idea is (3rd quarter) - (1st quarter). A semi interqualtile range is the ________________ which is defined by the middle 50%. (10.02.2014, stats lec 4 variability)
middle of distribution to the boundary
Describe sampling error
occurs where there is a difference between sample statistics and the value in relation to the actual population.
A researcher is interested in the texting habits of UofT. If he measures the number of text message that each individual sends each day and calculates the average number of the entire group, the average number would be an example of
parameter
The third is or called the standard variation which is the average squared distance from the mean. This involves finding the distance of each integer of the mean, square each distance, find the average of the result and then square rooting it. In the situation of a sample you would _______________________ because of a concept called degrees of freedom which mainly is that there is a tendency for deviations to be undervalues and needs to be accounted for. With this method you can achieve what is called one standard deviation from the mean which is 68% of the score, while 2 standard deviation involves 95%. (10.02.2014, stats lec 4 variability)
subtract 1 from the number dividing for the mean of the squared numbers
A sample statistic is _______ if the average value of the statistic is equal to the population parameter. (The average value of the statistic is obtained from all the possible samples for a specific sample size, n.) A sample statistic is biased if the average value of the statistic either underestimates or overestimates the corresponding population parameter. (10.03.2014 Stats ch 4 textbook extraction/notes)
unbiased
The third is or called the standard variation which is the average squared distance from the mean. This involves finding the distance of each integer of the mean, square each distance, find the average of the result and then square rooting it. In the situation of a sample you would subtract 1 from the number dividing for the mean of the squared numbers because of a concept called degrees of freedom which mainly is that there is a tendency for deviations to be __________________. With this method you can achieve what is called one standard deviation from the mean which is 68% of the score, while 2 standard deviation involves 95%. (10.02.2014, stats lec 4 variability)
undervalues and needs to be accounted for
Though central tendencies tell us good information, it is missing deacriptions of how many people are close to the average or whether they are scattered throughout the scores. Measures of ________ therefore is used to find how variable terms alongside basic descriptive statistics such as mean, medium, or mode. (10.02.2014, stats lec 4 variability)
variability
What is three major limitations of what a hypothesis test can tell us and what measure is used to try to make up for it?
(1) Hyp test focuses on data rather than on the hypothesis (2) Significant effect does not always equal substantial (3) Given enough researchers there will always be the probability that some samples commit type 1 error leading to a waste of resources when studies are done on those studies.
3 Assumptions of an independent t-test
(1) Observation of each sample must be independent (2) The two populations from which the sample is selected must be normal (3) The population from which the samples are selected must have equal variances "homogenize of variance""
1. The city school district is considering increasing class size in the elementary schools. However, some members of the school board are concerned that larger classes may have a negative effect on student learning. In words, what would the null hypothesis say about the effect of class size on student learning?1. The city school district is considering increasing class size in the elementary schools. However, some members of the school board are concerned that larger classes may have a negative effect on student learning. In words, what would the null hypothesis say about the effect of class size on student learning?
. The null hypothesis would say that class size has no effect on student learning.
1. Find the mean for the following sample of n = 5 scores: 1, 8, 7, 5, 9,
1. /X =30 and M= 6 2. a= 48
1. Define a Type I error.
1. A Type I error is rejecting a true null hypothesis—that is, saying that the treatment has an effect when, in fact, it does not
1. Under what circumstances is a t statistic used instead of a z-score for a hypothesis test?
1. A z-score is used when the population standard deviation (or variance) is known. The t statistic is used when the population variance or standard deviation is unknown. The t statistic uses the sample variance or standard deviation in place of the unknown population values.
1. Describe the basic characteristics of an independent measures, or a between-subjects, research study.
1. An independent-measures study uses a separate sample for each of the treatments or populations being compared.
1. If all other factors are held constant, an 80% confidence interval is wider than a 90% confidence interval. (True or false?)
1. False. Greater confidence requires a wider interval
1. Adding a new score to a distribution always changes the mean. (True or false?)
1. False. If the score is equal to the mean, it does not change the mean.
. Under what circumstances is the normal distribution an accurate approximation of the binomial distribution?
1. When pn and qn are both greater than 10
1. Find the median for each distribution of scores: a. 3, 4, 6, 7, 9, 10, 11 b. 8, 10, 11, 12, 14, 15
1. a. The median is X= 7. b. The median is X= 11.5.
1. A local hardware store has a "Savings Wheel" at the checkout. Customers get to spin the wheel and, when the wheel stops, a pointer indicates how much they will save. The wheel can stop in any one of 50 sections. Of the sections, 10 produce 0% off, 20 sections are for 10% off, 10 sections for 20%, 5 for 30%, 3 for 40%, 1 for 50%, and 1 for 100% off. Assuming that all 50 sections are equally likely, a. What is the probability that a customer's purchase will be free (100% off)? b. What is the probability that a customer will get no savings from the wheel (0% off)? c. What is the probability that a customer will get at least 20% off?
1. a. p = 1/50 = 0.02 b. p = 10/50 = 0.20 c. p = 20/50 = 0.40
1. A survey of the students in a psychology class revealed that there were 19 females and 8 males. Ot the 19 females, only 4 had no brothers or sisters, and 3 of the males were also the only child in the household. If a student is randomly selected from this class, a. What is the probability of obtaining a male? b. What is the probability of selecting a student who has at least one brother or sister? c. What is the probability of selecting a female who has no siblings?
1. b. 20/27 c. 4/27
1. Which measure of central tendency is most affected if one extremely large score is added to a distribution? (mean, median, mode)
1. mean
1. For a sample with a mean of M = 40 and a standard deviation of s = 12, find the z-score corresponding to each of the following X values. X = 43 X = 58 X = 49 X = 34 X = 28 X= 16 (10.13.2014 stats ch 6, zscores)
1. z = 0.25 z = 1.50 z = 0.75 z= —0.50 z= —1.00 z= —2.00
10. A student was asked to compute the mean and standard deviation for the following sample of n = 5 scores: 81, 87, 89, 86, and 87. To simplify the arithmetic, the student first subtracted 80 points from each score to obtain a new sample consisting of 1, 7, 9, 6, and 7. The mean and standard deviation for the new sample were then calculated to be M = 6 and s = 3. What are the values of the mean and standard deviation for the original sample? (10.03.2014 Stats ch 4 textbook extraction/notes)
10. The mean is M = 86 and the standard deviation is s = 3.
10. A sample of n = 8 scores has a mean of M = 10. If one new person with a score of X = 1 is added to the sample, what is the value for the new mean?
10. The original sample has n = 8 and ΣX = 80. The new sample has n = 9 and ΣX = 81. The new mean is M = 9.
13. A score that is 12 points above the mean corresponds to a z-score of z = 3.00. What is the population standard deviation? (10.13.2014 stats ch 6, zscores)
13. σ = 4
14. A psychologist is investigating the hypothesis that children who grow up as the only child in the household develop different personality characteristics than those who grow up in larger families. A sample of n = 30 only children is obtained and each child is given a standardized personality test. For the general population, scores on the test from a normal distribution with a mean of p. = 50 and a standard deviation of if = 15. If the mean for the sample is M = 58, can the researcher conclude that there is a significant difference in personality between only children and the rest of the population? Use a twotailed test with a = .05.
14. H0: μ = 50. The critical region consists of z-scores beyond z = ±1.96. For these data, σM = 2.74 and z = 2.92. Reject H0 and conclude that only children are significantly different.
14. Researchers have noted a decline in cognitive functioning as people age (Bartus, 1990). However, the results from other research suggest that the antioxidants in foods such as blueberries may reduce and even reverse these age-related declines (Joseph et al., 1999). To examine this phenomenon, suppose that a researcher obtains a sample of n = 16 adults who are between the ages of 65 and 75. The researcher uses a standardized test to measure cognitive performance for each individual. The participants then begin a 2-month program in which they receive daily doses of a blueberry supplement. At the end of the 2-month period, the researcher again measures cognitive performance for each participant. The results show an average increase in performance of MD = 7.4 with SS = 1215. a. Does this result support the conclusion that the antioxidant supplement has a significant effect on cognitive performance? Use a two-tailed test with a = .05. b. Construct a 95% confidence interval to estimate the average cognitive performance improvement for the population of older adults.
14. a. The null hypothesis states that the antioxidant has no effect on cognitive performance. For these data, the estimated standard error is 2.25, and t(15) = 3.29. With df = 15 the critical value is 2.131. Reject the null hypothesis, the antioxidant has a significant effect. b. For 95% confidence use t = 2.131 and the interval extends from 2.61 to 12.19.
15. The distribution of scores on the SAT is approximately normal with a mean of p = 500 and a standard deviation of cr = 100. For the population of students who have taken the SAT, a. What proportion have SAT scores greater than 700? b. What proportion have SAT scores greater than 550? c. What is the minimum SAT score needed to be in the highest 10% of the population? d. If the state college only accepts students from the top 60% of the SAT distribution, what is the minimum SAT score needed to be accepted?
15. a. z = 2.00, p = 0.0228 b. z = 0.50, p = 0.3085 c. z = 1.28, X = 628 d. z = -0.25, X = 475
15. For the following set of scores Scores: 5, 6,8, 9,5,5,7,5,6,4,6,6,5,7,7,5,4,7,6, 5 a. Place the scores in a frequency distribution table. b. Identify the shape of the distribution.
15. a. ──── X f ──── 9 1 8 1 7 4 6 5 5 7 4 2 ──── b. positively skewed
16. A population of scores forms a normal distribution with a mean of p = 40 and a standard deviation of o- = 12. a. What is the probability of randomly selecting a score less than X = 34? b. What is the probability of selecting a sample of n = 9 scores with a mean less than M = 34? c. What is the probability of selecting a sample of n = 36 scores with a mean less than M = 34?
16. a. z = -0.50 and p = 0.3085 b. σM = 4, z = -1.50 and p = 0.0668 c. σM = 2, z = -3.00 and p = 0.0013
16. The distribution of SAT scores is normal with |i = 500 and ct = 100. a. What SAT score, X value, separates the top 15% of the distribution from the rest? b. W hat SAT score, X value, separates the top 10% of the distribution from the rest? c. What SAT score, X value, separates the top 2% of the distribution from the rest?
16. a. z = 1.04, X = 604 b. z = 1.28, X = 628 c. z = 2.05, X = 705
16. For a sample with a mean of p, = 45, a score of X = 59 corresponds to z = 2.00. What is the sample standard deviation? (10.13.2014 stats ch 6, zscores)
16. s = 7
16. Place the following scores in a frequency distribution table. Based on the frequencies, what is the shape of the distribution? 5, 6, 4, 7, 7, 6, 8, 2, 5, 6 3, 1, 7, 4, 6, 8, 2, 6, 5, 7
16. ──── X f ──── 8 2 7 4 . 6 5 5 3 4 2 3 1 2 2 1 1 ──── Negatively skewed
17. A population of N = 16 scores has a mean of p. = 20. After one score is removed from the population, the new mean is found to be p. = 19. What is the value of the score that was removed? (Hint: Compare the values for /X before and after the score was removed.)
17. The original sample has n= 16 and ΣX= 320. The new sample has n= 15 and ΣX= 285. The score that was removed must be X= 35.
18. One of the primary advantages of a repeated-measures design, compared to independent-measures, is that it reduces the overall variability by removing variance caused by individual differences. The following data are from a research study comparing two treatment conditions.
18. a. The pooled variance is 6.4 and the estimated standard error is 1.46. b. For the difference scores the variance is 2 and the estimated standard error is 0.58.
18. At the end of the spring semester, the Dean of Students sent a survey to the entire freshman class. One question asked the students how much weight they had gained or lost since the beginning of the school year. The average was a gain of p = 9 pounds with a standard deviation of o- = 6. The distribution of scores was approximately normal. A sample of n = 4 students is selected and the average weight change is computed for the sample. a. What is the probability that the sample mean will be greater than M = 10 pounds? In symbols, what is p(M > 10)? b. Of all of the possible samples, what proportion will show an average weight loss? In symbols, what is p(M < 0)? c. What is the probability that the sample mean will be a gain of between M = 9 and M = 12 pounds? In symbols, what is p(9 < M < 12)?
18. a. z = 0.33 and p = 0.3707 b. z = 3.00 and p = 0.0013 c. p (0 < z < 1.00) = 0.3413
19. Calculate SS, variance, and standard deviation for the following sample of n = 5 scores: 9, 6, 2, 2, 6. (Note: The definitional formula works well with these scores.) (10.03.2014 Stats ch 4 textbook extraction/notes)
19. SS = 36, the sample variance is 9, and the standard deviation is 3.
2. As the power of a test increases, what happens to the probability of a Type II error?
2. As power increases, the probability of a Type II error decreases
2. If the alpha level is increased from a = .01 to a = .05, then the boundaries for the critical region move farther away from the center of the distribution. (True or false?)
2. False. A larger alpha means that the boundaries for the critical region move closer to the center of the distribution.
2. The homogeneity of variance assumption requires that the two sample variances be equal. (True or false?)
2. False. The assumption is that the two population variances are equal.
2. How does sample size influence the outcome of a hypothesis test and measures of effect size? How does the standard deviation influence the outcome of a hypothesis test and measures of effect size?
2. Increasing sample size increases the likelihood of rejecting the null hypothesis but has little or no effect on measures of effect size. Increasing the sample variance reduces the likelihood of rejecting the null hypothesis and reduces measures of effect size.
2. Why is it necessary to have more than one method for measuring central tendency?
2. No single method for computing a measure of central tendency works well in all situations. With three different methods, however, at least one usually works well
2. Can the value of the standard error ever be larger than the value of the population standard deviation? Explain your answer.
2. No. The standard error is computed by dividing the standard deviation by the square root of n. The standard error is always less than or equal to the standard deviation.
2. In a research report, the term significant is used when the null hypothesis is rejected. (True or false?)
2. True
2. A small value (near zero) for the z-score statistic is evidence that the sample data are consistent with the null hypothesis. (True or false?)
2. True. A z-score near zero indicates that the data support the null hypothesis.
2. If all other factors are held constant, a confidence interval computed from a sample ofn = 25 is wider than a confidence interval computed from a sample of n = 100. (True or false?)
2. True. The smaller sample produces a wider interval
2. What is the advantage of having a mean of p. = 0 for a distribution of z-scores? (10.13.2014 stats ch 6, zscores)
2. With a mean of zero, all positive scores are above the mean and all negative scores are below the mean.
2. In the game Rock-Paper-Scissors, the probability that both players will select the same response and tie is p = y, and the probability that they will pick different responses is p = §. If two people play 72 rounds of the game and choose their responses randomly, what is the probability that they will choose the same response (tie) more than 28 times?
2. With• p = j and1 q = y, the binomial distribution is normal with p = 24 and a = 4;2 p(X > 28.5) = p(z >1.13) = 0.1292.
2. For a sample with a mean of M = 80 and a standard deviation of s = 20, find the X value corresponding to each of the following z-scores. z = —1.00 z = —0.50 z = —0.20 z = 1.50 z = 0.80 z = 1.40 (10.13.2014 stats ch 6, zscores)
2. X = 60 X = 70 X = 76 X = 110 X = 96 X = 108
2. Construct a frequency distribution table for the following set of scores. Include columns for proportion and percentage in your table.
2. X f p % ──────────── 9 2 0.10 10% 8 3 0.15 15% 7 5 0.25 25% 6 4 0.20 20% 5 3 0.15 15% 4 2 0.10 10% 3 1 0.05 5% ────────────
2. A sample ofn = 25 scores has a mean of M = 83 and a standard deviation ofs = 15. a. Explain what is measured by the sample standard deviation. b. Compute the estimated standard error for the sample mean and explain what is measured by the standard error
2. a. The sample standard deviation describes the variability of the scores in the sample. In this case, the standard distance between a score and the sample mean is 15 points. b. The estimated standard error is 3 points. The standard error provides a measure of the standard distance between a sample mean and the population mean.
2. A population forms a normal shaped distribution with IA = 40 and u = 8. a. A sample of n = 16 scores from this population has a mean of M = 36. Would you describe this as a relatively typical sample, or is the sample mean an extreme value? Explain your answer. b. If the sample from part a had n = 4 scores, would it be considered typical or extreme?
2. a. With n = 16 the standard error is 2, and the sample mean corresponds to z = -2.00. This is an extreme value. b. With n = 4 the standard error is 4, and the sample mean corresponds to z = -1.00. This is a relatively typical value
2. A psychology class consists of 14 males and 36 females. If the professor selects names from the class list using random sampling, a. What is the probability that the first student selected will be a female? b. If a random sample of n = 3 students is selected and the first two are both females, what is the probability that the third student selected will be a male?
2. a. p = 36/50 = 0.72 b. p = 14/50 = 0.28 (Remember, a random sample requires replacement.)
2. Aj ar contains 10 red marbles and 30 blue marbles. a. If you randomly select 1 marble from the jar, what is the probability of obtaining a red marble? b. If you take a random sample of n = 3 marbles from the jar and the first two marbles are both blue, what is the probability that the third marble will be red?
2. a. p = f = 0.25 b- P = 45 = 0.25. Remember that random sampling requires sampling with replacement.
2. A sample of n = 9 scores has SS = 288. a. Compute the variance for the sample. b. Compute the estimated standard error for the sample mean.
2. a. s2 = 36 b. sm = 2
2. A sample of n = 6 scores has a mean of M = 8. What is the value of EX for this sample?
2. a= 48
2. A researcher selects a sample from a population with µ = 45 and if = 8. A treatment is administered to the sample and, after treatment, the sample mean is found to be M = 47. Compute Cohen's d to measure the size of the treatment effect.
2. d = 2/8 = 0.25
20. Explain why the mean is often not a good measure of central tendency for a skewed distribution.
20. With a skewed distribution, the extreme scores in the tail can displace the mean out toward the tail. The result is that the mean is often not a very representative value.
21. Rochester, New York, averages p = 21.9 inches of snow for the month of December. The distribution of snowfall amounts is approximately normal with a standard deviation of ct = 6.5 inches. This year, a local jewelry store is advertising a refund of 50% off of all purchases made in December, if Rochester finishes the month with more than 3 feet (36 inches) 198 CHAPTER 6 PROBABILITY of total snowfall. What is the probability that the jewelry store will have to pay off on its promise?
21. p(X > 36) = p(z > 2.17) = 0.0150 or 1.50%
22. The teacher from the previous problem also tried a different approach to answering the question of whether changing answers helps or hurts exam grades. In a separate class, students were encouraged to review their final exams and change any answers they wanted to before turning in their papers. However, the students had to indicate both the original answer and the changed answer for each question. The teacher then graded each exam twice, one using the set of original answers and once with the changes. In the class of n = 22 students, the average exam score improved by an average of MD = 2.5 points with the changed answers. The standard deviation for the difference scores was s = 3.1. Are the data sufficient to conclude that rethinking and changing answers can significantly improve exam scores? Use a one-tailed test at the .01 level of significance
22. The null hypothesis says that changing answers should not affect the student's scores. With df = 21 and α = .01, the one-tailed critical value is t = 2.518. For these data, the estimated standard error is 0.66, and t(21) = 3.79. Reject H0.
22. A researcher is investigating the effectiveness of a new medication for lowering blood pressure for individuals with systolic pressure greater than 140. For this population, systolic scores average p. = 160 with a standard deviation of a = 20, and the scores form a normal-shaped distribution. The researcher plans to select a sample ofn = 25 individuals, and measure their systolic blood pressure after they take the medication for 60 days. If the researcher uses a two-tailed test with a = .05, a. What is the power of the test if the medication has a 5-point effect? b. What is the power of the test if the medication has a 10-point effect?
22. a. The critical boundary, z = -1.96, corresponds to M = 152.16. With a 5-point effect, this mean is located at z = -0.71 and the power is 0.2389 or 23.89%. b. With a 10-point effect, M = 152.16 is located at z = +0.54 and the power is 0.7054 or 70.54%.
22. Find the requested percentiles and percentile ranks for the following distribution of quiz scores for a class of N = 40 students. X f cf c% 20 2 40 100.0 19 4 38 95.0 18 6 34 85.0 17 13 28 70.0 16 6 15 37.5 15 4 9 22.5 14 3 5 12.5 13 2 2 5.0 a. What is the percentile rank for X = 15? b. What is the percentile rank for X = 18? c. What is the 15th percentile? d. What is the 90th percentile?
22. a. The percentile rank for X = 15 is 17.5%. b. The percentile rank for X = 18 is 77.5%. c. The 15th percentile is X = 14.75. d. The 90th percentile is X = 19.
23. One question on a student survey asks: In a typical week, how many times do you eat at a fast-food restaurant? The following frequency distribution table summarizes the results for a sample of n = 20 students. x f 5 or more 2 4 2 3 3 2 6 1 4 0 3 a. Find the mode for this distribution. b. Find the median for the distribution. c. Explain why you cannot compute the mean using the data in the table.
23. a. Mode = 2 b. Median = 2 c. You cannot find the total number of fast-food visits (ΣX) for this sample.
23. A true/false test has 40 questions. If a students is simply guessing at the answers, a. What is the probability of guessing correctly for any one question? b. On average, how many questions would the student get correct for the entire test? c. What is the probability that the student would get more than 25 answers correct simply by guessing? d. What is the probability that the student would get 25 or more answers correct simply by guessing?
23. a. p = ½ b. μ = 20 c. σ = √10 = 3.16 and for X = 25.5, z = 1.74, and p = 0.0409 d. For X = 24.5, z = 1.42, and p = 0.0778
28. A national health organization predicts that 20% of American adults will get the flu this season. If a sample of 100 adults is selected from the population, a. What is the probability that at least 25 of the people will be diagnosed with the flu? (Be careful: "at least 25" means "25 or more.") b. What is the probability that fewer than 15 of the people will be diagnosed with the flu? (Be careful: "fewer than 15" means " 14 or less."
28. a. μ = 20 and σ = 4 For X = 24.5, z = 1.13, and p = 0.1292. b. For X = 14.5, z = -1.38, and p = 0.0838.
3. Explain the difference between a matched-subjects design and a repeated-measures design.
3. For a repeated-measures design the same subjects are used in both treatment conditions. In a matched-subjects design, two different sets of subjects are used. However, in a matched-subjects design, each subject in one condition is matched with respect to a specific variable with a subject in the second condition so that the two separate samples are equivalent with respect to the matching variable.
3. How does increasing sample size influence the power of a hypothesis test?
3. Increasing sample size increases the power of a test
3. In words, define the alpha level and the critical region for a hypothesis test.
3. The alpha level is a small probability value that defines the concept of "very unlikely." The critical region consists of outcomes that are very unlikely to occur if the null hypothesis is true, where "very unlikely" is defined by the alpha level.
3. One sample has n = 5 scores with a mean of M = 4. A second sample has n = 3 scores with a mean of M= 10. If the two samples are combined, what is the mean for the combined sample?
3. The combined sample has n = 8 scores that total MX = 50. The mean is M =6.25.
3. In words and in symbols, what is the null hypothesis for a repeated-measures t test?
3. The null hypothesis states that, for the general population, the average difference between the two conditions is zero. In symbols, H0 = 0.
3. In general, a distribution of t statistics is flatter and more spread out than the standard normal distribution. (True or false?)
3. True.
3. If you toss a balanced coin 36 times, you would expect, on the average, to get 18 heads and 18 tails. What is the probability of obtaining exactly 18 heads in 36 tosses?
3. X = 18 is an interval with real limits of 17.5 and 18.5. The real limits correspond to z = ±0.17, and a probability ofp = 0.1350.
3. In a distribution with p. = 50, a score of X = 42 corresponds to z = -2.00. What is the standard deviation for this distribution? (10.13.2014 stats ch 6, zscores) (10.13.2014 stats ch 6, zscores)
3. a = 4
3. A researcher plans to select a random sample from a population with a standard deviation of a = 12. a. How large a sample is needed to have a standard error of 6 points or less? b. How large a sample is needed to have a standard error of 4 points or less?
3. a. A sample of n = 4 or larger. b. A sample of n = 9 or larger.
4. What does it mean for a sample to have a standard deviation of zero? Describe the scores in such a sample. (10.03.2014 Stats ch 4 textbook extraction/notes)
4. A standard deviation of zero indicates there is no variability. In this case, all of the scores in the sample have exactly the same value.
4. The distribution of sample means is not always a normal distribution. Under what circumstances is the distribution of sample means not normal?
4. The distribution of sample means will not be normal when it is based on small samples (n < 30) selected from a population that is not normal.
4. A sample of n = 4 scores has a mean of 9. If one person with a score of X= 3 is removed from the sample, what is the value for the new sample mean?
4. The original sample has n = 4 and IX = 36. The new sample has n = 3 scores that total EX = 33. The new mean is M = 11.
4. If other factors are held constant, increasing the size of the sample increases the likelihood of rejecting the null hypothesis. (True or false?)
4. True. A larger sample produces a smaller standard error, which leads to a larger z-score.
4. Find the exact value of the power for the hypothesis test shown in Figure 8.13.
4. With n = 4, the critical boundary of z = 1.96 corresponds to a sample mean of M = 89.8, and the exact value for power is p = 0.3594 or 35.945%
5. Find the values for n, EX, and M for the sample that is summarized in the following frequency distribution table. x f 5 1 4 2 3 3 2 5 1 1
5. For this sample, n = 12, /X = 33, and M = 33/12=2.75.
5. Find the mean, median, and mode for the scores in the following frequency distribution table: x f 8 1 7 4 6 2 5 2 4 2 3 I
5. The mean is 69/12 = 5.75, the median is 6, and the mode i s 7.
5. Explain why the formulas for sample variance and population variance are different. (10.03.2014 Stats ch 4 textbook extraction/notes)
5. Without some correction, the sample variance underestimates the variance for the population. Changing the formula for sample variance (using n - 1 instead of N) is the necessary correction
5. If other factors are held constant, are you more likely to reject the null hypothesis with a standard deviation of a = 2 or with o• = 10?
5. a = 2. A smaller standard deviation produces a smaller standard error, which leads to larger z-score.
5. For df = 15, find the value(s) of t associated with each of the following: a. The top 5% of the distribution. b. The middle 95% of the distribution. c. The middle 99% of the distribution
5. a. r = +1.753 b. t = ±2.131 c. t = ±2.947
5. Find the t values that form the boundaries of the critical region for a two-tailed test with a = .05 for each of the following sample sizes: a. n = 6 b. n = 12 c. n = 24
5. a. t = ±2.571 b. t = ±2.201 c. t = ±2.069
5. Draw a vertical line through a normal distribution for each of the following z-score locations. Determine whether the tail is on the right or left side of the line and find the proportion in the tail. a. z = 2.00 b. z = 0.60 c. z = -1.30 d. z = -0.30
5. a. tail to the right, p = 0.0228 b. tail to the right, p = 0.2743 c. tail to the left, p = 0.0968 d. tail to the left, p = 0.3821
6. Find the mean, median, and mode for the scores in the following frequency distribution table: X f 10 1 9 2 8 3 7 3 6 4 5 2
6. The mean is 107/15 = 7.13, the median is 7, and the mode is 6.
6. The following sample of n = 6 scores was obtained from a population with unknown parameters. Scores: 7, 1, 6, 3, 6, 7 a. Compute the sample mean and standard deviation. (Note that these are descriptive values that summarize the sample data.) b. Compute the estimated standard error for M. (Note that this is an inferential value that describes how accurately the sample mean represents the unknown population mean.)
6. a. M = 5 and s = 6 = 2.45. b. sM = 1.
7. For each of the following samples, determine the interval width that is most appropriate for a grouped frequency distribution and identify the approximate number of intervals needed to cover the range of scores. a. Sample scores range from X = 24 to X = 41 b. Sample scores range from X = 46 to X = 103 c. Sample scores range from X = 46 to X = 133
7. a. 2 points wide and around 8 intervals b. 5 points wide and around 12 intervals or 10 points wide and around 6 intervals c. 10 points wide and around 9 intervals
7. For a population with a standard deviation of a = 20, how large a sample is necessary to have a standard error that is: a. less than or equal to 5 points? b. less than or equal to 2 points? c. less than or equal to 1 point?
7. a. n ≥ 16 b. n ≥ 100 c. n ≥ 400
7. Find each of the following probabilities for a normal distribution. a. p(z > 0.25) b. p(z > -0.75) c. p(z < 1.20) d. p(z < — 1.20)
7. a. p(z > 0.25) = 0.4013 c. p(z < 1.20) = 0.8849 b. p(z > -0.75) = 0.7734 d. p(z < -1.20) = 0.1151
8. What information can you obtain about the scores in a regular frequency distribution table that is not available from a grouped table?
8. A regular table reports the exact frequency for each category on the scale of measurement. After the categories have been grouped into class intervals, the table reports only the overall frequency for the interval but does not indicate how many scores are in each of the individual categories.
8. A sample of n = 7 scores has a mean of M = 9. What is the value of XX for this sample?
8. X = 63.
8. What proportion of a normal distribution is located between each of the following z-score boundaries? a. z = -0.50 and z = +0.50 b. z = -0.90 and z = +0.90 c. z = - 1.50 and z
8. a. p = 0.3830 b. p = 0.6318 c. p = 0.8664
9. As mentioned in Chapters 2 and 3 (pp. 38 and 81), Steven Schmidt (1994) reported a series of studies examining the effect of humor on memory. In one part of the study, participants were presented with a list containing a mix of humorous and nonhumorous sentences, and were then asked to recall as many sentences as possible. Schmidt recorded the number of humorous and the number of nonhumorous sentences recalled by each individual. Notice that the data consist of two memory scores for each participant. Suppose that a difference score is computed for each individual in a sample of n = 16 and the resulting data show that participants recalled an average of = 3.25 more humorous sentences than nonhumorous, with SS = 135. Are these results sufficient to conclude that humor has a significant effect on memory? Use a two-tailed test with a = .05.
9. The sample variance is 9, the estimated standard error is 0.75, and t(15) = 4.33. With critical boundaries of ±2.131, reject H0
9. To evaluate the effect of a treatment, a sample ofn = 9 is obtained from a population with a mean of p = 40, and the treatment is administered to the individuals in the sample. After treatment, the sample mean is found to be M = 33. a. If the sample has a standard deviation ofs = 9, are the data sufficient to conclude that the treatment has a significant effect using a two-tailed test with a = .05? b. If the sample standard deviation is s = 15, are the data sufficient to conclude that the treatment has a significant effect using a two-tailed test with a = .05? c. Comparing your answer for parts a and b, how does the variability of the scores in the sample influence the outcome of a hypothesis test?
9. a. With s = 9, sM = 3 and t = -7/3 = -2.33. This is beyond the critical boundaries of ±2.306, so we reject the null hypothesis and conclude that there is a significant treatment effect. b. With s = 15, sM = 5 and t = -7/5 = -1.40. This value is not beyond the critical boundaries, so there is no significant effect. c. As the sample variability increases, the likelihood of rejecting the null hypothesis decreases.
What's the difference between between subject design and repeated measures design?
Between subject design is the study of two different groups while repeated measures research/within subject design is two sets of data which is taken from the same group of participants
What are the arguments for and against the one tail hypothesis test?
For: Is more sensitive to treatment Against: Lowers the expectations
When is it best to use the mode? (09.28.2014 Stats 3 Ch 3 - Central tendency (notes + questions)
If there there unspecific values, nominal values or discrete Can also contain major and minor modes
If a data set is highly skewed on the left but there are some distribution on the right what type of central tendency should be used? (09.28.2014 Stats 3 Ch 3 - Central tendency (notes + questions)
Medium
If there is an open ended group like 5 or more people ate pizza what central tendency should be used? (09.28.2014 Stats 3 Ch 3 - Central tendency (notes + questions)
Medium
In a study measuring time it takes for people to finish a test if someone never finishes what central tendency should be used? (09.28.2014 Stats 3 Ch 3 - Central tendency (notes + questions)
Medium
What happens if you ad a new score to the distribution? What happens if the score is exactly the same as the mean? (09.28.2014 Stats 3 Ch 3 - Central tendency (notes + questions)
New score will change the mean. If exactly the same though there will be no change.
1. Under what circumstances is a t statistic used instead of a z-score for a hypothesis test?
S 1. A t statistic is used instead of a z-score when the population standard deviation and variance are not known.
1. A researcher selects a sample of n = 16 individuals from a normal population with a mean of p.. = 40 and o- = 8. A treatment is administered to the sample and, after treatment, the sample mean is M = 43. If the researcher uses a hypothesis test to evaluate the treatment effect, what z-score would be obtained for this sample?
S 1. The standard error is 2 points and z = 3/2 = 1.50
What is statistical power and how do you use it?
Statistical power is the likelihood a study will commit Type 2 error. Based on the concept that 1 - β should be 0. Step 1: Get Om Step 2: multiply α-Zscore by 2 (Only if one tail) Step 3: Add M + Step 2 result Step 4: Use step 3 as M and apply to Z-score formula
homogenize of variance
The population from which the samples are selected must have equal variances
What is X, N, n describe in statistics
X is scores for a variable, N is symbol for number of scores in a population, n is the number of scores in a sample.
What does the effect size tell you? What is the equation? How much would be considered a small, medium, and large effect size?
You get the Cohen's D provides a a measure in terms of SD and identifies the distance at which exists between the untreated and treated population. The equation for estimated Cohn's D = (Utreatment - Unotreatment)/SD d=0.2 Small effect d=0.5 medium effect d=0.8 large effect
The third is or called the standard variation which is the average squared distance from the mean. This involves finding the distance of each integer of the mean, square each distance, find the average of the result and then square rooting it. In the situation of ________ you would subtract 1 from the number dividing for the mean of the squared numbers because of a concept called degrees of freedom which mainly is that there is a tendency for deviations to be undervalues and needs to be accounted for. With this method you can achieve what is called one standard deviation from the mean which is 68% of the score, while 2 standard deviation involves 95%. (10.02.2014, stats lec 4 variability)
a sample
Datum
a single measurement or oberservation (score or raw score)
The median is the midpoint of a distribution of scores. The median is the preferred measure of central tendency when a distribution has ____a___. The median also is used for ___b___ scores that make it impossible to compute a mean. Finally, the median is the preferred measure of central tendency for data from an __c___ (09.28.2014 Stats 3 Ch 3 - Central tendency (notes + questions)
a. a few extreme scores that displace the value of the mean b. open-ended distributions and when there are undetermined (infinite) c. open-ended distributions and when there are undetermined (infinite)
For symmetrical distributions, the mean is __a__ to the median. If there is only one mode, then it has the __b__ For skewed distributions, the mode is located toward the side where the __c__, and the mean is pulled toward the ___d___. The median is usually located ___e_____. (09.28.2014 Stats 3 Ch 3 - Central tendency (notes + questions)
a. equal b. same value, too. c. scores pile up d. extreme scores in the tail e. between these two values
Changing any score in the distribution causes the mean to be changed. When a constant value is added to (or subtracted from) every score in a distribution, the same constant value is added to (or subtracted from) the mean. If every score is multiplied by a constant, the mean is _____a_____(09.28.2014 Stats 3 Ch 3 - Central tendency (notes + questions)
a. multiplied by the same constant
The purpose of central tendency is to determine the _______a________. The three standard measures of central tendency are the mode, the median, and the mean.
a. single value that identifies the center of the distribution and best represents the entire set of scores
The purpose of central tendency is to determine the _______a________. The three standard measures of central tendency are the mode, the median, and the mean. (09.28.2014 Stats 3 Ch 3 - Central tendency (notes + questions)
a. single value that identifies the center of the distribution and best represents the entire set of scores
The mode is the most frequently occurring score in a distribution. It is easily located by finding ____a___. For data measured on a nominal scale, the mode is the appropriate measure of central tendency. It is possible for a distribution to have more than one mode. (09.28.2014 Stats 3 Ch 3 - Central tendency (notes + questions)
a. the peak in a frequency distribution graph
ordinal scales
are a set of categories that are organized in an ordered sequence and ar ebest used in obersvations in terms of size or magnitude. This does not allow you to determine the size difference however.
A researchers measures the arousal level in penis size that an attractive blond girl in a tight skirt produces from UofT male heterosexual students. Would this be a discrete or continuous variable and why?
continuous since penis size when hard can vary by small degrees.
Deviation from the mean is the ____________. It measures the deviation from a common central tendency. It can tell us how an individual stands in relation to the other scores as well us how accurate this set of scores is to a population. When there is a small vulnerability then it is a good representation because it means that more scores are contained in an area. (10.02.2014, stats lec 4 variability)
conventional form of variation
Describe the two types of statistics
descriptive statistics makes data more presentable while inferential statistics seeks to prove some things with the numbers computed.
The mean of the distribution of sample means is equal to the mean of the population of scores, p, and is called the
expected value of M.
If H0 is U1-U2=0 and the confidence interval is 3.798 to 12.202 and 0 is concluded as accepted in the confidence interval then it is the same saying __________
failure to reject H0
3. For a sample with a mean of M = 85, a score of X = 80 corresponds to z = —0.50. What is the standard deviation for the sample? (10.13.2014 stats ch 6, zscores)
3. s = 10
What are three common assumptions made with Hypothesis test and Z-score?
(1) Random selection (2) Not consistent observation between 1 and 2 (3) SD is not changed for a population before and after treatment but rather a constant is added.
What are the two characteristics required for something to be considered an experimental study?
(1) manipulation where one variable is manipulated by changing it's value and a second variable is observed. (2) control where the situation is set up to ensure that there is as little influence as possible.
2. If you have a score of 52 on an 80-point exam, then you definitely scored above the median. (True or false?)
2. False. The value of the median would depend on where all of the scores are located.
4. In a distribution with o- = 12, a score of X = 56 corresponds to z = -0.25. What is the mean for this distribution? (10.13.2014 stats ch 6, zscores)
4. p. = 59
9. Find each of the following probabilities for a normal distribution. a. pi-0.25 < z < 0.25) b. p(-2.00 < z < 2.00) c. p (-0.30 < z < 1.00) d . p ( - 1.25 < z < 0.25)
9. a. p = 0.1974 c. p = 0.4592 b. p = 0.9544 d. p = 0.4931
The distribution of sample means is the
collection of sample means for all of the possible random samples of a particular size (n) that can be obtained from a population.
What are the four steps of hypothesis testing?
(1) Null hypothesis State the null hypothesis (2) Set the criteria (Alpha Level) (3) Collect data and compute sample statistics (z=(samplemean - hypothesizedpopmean)/standarderrorUandM (4) Make decision whether the sample data has reached beyond the critical zone
1. A population forms a normal distribution with a mean of ii = 80 and a standard deviation of o = 20. a. If single score is selected from this population, how much distance would you expect, on average, between the score and the population mean? b. If a sample of n = 100 scores is selected from this population, how much distance would you expect, on average, between the sample mean and the population mean?
1. a. For a single score, the standard distance from the mean is the standard deviation, if = 20. b. For a sample of n = 100 scores, the average distance between the sample mean and the population mean is the standard error, um = 20A/TE10 =2
1. In the z-score formula as it is used in a hypothesis test, a. Explain what is measured by M — IL in the numerator. b. Explain what is measured by the standard error in the denominato
1. a. M - μ measures the difference between the sample mean and the hypothesized population mean. b. A sample mean is not expected to be identical to the population mean. The standard error measures how much difference, on average, is reasonable to expect between M and μ.
1. After years of teaching driver's education, an instructor knows that students hit an average of p. = 10.5 orange cones while driving the obstacle course in their final exam. The distribution of run-over cones is approximately normal with a standard deviation of a = 4.8. To test a theory about text messaging and driving, the instructor recruits a sample of n = 16 student drivers to attempt the obstacle course while sending a text message. The individuals in this sample hit an average of M = 15.9 cones. a. Do the data indicate that texting has a significant effect on driving? Test with a = .01. b. Write a sentence describing the outcome of the hypothesis test as it would appear in a research report.
1. a. With a = .01, the critical region consists of z-scores in the tails beyond z = ± 2.58. For these data, the standard error is 1.2 and z = 4.50. Reject the null hypothesis and conclude that texting has a significant effect on driving. b. Texting while driving had a significant effect on the number of cones hit by the participants, z = 4.50, p < .01.
1. A population has a standard deviation of a = 10. a. On average, how much difference should there be between the population mean and a single score selected from this population? b. On average, how much difference should there be between the population mean and the sample mean for n = 4 scores selected from this population? c. On average, how much difference should there be between the population mean and the sample mean for n = 25 scores selected from this population?
1. a. o = 10 points b. crAi = 5 points c. QM = 2 points
10. Miller (2008) examined the energy drink consumption of college undergraduates and found that males use energy drinks significantly more often than females. To further investigate this phenomenon, suppose that a researcher selects a random sample of n = 36 male undergraduates and a sample of n = 25 females. On average, the males reported consuming M = 2.45 drinks per month and females had an average of M = 1.28. Assume that the overall level of consumption for college undergraduates averages g. = 1.85 energy drinks per month, and that the distribution of monthly consumption scores is approximately normal with a standard deviation of a = 1.2. a. Did this sample of males consume significantly more energy drinks than the overall population average? Use a one-tailed test with a = .01. b. Did this sample of females consume significantly fewer energy drinks than the overall population average? Use a one-tailed test with a = .01
10. a. H0: μ ≤ 1.85 (not more than average) For the males, the standard error is 0.2 and z = 3.00. With a critical value of z = 2.33, reject the null hypothesis. b. H0: μ ≥ 1.85 (not fewer than average) For the females, the standard error is 0.24 and z = -2.38. With a critical value of z = -2.33, reject the null hypothesis.
12. A sample of n = 11 scores has a mean of M = 4. One person with a score of X = 16 is added to the sample. What is the value for the new sample mean?
12. With the new score, n = 12, Σ X = 60, and M = 5.
10. For the following set of quiz scores: 3, 5, 4, 6, 2, 3, 4, 1, 4, 3 7, 7, 3, 4, 5, 8, 2, 4, 7, 10 a. Construct a frequency distribution table to organize the scores. b. Draw a frequency distribution histogram for these data
10. a. X f 10 1 9 0 8 1 7 3 6 1 5 2 4 5 3 4 2 2 1 1
11. Strack, Martin, and Stepper (1988) reported that people rate cartoons as funnier when holding a pen in their teeth (which forced them to smile) than when holding a pen in their lips (which forced them to frown). A researcher attempted to replicate this result using a sample of n = 25 adults between the ages of 40 and 45. For each person, the researcher recorded the difference between the rating obtained while smiling and the rating obtained while frowning. On average the cartoons were rated as funnier when the participants were smiling, with an average difference of MD = 1.6 with SS = 150. a. Do the data indicate that the cartoons are rated significantly funnier when the participants are smiling? Use a one-tailed test with a = .01. b. Compute r2 to measure the size of the treatment effect. c. Write a sentence describing the outcome of the hypothesis test and the measure of effect size as it would appear in a research report
11. a. The null hypothesis says that there is no difference in judgments for smiling versus frowning. For these data, the sample variance is 6.25, the estimated standard error is 0.5, and t = 1.6/0.5 = 3.20. For a one-tailed test with df = 24, the critical value is 2.492. Reject the null hypothesis. b. r2 = 10.24/34.24 = 0.299 (29.9%) c. The cartoons were rated significantly funnier when people held a pen in their teeth compared to holding a pen in their lips, t(24) = 3.20, p < .01, one tailed, r2 = 0.299.
12. There are two different formulas or methods that can be used to calculate SS. a. Under what circumstances is the definitional formula easy to use? b. Under what circumstances is the computational formula preferred? (10.03.2014 Stats ch 4 textbook extraction/notes)
12. a. The definitional formula is easy to use when the mean is a whole number and there are relatively few scores. b. The computational formula is preferred when the mean is not a whole number.
13. Hallam, Price, and Katsarou (2002) investigated the influence of background noise on classroom performance for children aged 10 to 12. In one part of the study, calming music led to better performance on an arithmetic task compared to a no-music condition. Suppose that a researcher selects one class of n = 18 students who listen to calming music each day while working on arithmetic problems. A second class of n = 18 serves as a control group with no music. Accuracy scores are measured for each child and the average for students in the music condition is M = 86.4 with SS = 1550 compared to an average of M = 78.8 with SS = 1204 for students in the nomusic condition. a. Is there a significant difference between the two music conditions? Use a two-tailed test with a = .05. b. Compute the 90% confidence interval for the population mean difference. c. Write a sentence demonstrating how the results from the hypothesis test and the confidence interval would appear in a research report
13. a. Using df = 30, , because 34 is not listed in the table, and α = .05, the critical region consists of t values beyond 2.042. The pooled variance is 81, the estimated standard error is 3, and t(34) = 7.6/3 = 2.53. The t statistic is in the critical region. Reject H0 and conclude that there is a significant difference. b. For 90% confidence, the t values are 1.697 (using df = 30), and the interval extends from 2.509 to 12.691 points higher with the calming music. c. Classroom performance was significantly better with background music, t(34) = 2.53, p < .05, 95% CI [2.509, 12.691].
13. A random sample is obtained from a normal population with a mean of p = 30 and a standard deviation of a = 8. The sample mean is M = 33. a. Is this a fairly typical sample mean or an extreme value for a sample of n = 4 scores? b. Is this a fairly typical sample mean or an extreme value for a sample of n = 64 scores?
13. a. With a standard error of 4, M = 33 corresponds to z = 0.75, which is not extreme. b. With a standard error of 1, M = 33 corresponds to z = 3.00, which is extreme.
13. Standardized measures seem to indicate that the average level of anxiety has increased gradually over the past 50 years (Twenge, 2000). In the 1950s, the average score on the Child Manifest Anxiety Scale was p, = 15.1. A sample of n = 16 of today's children produces a mean score of M = 23.3 with SS = 240. a. Based on the sample, has there been a significant change in the average level of anxiety since the 1950s? Use a two-tailed test with a = .01. b. Make a 90% confidence interval estimate of today's population mean level of anxiety. c. Write a sentence that demonstrates how the outcome of the hypothesis test and the confidence interval would appear in a research report.
13. a. With df = 15, the critical values are ±2.947. For these data, the sample variance is 16, the estimated standard error is 1, and t = 78.2/1 = 8.20. Reject the null hypothesis and conclude that there has been a significant change in the level of anxiety. b. With df = 15, the t values for 90% confidence are 1.753, and the interval extends from 21.547 to 25.053. c. The data indicate a significant change in the level of anxiety, t(16) = 8.20, p < .01, 95% CI [21.547, 25.053].
15. A researcher is testing the hypothesis that consuming a sports drink during exercise improves endurance. A sample of n = 50 male college students is obtained and each student is given a series of three endurance tasks and asked to consume 4 ounces of the drink during each break between tasks. The overall endurance score for this sample is M = 53. For the general population of male college students, without any sports drink, the scores for this task average p. = 50 with a standard deviation of a = 12. a. Can the researcher conclude that endurance scores with the sports drink are significantly higher than scores without the drink? Use a one-tailed test with a = .05. b. Can the researcher conclude that endurance scores with the sports drink are significantly different than scores without the drink? Use a two-tailed test with a = .05. c. You should find that the two tests lead to different conclusions. Explain why.
15. a. H0: μ < 50 (endurance is not increased). The critical region consists of z-scores beyond z = +1.65. For these data, σM = 1.70 and z = 1.76. Reject H0 and conclude that endurance scores are significantly higher with the sports drink. b. H0: μ = 50 (no change in endurance). The critical region consists of z-scores beyond z = ±1.96. Again, σM = 1.70 and z = 1.76. Fail to reject H0 and conclude that the sports drink does not significantly affect endurance scores. c. The two-tailed test requires a larger z-score for the sample to be in the critical region.
16. In a classic study of infant attachment, Harlow (1959) placed infant monkeys in cages with two artificial surrogate mothers. One "mother" was made from bare wire mesh and contained a baby bottle from which the infants could feed. The other mother was made from soft terry cloth and did not provide any access to food. Harlow observed the infant monkeys and recorded how much time per day was spent with each mother. In a typical day, the infants spent a total of 18 hours clinging to one of the two mothers. If there were no preference between the two, you would expect the time to be divided evenly, with an average of p = 9 hours for each of the mothers. However, the typical monkey spent around 15 hours per day with the terry-cloth mother, indicating a strong preference for the soft, cuddly mother. Suppose a sample ofn = 9 infant monkeys averaged M = 15.3 hours per day with SS = 216 with the terry-cloth mother. Is this result sufficient to conclude that the monkeys spent significantly more time with the softer mother than would be expected if there were no preference? Use a two-tailed test with a = .0
16. The sample variance is 27, the estimated standard error is √3 = 1.73, and t = 6.3/1.73 = 3.64. For a two tailed test, the critical value is 2.306. Reject the null hypothesis, there is a significant preference for the terry cloth mother.
16. A researcher for a cereal company wanted to demonstrate the health benefits of eating oatmeal. A sample of 9 volunteers was obtained and each participant ate a fixed diet without any oatmeal for 30 days. At the end of the 30-day period, cholesterol was measured for each individual. Then the participants began a second 30-day period in which they repeated exactly the same diet except that they added 2 cups of oatmeal each day. After the second 30-day period, cholesterol levels were measured again and the researcher recorded the difference between the two scores for each participant. For this sample, cholesterol scores averaged MD = 16 points lower with the oatmeal diet with SS = 538 for the difference scores. a. Are the data sufficient to indicate a significant change in cholesterol level? Use a two-tailed test with a = .01. b. Compute r2, the percentage of variance accounted for by the treatment, to measure the size of the treatment effect. c. Write a sentence describing the outcome of the hypothesis test and the measure of effect size as it would appear in a research report.
16. a. The null hypothesis states that the oatmeal has no effect on cholesterol level. For these data, the sample variance is 67.25, the estimated standard error is 2.73, and t(8) = 5.86. With df = 8 and α = .01, the critical values are t = ±3.355. Reject the null hypothesis. b. r2 = 5.862/(5.862 + 8) = 0.811 c. The results show that the oatmeal has a significant effect on cholesterol levels, t(8) = 5.86, p < .01, r2 = 0.811.
17. In 1974, Loftus and Palmer conducted a classic study demonstrating how the language used to ask a question can influence eyewitness memory. In the study, college students watched a film of an automobile accident and then were asked questions about what they saw. One group was asked, "About how fast were the cars going when they smashed into each other?" Another group was asked the same question except the verb was changed to "hit" instead of "smashed into." The "smashed into" group reported significantly higher estimates of speed than the "hit" group. Suppose a researcher repeats this study with a sample of today's college students and obtains the following results. Estimated Speed Smashed into n = 15 M = 40.8 SS = 510 a. Do the results indicate a significantly higher estimated speed for the "smashed into" group? Use a one-tailed test with a = .01. b. Compute the estimated value for Cohen's d to measure the size of the effect. c. Write a sentence demonstrating how the results of the hypothesis test and the measure of effect size ld appear in a research report Hit = IS M = 34.0 SS = 414
17. a. The research prediction is that participants who hear the verb "smashed into" will estimate higher speeds than those who hear the verb "hit." For these data, the pooled variance is 33, the estimated standard error is 2.10, and t(28) = 3.24. With df = 28 and α = .01, the critical value is t = 2.467. The sample mean difference is in the right direction and is large enough to be significant. Reject H0. b. The estimated Cohen's d = 6.8/√33 = 1.18. c. The results show that participants who heard the verb "smashed into" estimated significantly higher speeds than those who heard the verb "hit," t(28) = 3.24, p < .01, d = 1.18.
18. Numerous studies have found that males report higher self-esteem than females, especially for adolescents (Kling, Hyde, Showers, & Buswell, 1999). Typical results show a mean self-esteem score ofM = 39.0 with SS = 60.2 for a sample ofn = 10 male adolescents and a mean ofM = 35.4 withSS = 69.4 for a sample of n = 10 female adolescents. a. Do the results indicate that self-esteem is significantly higher for males? Use a one-tailed test with a = .01. b. Use the data to make a 95% confidence interval estimate of the mean difference in self-esteem between male and female adolescents. c. Write a sentence demonstrating how the results from the hypothesis test and the confidence interval would appear in a research report.
18. a. The pooled variance is 7.2, the estimated standard error is 1.2, and t(18) = 3.00. For a one-tailed test with df = 18 the critical value is 2.552. Reject the null hypothesis. There is a significant difference between the two groups. b. For 95% confidence, the t values are 2.101, and the interval extends from 1.079 to 6.121 points higher for boys. c. The results indicate that adolescent males have significantly higher self-esteem than girls, t(18) = 3.00, p < .01, one tailed, 95% CI [1.079, 6.121].
19. A random sample of n = 25 scores is obtained from a population with a mean of p = 45. A treatment is administered to the individuals in the sample and, after treatment, the sample mean is M = 48. a. Assuming that the sample standard deviation is s = 6 compute r2 and the estimated Cohen's d to measure the size of the treatment effect. b. Assuming that the sample standard deviation is s = 15, compute r2 and the estimated Cohen's d to measure the size of the treatment effect. c. Comparing your answers from parts a and b, how does the variability of the scores in the sample influence the measures of effect size?
19. a. Cohen's d = 3/6 = 0.50. With s = 6, the estimated standard error is 1.2 and t = 3/1.2 = 2.50. r2 = 6.25/30.25 = 0.207. b. Cohen's d = 3/15 = 0.20. With s = 15, the estimated standard error is 3 and t = 3/3 = 1.00. r2 = 1.00/25.00 = 0.04. c. Measures of effect size tend to decrease as sample variance increases.
19. One sample has a mean of M = 5 and a second sample has a mean of M = 10. The two samples are combined into a single set of scores. a. What is the mean for the combined set if both of the original samples have n = 5 scores? b. What is the mean for the combined set if the first sample has n = 4 scores and the second sample has n = 6? c. What is the mean for the combined set if the first sample has n = 6 scores and the second sample has n = 4?
19. a. The new mean is 75/10 = 7.5. b. The new mean is (20 + 60)/10 = 8 c. The new mean is (30 + 40)/10 = 7
19. A researcher is comparing the effectiveness of two sets of instructions for assembling a child's bike. A sample of eight fathers is obtained. Half of the fathers are given one set of instructions and the other half receives the second set. The researcher measures how much time is needed for each father to assemble the bike. The scores are the number of minutes needed by each participant. Instruction Set I Instruction Set II 8 14 4 10 8 6 4 10 a. Is there a significant difference in time for the two sets of instructions? Use a two-tailed test at the .05 level of significance. b. Calculate the estimated Cohen's d and r2 to measure effect size for this study.
19. a. The null hypothesis states that there is no difference between the two sets of instructions, H0: μ1 - μ2 = 0. With df = 6 and α = .05, the critical region consists of t values beyond ±2.447. For the first set, M = 6 and SS = 16. For the second set, M = 10 with SS = 32. For these data, the pooled variance is 8, the estimated standard error is 2, and t(6) = 2.00. Fail to reject H0. The data are not sufficient to conclude that there is a significant difference between the two sets of instructions. b. For these data, the estimated d = 4/√8 = 1.41 (a very large effect) and r2 = 4/10 = 0.40 (40%).
19. The machinery at a food-packing plant is able to put exactly 12 ounces of juice in every bottle. However, some items such as apples come in variable sizes so it is almost impossible to get exactly 3 pounds of apples in a bag labeled "3 lbs." Therefore, the machinery is set to put an average of p = 50 ounces (3 pounds and 2 ounces) in each bag. The distribution of bag weights is approximately normal with a standard deviation of a = 4 ounces. a. What is the probability of randomly picking a bag of apples that weighs less than 48 ounces (3 pounds)? b. What is the probability of randomly picking n = 4 bags of apples that have an average weight less than M = 48 ounces?
19. a. p(z < -0.50) = 0.3085 b. p(z < -1.00) = 0.1587
2. Define a Type II error.
2. A Type II error is the failure to reject a false null hypothesis. In terms of a research study, a Type II error occurs when a study fails to detect a treatment effect that really exists.
2. Describe what is measured by the estimated standard error in the bottom of the independent-measures t statistic
2. The standard error for the independent measures t provides an estimate of the standard distance between a sample mean difference (M1 - M2) and the population mean difference (μ1 - μ2). When the two samples come from populations with the same mean (when H0 is true), the standard error indicates the standard amount of error (distance) between two sample means.
2. What is the probability of obtaining a sample mean greater than M = 60 for a random sample of n = 16 scores selected from a normal population with a mean of p = 65 and a standard deviation of a = 20?
2. The standard error is om = 5, and M = 60 corresponds to z = —1.00, p(M > 60) = p(z > —1.00) = 0.8413 (or 84.13%)
2. The value of the z-score in a hypothesis test is influenced by a variety of factors. Assuming that all other variables are held constant, explain how the value of z is influenced by each of the following: a. Increasing the difference between the sample mean and the original population mean. b. Increasing the population standard deviation. c. Increasing the number of scores in the sample.
2. a. A larger difference will produce a larger value in the numerator which will produce a larger z-score. b. A larger standard deviation will produce larger standard error in the denominator which will produce a smaller z-score. c. A larger sample will produce a smaller standard error in the denominator which will produce a larger z-score.
2. Describe the location in the distribution for each of the following z-scores. (For example, z = +1.00 is located above the mean by 1 standard deviation.) a. z = — 1.50 b. z = 0.25 c. z = — 2.50 d. z = 0.50 (10.13.2014 stats ch 6, zscores)
2. a. Below the mean by 12 standard deviations. b. Above the mean by y standard deviation. c. Below the mean by 2Z standard deviations. d. Above the mean by 1- standard deviation.
2. A researcher report states that there is a significant difference between treatments for an independent-measures design with t(28) = 2.27. a. How many individuals participated in the research study? (Hint: Start with the dfvalue.) b. Should the report state that p > .05 or p < .05?
2. a. The df = 28, so the total number of participants is 30. b. A significant result is indicated by p < .05.
2. A population with a mean of tL = 44 and a standard deviation of a = 6 is standardized to create a new distribution with μ = 50 and a = 10. a. What is the new standardized value for a score of X = 47 from the original distribution? b. One individual has a new standardized score of X = 65. What was this person's score in the original distribution? (10.13.2014 stats ch 6, zscores)
2. a. X = 47 corresponds to z = +0.50 in the original distribution. In the new distribution, the corresponding score is X = 55. b. In the new distribution, X = 65 corresponds to z = +1.50. The corresponding score in the original distribution is X = 53.
20. Briefly explain how increasing sample size influences each of the following. Assume that all other factors are held constant. a. The size of the z-score in a hypothesis test. b. The size of Cohen's d. c. The power of a hypothesis test.
20. a. The z-score increases (farther from zero). b. Cohen's d is not influenced by sample size. c. Power increases.
20. A researcher uses a matched-subjects design to investigate whether single people who own pets are generally happier than singles without pets. A mood inventory questionnaire is administered to a group of 20- to 29-year-old non—pet owners and a similar age group of pet owners. The pet owners are matched one to one with the non—pet owners for income, number of close friendships, and general health. The data are as follows: Matched Non-Pet Pet Pair Owner Owner A 12 14 B 8 7 C 10 13 D 9 9 E 7 13 F 10 12 a. Is there a significant difference in the mood scores for non—pet owners versus pet owners? Test with a = .05 for two tails. b. Construct the 95% confidence interval to estimate the size of the mean difference in mood between the population of pet owners and the population of non—pet owners. (You should find that a mean difference of IID = 0 is an acceptable value, which is consistent with the conclusion from the hypothesis test.) 19. The previous problem demonstrates that removing individual differences can substantially reduce variance and lower the standard error. However, this benefit only occurs if the individual differences are consistent across treatment conditions. In problem 18, for example, the first two participants (top two rows) consistently had the highest scores in both treatment conditions. Similarly, the last two participants consistently had the lowest scores in both treatments. To construct the following data, we started with the scores in problem 18 and scrambled the scores in treatment 1 to eliminate the consistency of the individual differences. a. Assume that the data are from an independentmeasures study using two separate samples, each with n = 6 participants. Compute the pooled variance and the estimated standard error for the mean difference. b. Now assume that the data are from a repeatedmeasures study using the same sample of n = 6 participants in both treatment conditions. Compute the variance for the sample of difference scores and the estimated standard error for the mean difference. (This time you should find that removing the individual differences does not reduce the variance or the standard error.) reatment 1 Treatment 2 Difference 6 13 7 7 12 5 8 10 2 10 10 0 5 6 1 12 9 -3
20. a. The null hypothesis says that pets do not have an effect on mood, H 0: μD = 0. With df = 5 and α = .05, the critical values are t = ±2.571. For these data, MD = 2, SS = 30, the estimated standard error is 1.00, and t(5) = 2.00. Fail to reject H0. b. For 95% confidence, use t = ±2.571. The interval extends from 0.571 to 4.571. (Note that a value of zero is contained in the interval.)
21. An example of the vertical-horizontal illusion is shown in the figure below. Although the two lines are exactly the same length, the vertical line appears to be much longer. To examine the strength of this illusion, a researcher prepared an example in which both lines were exactly 10 inches long. The example was shown to individual participants who were told that the horizontal line was 10 inches long and then were asked to estimate the length of the vertical line. For a sample of n = 25 participants, the average estimate was M = 12.2 inches with a standard deviation ofs = 1.00. An example of the verticalhorizontal illusion a. Use a one-tailed hypothesis test with a = .01 to demonstrate that the individuals in the sample significantly overestimate the true length of the line. (Note: Accurate estimation would produce a mean of p = 10 inches.) b. Calculate the estimated d and r2, the percentage of variance accounted for, to measure the size of this effect. c. Construct a 95% confidence interval for the population mean estimated length of the vertical line.
21. a. The estimated standard error is 0.20 and t = 2.2/0.2 = 11.00. The t value is well beyond the critical value of 2.492. Reject the null hypothesis. b. Cohen's d = 2.2/1 = 2.20 and r2 = 121/145 = 0.8345
22. Downs and Abwender (2002) evaluated soccer players and swimmers to determine whether the routine blows to the head experienced by soccer players produced long-term neurological deficits. In the study, neurological tests were administered to mature soccer players and swimmers and the results indicated significant differences. In a similar study, a researcher obtained the following data. Swimmers Soccer players 10 7 8 4 7 9 9 3 13 7 7 6 12 a. Are the neurological test scores significantly lower for the soccer players than for the swimmers in the control group? Use a one-tailed test with a = .05. b. Compute the value of r2 (percentage of variance accounted for) for these data.
22. a. The null hypothesis states that the type of sport does not affect neurological performance. For a one-tailed test, the critical boundary is t = 1.796. For the swimmers, M = 9 and SS = 44. For the soccer players, M = 6 and SS = 24. The pooled variance is 6.18 and t(11) = 2.11. Reject H0. The data show that the soccer players have significantly lower scores. b. For these data, r2 = 0.288 (28.8%).
22. For each of the following, identify the exam score that should lead to the better grade. In each case, explain your answer. a. A score of X = 56, on an exam with p, = 50 and Cr = 4; or a score of X = 60 on an exam with p, = 50 and cr = 20. b. A score of X = 40, on an exam with 1.1, = 45 and Cr = 2; or a score of X = 60 on an exam with u= 70 and u= 20. c. A score of X = 62, on an exam with p. = 50 and Cr = 8; or a score of X = 23 on an exam with = 20 and a = 2. (10.13.2014 stats ch 6, zscores)
22. a. X = 56 corresponds to z = 1.50 (better grade), and X = 60 corresponds to z = 0.50. b. X = 60 corresponds to z = -0.50 (better grade), and X = 40 corresponds to z = -2.50. c. X = 62 corresponds to z = 1.50, and X = 23 also corresponds to z = 1.50. The two scores have the same relative position and should receive the same grade.
22. A multiple-choice test has 48 questions, each with four response choices. If a student is simply guessing at the answers, a. What is the probability of guessing correctly for any question? b. On average, how many questions would a student get correct for the entire test? c. What is the probability that a student would get more than 15 answers correct simply by guessing? d. What is the probability that a student would get 15 or more answers correct simply by guessing?
22. a. p = ¼ b. μ = 12 c. σ = √9 = 3 and for X = 15.5, z = 1.17, and p = 0.1210 d. For X = 14.5, z = 0.83, and p = 0.2033
23. In the Preview section for this chapter, we discussed a research study demonstrating that 8-month-old infants appear to recognize which samples are likely to be obtained from a population and which are not. In the study, the infants watched as a sample of n = 5 ping pong balls was selected from a large box. In one condition, the sample consisted of 1 red ball and 4 white balls. After the sample was selected, the front panel of the box was removed to reveal the contents. In the expected condition, the box contained primarily white balls like the sample and the infants looked at it for an average of M = 7.5 seconds. In the unexpected condition, the box had primarily red balls, unlike the sample, and the infants looked at it for M = 9.9 seconds. Assuming that the standard error for both means is (am = 1 second, draw a bar graph showing the two sample means using brackets to show the size of the standard error for each mean.
23. 10 │ │ Mean 8 │ Looking │ Time 6 │ (seconds) │ 4 │ │ 2 │ │ └────────────────────────── Expected Unexpected
23. A researcher is evaluating the influence of a treatment using a sample selected from a normally distributed population with a mean of p. = 80 and a standard deviation ofa = 20. The researcher expects a 12-point treatment effect and plans to use a two-tailed hypothesis test with a = .05. a. Compute the power of the test if the researcher uses a sample ofn = 16 individuals. (See Example 8.6.) b. Compute the power of the test if the researcher uses a sample of n = 25 individuals.
23. a. For a sample of n = 16 the standard error would be 5 points, and the critical boundary for z = 1.96 corresponds to a sample mean of M = 89.8. With a 12-point effect, the distribution of sample means would be centered at = 92. In this distribution, the critical boundary of M = 89.8 corresponds to z = -0.44. The power for the test is p(z > -0.44) = 0.6700 or 67%. b. For a sample of n = 25 the standard error would be 4 points, and the critical boundary for z = 1.96 corresponds to a sample mean of M = 87.84. With a 12-point effect, the distribution of sample means would be centered at = 92. In this distribution, the critical boundary of M = 87.84 corresponds to z = -1.04. The power for the test is p(z > -1.04) = 0.8508 or 85.08%.
25. Construct a stem and leaf display for the data in problem 6 using one stem for the scores in the 60s, one for scores in the 50s, and so on.
25. 1│796 2│0841292035826 3│094862 4│543 5│3681 6│4
26. A trick coin has been weighted so that heads occurs with a probability ofp = j , and /j(tails) = j . If you toss this coin 72 times, a. How many heads would you expect to get on average? b. What is the probability of getting more than 50 heads? c. What is the probability of getting exactly 50 heads?
26. a. = pn = 48 b. z = 0.63, p = 0.2643 c. p(0.38 < z < 0.63) = 0.0877
3. Under what circumstances is a Type II error likely to occur?
3. A Type II error is likely to occur when the treatment effect is very small. In this case, a research study is more likely to fail to detect the effect.
3. If a researcher conducted a hypothesis test with an alpha level of a = .02, what z-score values would form the boundaries for the critical region?
3. The .02 would be split between the two tails, with .01 in each tail. The z-score boundaries would be z = +2.33 and z = —2.33
3. Find the mean, median, and mode for the following sample of scores: 6, 2, 4, 1, 2, 2, 3, 4, 3, 2
3. The mean is 29/10 = 2.9, the median is 2.5, and the mode is 2.
3. When you are using an F-max test to evaluate the homogeneity of variance assumption, you usually do not want to find a significant difference between the variances. (True or false?)
3. True. If there is a significant difference between the two variances, you cannot do the t test with pooled variance
3. A researcher obtains z = 2.43 for a hypothesis test. Using a = .01, the researcher should reject the null hypothesis for a one-tailed test but fail to reject for a twotailed test. (True or false?)
3. True. The one-tailed critical value is z = 2.33 and the two-tailed value is z = 2.58.
3. What is the probability of selecting a score greater than 45 from a positively skewed distribution with p = 40 and ct = 10? (Be careful.)
3. You cannot obtain the answer. The unit normal table cannot be used to answer this question because the distribution is not normal.
3. A population has a mean of p. = 40. a. If 5 points were added to every score, what would be the value for the new mean? b. If every score were multiplied by 3, what would be the value for the new mean?
3. a. The new mean would be 45. b. The new mean would be 120.
3. Find the estimated standard error for the sample mean for each of the following samples. a. n = 4 with SS = 48 b. n = 6 with SS = 270 c. n= 12 with SS = 132
3. a. The sample variance is 16 and the estimated standard error is 2. b. The sample variance is 54 and the estimated standard error is 3. c. The sample variance is 12 and the estimated standard error is 1.
4. A distribution with a mean of 70 and a median of 75 is probably positively skewed. (True or false?)
4. False. The mean is displaced toward the tail on the left-hand side.
4. If a sample mean is in the critical region with a = .05, it would still (always) be in the critical region if alpha were changed to a = .01. (True or false?)
4. False. With a = .01, the boundaries for the critical region move farther out into the tails of the distribution. It is possible that a sample mean could be beyond the .05 boundary but not beyond the .01 boundary
4. Find the mean, median, and mode for the following sample of scores: 8, 7, 8, 8, 4, 9, 10, 7, 8, 8, 9, 8
4. For this sample, the mean is 94/12 = 7.83, the median is 8, and the mode is 8.
4. For a sample with a standard deviation of s = 12, a score of X = 83 corresponds to z = 0.50. What is the mean for the sample? (10.13.2014 stats ch 6, zscores)
4. M = 77
4. An automobile manufacturer claims that a new model will average p = 45 miles per gallon with a = 4. A sample ofn = 16 cars is tested and averages only M = 43 miles per gallon. Is this sample mean likely to occur if the manufacturer's claim is true? Specifically, is the sample mean within the range of values that would be expected 95% of the time? (Assume that the distribution of mileage scores is normal.)
4. With n = 16, the standard error is am = I. If the real mean is p = 45, then 95% of all sample means should be within 1.96(1) = 1.96 points of N = 45. This is a range of values from 43.04 to 46.96. Our sample mean is outside this range, so it is not the kind of sample that ought to be obtained if the manufacturer's claim is true
4. A researcher conducts an experiment comparing two treatment conditions and obtains data with 10 scores for each treatment condition. a. If the researcher used an independent-measures design, how many subjects participated in the experiment? b. If the researcher used a repeated-measures design, how many subjects participated in the experiment? c. If the researcher used a matched-subjects design, how many subjects participated in the experiment?
4. a. An independent-measures design would require two separate samples, each with 10 subjects, for a total of 20 subjects. b. A repeated-measures design would use the same sample of n = 10 subjects in both treatment conditions. c. A matched-subjects design would require two separate samples with n = 10 in each, for a total of 20 subjects.
5. True. With a = .01, the boundaries for the critical region are farther out into the tails of the distribution than for a = .05. If a sample mean is beyond the .01 boundary it is definitely beyond the .05 boundary.
5. True. With a = .01, the boundaries for the critical region are farther out into the tails of the distribution than for a = .05. If a sample mean is beyond the .01 boundary it is definitely beyond the .05 boundary
5. A population has a standard deviation of r = 30. a. On average, how much difference should exist between the population mean and the sample mean for n = 4 scores randomly selected from the population? b. On average, how much difference should exist for a sample of n = 25 scores? c. On average, how much difference should exist for a sample of n = 100 scores?
5. a. Standard error = 30/4 = 15 points b. Standard error = 30/25 = 6 points c. Standard error = 30/100 = 3 points
5. One sample has SS = 48 and a second sample has SS = 32. a. If n = 5 for both samples, find each of the sample variances and compute the pooled variance. Because the samples are the same size, you should find that the pooled variance is exactly halfway between the two sample variances. b. Now assume that n = 5 for the first sample and n = 9 for the second. Again, calculate the two sample variances and the pooled variance. You should find that the pooled variance is closer to the variance for the larger sample
5. a. The first sample has s2 = 12 and the second has s2 = 8. The pooled variance is 80/8 = 10 (halfway between). b. The first sample has s2 = 12 and the second has s2 = 4. The pooled variance is 80/12 = 6.67 (closer to the variance for the larger sample).
6. For a population with a mean of p = 70 and a standard deviation of o = 20, how much error, on average, would you expect between the sample mean (M) and the population mean for each of the following sample sizes? a. n = 4 scores b. n = 16 scores c. n = 25 scores
6. a. 20/4 = 10 points b. 20/16 = 5 points c. 20/25 = 4 points
6. a. A repeated-measures study with a sample of n = 25 participants produces a mean difference ofMD = 3 with a standard deviation of s = 4. Based on the mean and standard deviation, you should be able to visualize (or sketch) the sample distribution. Use a two-tailed hypothesis test with a = .05 to determine whether it is likely that this sample came from a population with p, = 0. b. Now assume that the sample standard deviation is s = 12, and once again visualize the sample distribution. Use a two-tailed hypothesis test with a = .05 to determine whether it is likely that this sample came from a population with pp = 0. Explain how the size of the sample standard deviation influences the likelihood of finding a significant mean difference
6. a. The estimated standard error is 0.8 points and t(8) = 3.75. With a critical boundary of ±2.064, reject the null hypothesis. b. The estimated standard error is 2.4 point and t(8) = 1.25. With a critical boundary of ±2.064, fail to reject the null hypothesis. c. A larger standard deviation (or variance) reduces the likelihood of finding a significant difference.
When the RA measured your penis size they determined it was 68 inches long. She is unsure what category to put you in when considering real limits. Aside from your unreal size what would be the section?
67.5 and 68.5 inches
9. Describe the difference in appearance between a bar graph and a histogram and describe the circumstances in which each type of graph is used.
9. A bar graph leaves a space between adjacent bars and is used with data from nominal or ordinal scales. In a histogram, adjacent bars touch at the real limits. Histograms are used to display data from interval or ratio scales.
9. A population with a mean of p. = 10 has /X = 250. How many scores are in the population?
9. N = 25
9. a. After 3 points have been added to every score in a sample, the mean is found to be M = 83 and the standard deviation is s = 8. What were the values for the mean and standard deviation for the original sample? b. After every score in a sample has been multiplied by 4, the mean is found to be M = 48 and the standard deviation is s = 12. What were the values for the mean and standard deviation for the original sample? (10.03.2014 Stats ch 4 textbook extraction/notes)
9. a. The original mean is M = 80 and the standard deviation is s = 8. b. The original mean is M = 12 and the standard deviation is s = 3.
9. For a sample of n = 25 scores, what is the value of the population standard deviation (cr) necessary to produce each of the following a standard error values? a. um = 10 points? b. cr,w = 5 points? c. crAi = 2 points?
9. a. σ = 50 b. σ = 25 c. σ = 10
A study investigating the effect on your attractiveness on Tinder when you post a shirtless photo of yours abs finds that out of a sample of n=6 profiles the null hypothesis could not be rejected. What type of statistical error could occur here?
In this case it is possible that the null hypothesis should be rejected but the test was not sensitive enough to detect it.
27. Use a stem and leaf display to organize the following distribution of scores. Use seven stems with each stem corresponding to a 10-point interval. Scores: 28, 54, 65, 53, 81 45, 44, 51, 72, 34 43, 59, 65, 39, 20 53, 74, 24, 30, 49 36, 58, 60, 27, 47 22, 52, 46, 39, 65 hj
Ni j2b n997. 2 │80472 3 │49069 4 │543976 5 │4319382 6 │5505 7 │24 8 │1
What happens to the mean if you multiply every distribution by the same amount? What happens if you divide? (09.28.2014 Stats 3 Ch 3 - Central tendency (notes + questions)
Nothing changes
Survey asks to identify the number of cats they have, annual income, and martial status. Which forms of scale of measurement should be used for each?
Number of cats: ratio, annual income: ratio, and martial status: nominal.
26. A set of scores has been organized into the following stem and leaf display. For this set of scores: a. How many scores are in the 70s? b. Identify the individual scores in the 70s. c. How many scores are in the 40s? d. Identify the individual scores in the 40s. 3 8 4 60 5 734 6 81469 7 2184 8 247
Thanks so that the only thing26. a. 4 b. 72, 71, 78, and 74 c. 2 d. 46 and 40
What are the independent variables for an experiment where boys and girls of the same age are evaluated for their IQ scores? Why is this a non experimental quasi-independent variable?
The quasi-independent variable is the boys and girls while the dependent is IQ score. This is not a true manipulation of variables.
What does Hartey's F-Max test do?
This test tries to get an understanding of how different the sample variances are. If it fails to reject H0 then it is good news because the differences are not extreme enough to be a problem
3. The tail will be on the right-hand side of a normal distribution for any positive z-score. (True or false)
True
Though central tendencies tell us good information, it is missing deacriptions of how many people are close to the average or whether they are scattered throughout the scores. Measures of variability therefore is used to find how variable terms alongside _______ statistics such as mean, medium, or mode. (10.02.2014, stats lec 4 variability)
basic descriptive statistics
Though central tendencies tell us good information, it is missing deacriptions of ____________. Measures of variability therefore is used to find how variable terms alongside basic descriptive statistics such as mean, medium, or mode. (10.02.2014, stats lec 4 variability)
how many people are close to the average or whether they are scattered throughout the scores
The standard deviation of the distribution of sample means, gm, is called the standard error of M. The standard error provides a measure of
how much distance is expected on average between a sample mean (M) and the population mean (p).
Deviation from the mean is the conventional form of variation. It measures the deviation from a common central tendency. It can tell us how an individual stands in relation to the other scores as well as how accurate this set of scores is to a population. When there is a small vulnerability then it is a good representation because _________________ (10.02.2014, stats lec 4 variability)
it means that more scores are contained in an area.
Degrees of freedom describe the
number of scores in a sample that are independent and free to vary. Because the sample mean places a restriction on the value of one score in the sample, there are n — 1 degrees of freedom for a sample with n scores
Professor Morgan from Abnormal psychology uses letter grades to evaluate a set of student essays. What kind of scale is being used to measure the quality of the essays?
ordinal scale
The second is the Interquartile range of 25th to 75th percentile. This is a better representation of the outliers, but may not give the entire picture of the variables. To achieve this you look for the the_______________. The general idea is (3rd quarter) - (1st quarter). A semi interqualtile range is the middle of distribution to the boundary which is defined by the middle 50%. (10.02.2014, stats lec 4 variability)
second quarter of the value
There are three forms of verifiability. The first is called the range difference .It is the simplest type but can be influenced by outliers or explain values which make it less useful. The second is the Interquartile range of 25th to 75th percentile. This is a better representation of the outliers, but may not give the entire picture of the variables. The third is or called the __________ which is the average squared distance from the mean. This involves finding the distance of each integer of the mean, square each distance, find the average of the result and then square rooting it. (10.02.2014, stats lec 4 variability)
standard variation
What is the order of operations when you consider SIGMA?
1. Parenthes, 2. Squaring, 3. Multiply, 4. Summation, 5. Addition/subtraction
standard error and it's relationship to sample varience and sample size (a) (Increase) Sample varience
(a) (Increase) standard error (increase) Sample error (b) (Increase) sample size (decrease) Sample errorq
What are the zscore (a) α=0.5 (b) 0.01 (c) 0.001
(a) 1.96 (b) 2.58 (c) 3.3
How does the following variables effect (a) Sample (b) alpha level change (c) 2 tail to 1 tail
(a) Smalle n smaller O, and a large B (b) Making the alpha more precise lowers the hypothesis (c) Going from two tailed to 1 trailed inceases the power of the test.
When you add a constant number to the variables the mean ___(a)___ but the standard deviation ____(b)____. On the other hand if you multiply by a constant then the standard deviation will increase by that change multiple. (10.02.2014, stats lec 4 variability)
(a) mean changes (b) standard deviation does not change
What are the alpha levels for (a) 1.96 (b) 2.58 (c) 3.3
(a) α=0.5 (b) 0.01 (c) 0.001
1. For a population with a mean of p = 40 and a standard deviation of o- = 8, find the z-score corresponding to a sample mean of M = 44 for each of the following sample sizes. a. n = 4 b. n = 16
. a. The standard error is um = 4, and z =1.00. b. The standard error is um = 2, and z = 2.00.
1. Identify the z-score value corresponding to each of the following locations in a distribution. a. Below the mean by 2 standard deviations. b. Above the mean by z standard deviation. c. Below the mean by 1.50 standard deviations. (10.13.2014 stats ch 6, zscores)
. a. z = —2.00 b. z = +0.50 C. z = —1.50
1. For a particular hypothesis test, the power is .50 (50%) for a 5-point treatment effect. Will the power be greater or less than .50 for a 10-point treatment effect?
1. The hypothesis test is more likely to detect a 10-point effect, so power will be greater
1. What assumptions must be satisfied for repeated-measures t tests to be valid?
1. The observations within a treatment are independent. The population distribution of D scores is assumed to be normal.
Three assumptions that should be satisfied before you use the independentmeasures t formula for hypothesis testing:
1. The observations within each sample must be independent (see p. 254). 2. The two populations from which the samples are selected must be normal. 3. The two populations from which the samples are selected must have equal variances.
1. What general purpose is served by a good measure of central tendency?
1. The purpose of central tendency is to identify a single score that serves as the best representative for an entire distribution, usually a score from the center of the distribution.
1. What information is provided by the sign (+1-) of a z-score? What information is provided by the numerical value of the z-score? (10.13.2014 stats ch 6, zscores)
1. The sign of the z score tells whether the location is above (+) or below (-) the mean, and the magnitude tells the distance from the mean in terms of the number of standard deviations.
1. A normal-shaped distribution with II = 40 and a = 8 is transformed into z-scores. Describe the shape, the mean, and the standard deviation for the resulting distribution of z-scores. (10.13.2014 stats ch 6, zscores)
1. The z-score distribution would be normal with a mean of 0 and a standard deviation of 1.
1. If a researcher predicts that a treatment will increase scores, then the critical region for a one-tailed test would be located in the right-hand tail of the distribution. (True or false?)
1. True. A large sample mean, in the right-hand tail, would indicate that the treatment worked as predicted.
1. Place the following sample of n = 20 scores in a frequency distribution table. 6, 9, 9, 10, 8, 9, 4, 7, 10, 9 5, 8, 10, 6, 9, 6, 8, 8, 7, 9
1. X f ──── 10 3 9 6 8 4 7 2 6 3 5 1 4 1 ────
1. For a distribution with p. = 40 and o- = 12, find the z-score for each of the following scores. a. X = 36 b. X = 46 c. X = 56 (10.13.2014 stats ch 6, zscores) (10.13.2014 stats ch 6, zscores)
1. a) z= -0.33 b) 0.5 c) z=1.33
1. A new over-the-counter cold medication includes a warning label stating that it "may cause drowsiness." A researcher would like to evaluate this effect. It is known that under regular circumstances the distribution of reaction times is normal with p = 200. A sample of n = 9 participants is obtained. Each person is given the new cold medication, and, 1 hour later, reaction time is measured for each individual. The average reaction time for this sample is M = 206 with SS = 648. The researcher would like to use a hypothesis test with a = .05 to evaluate the effect of the medication. a. Use a two-tailed test with a = .05 to determine whether the medication has a significant effect on reaction time. b. Write a sentences that demonstrates how the outcome of the hypothesis test would appear in a research report. c. Use a one-tailed test with a = .05 to determine whether the medication produces a significant increase in reaction time. d. Write a sentence that demonstrates how the outcome of the one-tailed hypothesis test would appear in a research report.
1. a. For the two-tailed test, Ho: p = 200. The sample variance is 81, the estimated standard error is 3, and t = 6/3 = 2.00. With df = 8, the critical boundaries are ± 2.306. Fail to reject the null hypothesis. b. The result indicates that the medication does not have a significant effect on reaction time, t(8) = 2.00, p > .05. c. For a one-tailed test, H0: p 5_ 200 (no increase). The data product t = 6/3 = 2.00. With df = 8, the critical boundary is 1.860. Reject the null hypothesis. d. The results indicate that the medication produces a significant increase in reaction time, t(8) = 2.00, p < .05, one tailed.
1. A researcher is investigating the effectiveness of acupuncture treatment for chronic back pain. A sample of n = 4 participants is obtained from a pain clinic. Each individual ranks the current level of pain and then begins a 6-week program of acupuncture treatment. At the end of the program, the pain level is rated again and the researcher records the amount of difference between the two ratings. For this sample, pain level decreased by an average of M = 4.5 points with SS = 27. a. Are the data sufficient to conclude that acupuncture has a significant effect on back pain? Use a two-tailed test with a = .05. b. Can you conclude that acupuncture significantly reduces back pain? Use a one-tailed test with a = .05.
1. a. For these data, the sample variance is 9, the standard error is 1.50, and t = 3.00. With df = 3, the critical values are t = ±3.182. Fail to reject the null hypothesis. b. For a one-tailed test, the critical value is t = 2.353. Reject the null hypothesis and conclude that acupuncture treatment significantly reduces pain
1. a. How does increasing sample size influence the outcome of a hypothesis test? b. How does increasing sample size influence the value of Cohen's d?
1. a. Increasing sample size increases the likelihood of rejecting the null hypothesis. b. Cohen's d is not influenced at all by the sample size
1. In words, explain what is measured by each of the following: a. SS b. Variance c. Standard deviation (10.03.2014 Stats ch 4 textbook extraction/notes)
1. a. SS is the sum of squared deviation scores. b. Variance is the mean squared deviation. c. Standard deviation is the square root of the variance. It provides a measure of the standard distance from the mean.
1. Briefly define each of the following: a. Distribution of sample means b. Expected value of M c. Standard error of M
1. a. The distribution of sample means consists of the sample means for all the possible random samples of a specific size (n) from a specific population. b. The expected value of M is the mean of the distribution of sample means (μ). c. The standard error of M is the standard deviation of the distribution of sample means (σM = σ/n).
1. A population has a mean of p = 50 and a standard deviation of a = 12. a. For samples of size n = 4, what is the mean (expected value) and the standard deviation (standard error) for the distribution of sample means? b. If the population distribution is not normal, describe the shape of the distribution of sample means based on n = 4. c. For samples of size n = 36, what is the mean (expected value) and the standard deviation (standard error) for the distribution of sample means? d. If the population distribution is not normal, describe the shape of the distribution of sample means based on n = 36.
1. a. The distribution of sample means would have a mean of p = 50 and a standard error of am = 12/VTI = 6. b. The distribution of sample means does not satisfy either criterion to be normal. It would not be a normal distribution. c. The distribution of sample means is normal and would have a mean of p = 50 and a standard error of om = 12/V = 2. d. Because the sample size is greater than 30, the distribution of sample means is a normal distribution.
1. A sample of n = 16 individuals is selected from a population with a mean of = 80. A treatment is administered to the sample and, after treatment, the sample mean is found to be M = 86 with a standard deviation of s = 8. a. Does the sample provide sufficient evidence to conclude that the treatment has a significant effect? Test with a = .05. b. Compute Cohen's d and r2 to measure the effect size. c. Find the 95% confidence interval for the population mean after treatment
1. a. The estimated standard error is 2 points and the data produce t = 6/2 = 3.00. With df = 15, the critical values are t = ±2.131, so the decision is to reject H0 and conclude that there is a significant treatment effect. b. For these data, d = 6/8 = 0.75 and r2 = 9/24 = 0.375 or 37.5%. c. For 95% confidence and df = 15, use t = ±2.131. The confidence interval is p. = 86 ±2.131(2) and extends from 81.738 to 90.262.
1. A researcher is using an independent-measures design to evaluate the difference between two treatment conditions with n = 8 in each treatment. The first treatment produces M = 63 with a variance of s2 = 18, and the second treatment has M = 58 with s2 = 14. a. Use a one-tailed test with a = .05 to determine whether the scores in the first treatment are significantly greater than the scores in the second. (Note: Because the two samples are the same size, the pooled variance is simply the average of the two sample variances.) b. Predict how the value for the t statistic would be affected if the two sample variances were increased to s2 = 68 and s2 — 60. Compute the new t to confirm your answer. c. Predict how the value for the t statistic for the original samples would be affected if each sample had n = 32 scores (instead of n = 8). Compute the new t to confirm your answer.
1. a. The pooled variance is 16, the estimated standard error is 2, and t(14) = 2.50. With a one-tailed critical value of 1.761, reject the null hypothesis. Scores in the first treatment are significantly higher than scores in the second. b. Increasing the variance should lower the value of t. The new pooled variance is 64, the estimated standard error is 4, and t(14) = 1.25. c. Increasing the sample sizes should increase the value of t. The pooled variance is still 16, but the new standard error is 1, and t(62) = 5.00.
. An educational psychologist would like to determine whether access to computers has an effect on grades for high school students. One group of n = 16 students has home room each day in a computer classroom in which each student has a computer. A comparison group of n = 16 students has home room in a traditional classroom. At the end of the school year, the average grade is recorded for each student. The data are as follows: Computer Traditional M = 86 M = 82.5 SS = 1005 SS = 1155 a. Is there a significant difference between the two groups? Use a two-tailed test with a = .05. b. Compute Cohen's d to measure the size of the difference. c. Write a sentence that demonstrates how the outcome of the hypothesis test and the measure of effect size would appear in a research report. d. Compute the 90% confidence interval for the population mean difference between a computer classroom and a regular classroom.
1. a. The pooled variance is 72, the standard error is 3, and t = 1.17. With a critical value of t = 2.042, fail to reject the null hypothesis. b. Cohen's d = 3.5A172 = 0.412 c. The results show no significant difference in grades for students with computers compared to students without computers, t(30) = 1.17, p > .05, d = 0.412. d. With df = 30 and 90% confidence, the t values for the confidence interval are ± 1.697. The interval is p., — p,2 = 3.5 ± 1.697(3). Thus, the population mean difference is estimated to be between —1.591 and 8.591. The fact that zero is an acceptable value (inside the interval) is consistent with the decision that there is no significant difference between the two population means
1. For a normal distribution with a mean of |a = 60 and a standard deviation of ct = 12, find each probability value requested. a. p(X > 66) b. p(X < 75) c. p(X < 57) d. p (48 < X < 72)
1. a. p = 0.3085 b. p = 0.8944 c. p = 0.4013 d. p = 0.6826
1. Find the proportion of a normal distribution that corresponds to each of the following sections: a. z < 0.25 b. z > 0.80 c. z < -1.50 d. z > -0.75
1. a. p ~ 0.5987 b. p = 0.2119 c. p = 0.0668 d. p = 0.7734
1. A population of scores has p. = 73 and a = 8. If the distribution is standardized to create a new distribution with 1.1, = 100 and a = 20, what are the new values for each of the following scores from the original distribution? a. X = 65 b. X = 71 c. X = 81 d. X = 83 (10.13.2014 stats ch 6, zscores)
1. a. z = —1.00, X = 80 b. z = —0.25, X = 95 c. z = 1.00, X = 120 d. z = 1.25, X = 125
10. A random sample ofn = 16 individuals is selected from a population with p = 70, and a treatment is administered to each individual in the sample. After treatment, the sample mean is found to be M = 76 with SS = 960. a. How much difference is there between the mean for the treated sample and the mean for the original population? (Note: In a hypothesis test, this value forms the numerator of the t statistic.) b. How much difference is expected just by chance between the sample mean and its population mean? That is, find the standard error for M. (Note: In a hypothesis test, this value is the denominator of the t statistic.) c. Based on the sample data, does the treatment have a significant effect? Use a two-tailed test with a = .05.
10. a. 6 points b. The sample variance is 64 and the estimated standard error is sM = 2. c. For these data, t = 3.00. With df = 15 the critical value is t = ±2.131. Reject H0 and conclude that there is a significant effect.
10. For each of the following, assume that the two samples are selected from populations with equal means and calculate how much difference should be expected, on average, between the two sample means. a. Each sample has n = 5 scores with s2 = 38 for the first sample and s2 = 42 for the second. (Note: Because the two samples are the same size, the pooled variance is equal to the average of the two sample variances.) b. Each sample has n = 20 scores with s2 = 38 for the first sample and s2 = 42 for the second. c. In part b, the two samples are bigger than in part a, but the variances are unchanged. How does sample size affect the size of the standard error for the sample mean difference?
10. a. The estimated standard error for the sample mean difference is 4 points. b. The estimated standard error for the sample mean difference is 2 points. c. Larger samples produce a smaller standard error.
10. Research has shown that losing even one night's sleep can have a significant effect on performance of complex tasks such as problem solving (Linde & Bergstroem, 1992). To demonstrate this phenomenon, a sample of n = 25 college students was given a problem-solving task at noon on one day and again at noon on the following day. The students were not permitted any sleep between the two tests. For each student, the difference between the first and second score was recorded. For this sample, the students averaged MD = 4.7 points better on the first test with a variance of s2 = 64 for the difference scores. a. Do the data indicate a significant change in problemsolving ability? Use a two-tailed test with a = .05. b. Compute an estimated Cohen's d to measure the size of the effect
10. a. The null hypothesis says that losing one night's sleep will have no effect on performance. For these data, the estimated standard error is 1.60, and t(24) = 2.94. With df = 24, the critical boundaries are ±2.064. Reject the null hypothesis. b. Cohen's d = 4.7/64 = 0.5875
10. Find the z-score corresponding to a score of X = 60 for each of the following distributions. a. μ = 50 and cr = 20 b. p. = 50 and a = 10 c. μ=50anda=5 d. μ= 50anda=2 (10.13.2014 stats ch 6, zscores)
10. a. z = +0.50 b. z = +1.00 c. z = +2.00 d. z = +5.00
10. Find the z-score location of a vertical line that separates a normal distribution as described in each of the following. a. 20% in the tail on the left b. 40% in the tail on the right c. 75% in the body on the left d . 99% in the body on the right
10. a. z = -0.84 c. z = 0.67 b. z = 0.25 d. z = -2.33
10. For a population with a mean of p = 80 and a standard deviation of a = 12, find the z-score corresponding to each of the following samples. a. M = 83 for a sample of n = 4 scores b. M = 83 for a sample of n = 16 scores c. M = 83 for a sample ofn = 36 scores
10. a. σM = 6 points and z = 0.50 b. σM = 3 points and z = 1.00 c. σM = 2 points and z = 1.50
11. A sample of n = 5 scores has a mean of M = 12. If one person with a score of X = 8 is removed from the sample, what is the value for the new mean?
11. The original sample has n= 5 and ΣX = 60. The new sample has n= 4 and ΣX= 52. The new mean is M = 13.
11. For each of the following, calculate the pooled variance and the estimated standard error for the sample mean difference a. The first sample has n = 4 scores and a variance of s2 = 55, and the second sample has n = 6 scores and a variance of s2 = 63. b. Now the sample variances are increased so that the first sample has n = 4 scores and a variance of s2 = 220, and the second sample has n = 6 scores and a variance of s2 = 252. c. Comparing your answers for parts a and b, how does increased variance influence the size of the estimated standard error?
11. a. The pooled variance is 60 and the estimated standard error is 5. b. The pooled variance is 240 and the estimated standard error is 10. c. Increasing the sample variance produces an increase in the standard error.
11. A random sample is selected from a normal population with a mean of p. = 40 and a standard deviation of o = 10. After a treatment is administered to the individuals in the sample, the sample mean is found to be M = 42. a. How large a sample is necessary for this sample mean to be statistically significant? Assume a two-tailed test with a = .05. b. If the sample mean were M = 41, what sample size is needed to be significant for a two-tailed test with a = .05?
11. a. With a 2-point treatment effect, for the z-score to be greater than 1.96, the standard error must be smaller than 1.02. The sample size must be greater than 96.12; a sample of n = 97 or larger is needed. b. With a 1-point treatment effect, for the z-score to be greater than 1.96, the standard error must be smaller than 0.51. The sample size must be greater than 384.47; a sample of n = 385 or larger is needed.
11. The spotlight effect refers to overestimating the extent to which others notice your appearance or behavior, especially when you commit a social faux pas. Effectively, you feel as if you are suddenly standing in a spotlight with everyone looking. In one demonstration of this phenomenon, Gilovich, Medvec, and Savitsky (2000) asked college students to put on a Barry Manilow T-shirt that fellow students had previously judged to be embarrassing. The participants were then led into a room in which other students were already participating in an experiment. After a few minutes, the participant was led back out of the room and was allowed to remove the shirt. Later, each participant was asked to estimate how many people in the room had noticed the shirt. The individuals who were in the room were also asked whether they noticed the shirt. In the study, the participants significantly overestimated the actual number of people who had noticed. a. In a similar study using a sample of n = 9 participants, the individuals who wore the shirt produced an average estimate of M = 6.4 with SS = 162. The average number who said they noticed was 3.1. Is the estimate from the participants significantly different from the actual number? Test the null hypothesis that the true mean is p = 3.1 using a two-tailed test with a = .05. b. Is the estimate from the participants significantly higher than the actual number (p = 3.1)? Use a one-tailed test with a = .05.
11. a. With a two tailed test, the critical boundaries are ±2.306 and the obtained value of t = 3.3/1.5 = 2.20 is not sufficient to reject the null hypothesis. b. For the one-tailed test the critical value is 1.860, so we reject the null hypothesis and conclude that participants significantly overestimated the number who noticed.
11. Find the X value corresponding to z = 0.25 for each of the following distributions. a. p, = 40 and = 4 b. p. = 40 and a = 8 c. μ= 40andC= 12 d. μ =40and a= 20 (10.13.2014 stats ch 6, zscores)
11. a. X = 41 b. X = 42 c. X = 43 d. X = 45
11. Find the z-score boundaries that separate a normal distribution as described in each of the following. a. The middle 20% from the 80% in the tails. b. The middle 50% from the 50% in the tails. c. The middle 95% from the 5% in the tails. d. The middle 99% from the 1 % in the tails
11. a. z = ±0.25 c. z = ±1.96 b. z = ±0.67 d. z = ±2.58
11. A sample of n = 4 scores has a mean of M = 75. Find the z-score for this sample: a. If it was obtained from a population with p = 80 and if = 10. b. If it was obtained from a population with p = 80 and if = 20. c. If it was obtained from a population with p = 80 and if = 40.
11. a. σM = 5 points and z = 1.00 b. σM = 10 points and z = 0.50 c. σM = 20 points and z = 0.25
12. How would you react to doing much worse on an exam than you expected? There is some evidence to suggest that most individuals believe that they can cope with this kind of problem better than their fellow students (Igou, 2008). In the study, participants read a scenario of a negative event and were asked to use a 10-point scale to rate how it would affect their immediate well-being (-5 strongly worsen to +5 strongly improve). Then they were asked to imagine the event from the perspective of an ordinary fellow student and rate how it would affect that person. The difference between the two ratings was recorded. Suppose that a sample of n = 25 participants produced a mean difference ofMD = 1.28 points (self rated higher) with a standard deviation of s = 1.50 for the difference scores. a. Is this result sufficient to conclude that there is a significant difference in the ratings for self versus others? Use a two-tailed test with a = .05. b. Compute r2 and estimate Cohen's d to measure the size of the treatment effect. c. Write a sentence describing the outcome of the hypothesis test and the measure of effect size as it would appear in a research report.
12. a. The null hypothesis states that self-ratings are not different from ratings of others. For these data the estimated standard error is 0.30 and t = 4.27. With df = 24, the critical value is 2.064. Reject the null hypothesis and conclude that self-ratings are significantly different from ratings of others. b. r2 = 4.272/(4.272 + 24) = 0.432 and Cohen's d = 1.28/1.50 = 0.853. c. Participants rated their own ability to cope significantly higher than the ability of others, t(24) = 4.27, p < .01, d = 0.853
12. There is some evidence that REM sleep, associated with dreaming, may also play a role in learning and memory processing. For example, Smith and Lapp (1991) found increased REM activity for college students during exam periods. Suppose that REM activity for a sample of n = 16 students during the final exam period produced an average score of M = 143. Regular REM activity for the college population averages p. = 110 with a standard deviation of cr = 50. The population distribution is approximately normal. a. Do the data from this sample provide evidence for a significant increase in REM activity during exams? Use a one-tailed test with a = .01. b. Compute Cohen's d to estimate the size of the effect. c. Write a sentence describing the outcome of the hypothesis test and the measure of effect size as it would appear in a research report.
12. a. The null hypothesis states that there is no increase in REM activity, µ ≤ 110. The critical region consists of z-scores beyond z = 2.33. For these data, the standard error is 12.5 and z = 33/12.5 = 2.64. Reject H0. There is a significant increase in REM activity. b. Cohen's d = 33/50 = 0.66. c. The results show a significant increase in REM activity for college students during exam periods, z = 2.64, p < .01, d = 0.66.
2. Can SS ever have a value less than zero? Explain your answer. (10.03.2014 Stats ch 4 textbook extraction/notes)
2. SS cannot be less than zero because it is computed by adding squared deviations. Squared deviations are always greater than or equal to zero.
12. A researcher conducts an independent-measures study comparing two treatments and reports the t statistic as t(30) = 2.085. a. How many individuals participated in the entire study b. Using a two-tailed test with a = .05, is there a significant difference between the two treatments? c. Compute r2 to measure the percentage of variance accounted for by the treatment effect.
12. a. The two samples combined have a total of 32 participants. b. With df = 30 and α = .05, the critical region consists of t values beyond 2.042. The t statistic is in the critical region. Reject H0 and conclude that there is a significant difference. c. r2 =4.35/34.35 = 0.127 or 12.7%
12. A population forms a normal distribution with a mean of p = 80 and a standard deviation of a = 15. For each of the following samples, compute the z-score for the sample mean and determine whether the sample mean is a typical, representative value or an extreme value for a sample of this size. a. M = 84 for n = 9 scores b. M = 84 for n = 100 scores
12. a. With a standard error of 5, M = 84 corresponds to z = 0.80, which is not extreme. b. With a standard error of 1.5, M = 84 corresponds to z = 2.67, which is extreme.
12. Many animals, including humans, tend to avoid direct eye contact and even patterns that look like eyes. Some insects, including moths, have evolved eye-spot patterns on their wings to help ward off predators. Scaife (1976) reports a study examining how eye-spot patterns affect the behavior of birds. In the study, the birds were tested in a box with two chambers and were free to move from one chamber to another. In one chamber, two large eye-spots were painted on one wall. The other chamber had plain walls. The researcher recorded the amount of time each bird spent in the plain chamber during a 60-minute session. Suppose the study produced a mean of M = 37 minutes in the plain chamber with SS = 288 for a sample of n = 9 birds. (Note: If the eye-spots have no effect, then the birds should spend an average of p = 30 minutes in each chamber.) a. Is this sample sufficient to conclude that the eyespots have a significant influence on the birds' behavior? Use a two-tailed test with a = .05. b. Compute the estimated Cohen's d to measure the size of the treatment effect. c. Construct the 95% confidence interval to estimate the mean amount of time spent on the plain side for the population of birds
12. a. With df = 8, the critical values are ±2.306. For these data, the sample variance is 36, the estimated standard error is 2, and t = 7/2 = 3.50. Reject the null hypothesis and conclude that the amount of time spent in the plain chamber is significantly different from chance. b. d = 7/6 = 1.17. c. With df = 8, the t values for 95% confidence are 2.306, and the interval extends from 2.388 to 11.612 seconds.
12. A survey given to a sample of 200 college students contained questions about the following variables. For each variable, identify the kind of graph that should be used to display the distribution of scores (histogram, polygon, or bar graph). a. number of pizzas consumed during the previous week b. size of T-shirt worn (S, M, L, XL) c. gender (male/female) d. grade point average for the previous semester e. college class (freshman, sophomore, junior, senior)
12. a. histogram or polygon (ratio scale) b. bar graph (ordinal scale) c. bar graph (nominal scale) d. histogram or polygon (ratio scale) e. bar graph (ordinal scale)
12. For a normal distribution with a mean of p = 80 and a standard deviation of ct = 20, find the proportion of the population corresponding to each of the following scores. a. Scores greater than 85. b. Scores less than 100. c. Scores between 70 and 90.
12. a. p(z > 0.25) = 0.4013 b. p(z < 1.00) = 0.8413 c. p(-0.50 < z < 0.50) = 0.3830
12. A score that is 6 points below the mean corresponds to a z-score of z = -0.50. What is the population standard deviation? (10.13.2014 stats ch 6, zscores)
12. σ = 12
13. A sample of n = 9 scores has a mean of M = 10. One person with a score of X = 2 is removed from the sample. What is the value for the new sample mean?
13. After the score is removed, n =8, ΣX= 88, and M = 11.
13. Research results indicate that physically attractive people are also perceived as being more intelligent (Eagly, Ashmore, Makhijani, & Longo, 1991). As a demonstration of this phenomenon, a researcher obtained a set of 10 photographs, 5 showing men who were judged to be attractive and 5 showing men who were judged to be unattractive. The photographs were shown to a sample of n = 25 college students and the students were asked to rate the intelligence of the person in the photo on a scale from 1 to 10. For each student, the researcher determined the average rating for the 5 attractive photos and the average for the 5 unattractive photos, and then computed the difference between the two scores. For the entire sample, the average difference was MD = 2.7 (attractive photos rated higher) with s = 2.00. Are the data sufficient to conclude that there was a significant difference in perceived intelligence for the two sets of photos? Use a two-tailed test at the .05 level of significance
13. The null hypothesis states that there is no difference in the perceived intelligence between attractive and unattractive photos. For these data, the estimated standard error is 0.4 and t = 2.7/0.4 = 6.75. With df = 24, the critical value is 2.064. Reject the null hypothesis.
13. There is some evidence indicating that people with visible tattoos are viewed more negatively than people without visible tattoos (Resenhoeft, Villa, & Wiseman, 2008). In a similar study, a researcher first obtained overall ratings of attractiveness for a woman with no tattoos shown in a color photograph. On a 7-point scale, the woman received an average rating of = 4.9, and the distribution of ratings was normal with a standard deviation of if = 0.84. The researcher then modified the photo by adding a tattoo of a butterfly on the woman's left arm. The modified photo was then shown to a sample of n = 16 students at a local community college and the students used the same 7-point scale to rate the attractiveness of the woman. The average score for the photo with the tattoo was M = 4.2. a. Do the data indicate a significant difference in rated attractiveness when the woman appeared to have a tattoo? Use a two-tailed test with a = .05. b. Compute Cohen's d to measure the size of the effect. c. Write a sentence describing the outcome of the hypothesis test and the measure of effect size as it would appear in a research report.
13. a. H0: μ = 4.9 and the critical values are ±1.96. The standard error is 0.21 and z = -3.33. Reject the null hypothesis. b. Cohen's d = 0.7/0.84 = 0.833 or 83.3% c. The results indicate that the presence of a tattoo has a significant effect on the judged attractiveness of a woman, z = -3.33, p < .01, d = 0.833.
2. Why is it usually considered inappropriate to compute a mean for scores measured on an ordinal scale?
2. The definition of the mean is based on distances (the mean balances the distances) and ordinal scales do not measure distance.
13. A normal distribution has a mean of p = 50 and a standard deviation of ct = 12. For each of the following scores, indicate whether the tail is to the right or left of the score and find the proportion of the distribution located in the tail. a. X = 53 b. X = 44 c. X = 68 d. X = 38
13. a. tail to the right, p = 0.4013 b. tail to the left, p = 0.3085 c. tail to the right, p = 0.0668 d. tail to the left, p = 0.1587
14. A population of N = 20 scores has a mean of p. = 15. One score in the population is changed from X = 8 to X = 28. What is the value for the new population mean?
14. After the score is changed, N = 20, ΣX= 320, and μ =16.
14. The librarian at the local elementary school claims that, on average, the books in the library are more than 20 years old. To test this claim, a student takes a sample of n = 30 books and records the publication date for each. The sample produces an average age of M = 23.8 years with a variance of s2 = 67.5. Use this sample to conduct a one-tailed test with a = .01 to determine whether the average age of the library books is significantly greater than 20 years (p > 20).
14. The null hypothesis states that the average age is not more than 20 years: H0: μ < 20 years. With df = 29, the one-tailed critical value is 2.462. For these data, the estimated standard error is 1.5, and t = 3.8/1.5 = 2.53. Reject the null hypothesis and conclude that the average age of the library books is significantly greater than 20 years.
14. Do you view a chocolate bar as delicious or as fattening? Your attitude may depend on your gender. In a study of American college students, Rozin, Bauer, and Catanese (2003) examined the importance of food as a source of pleasure versus concerns about food associated with weight gain and health. The following results are similar to those obtained in the study. The scores are a measure of concern about the negative aspects of eating. Males Females n = 9 n = 15 M = 33 M = 42 SS = 740 SS = 1240 a. Based on these results, is there a significant difference between the attitudes for males and for females? Use a two-tailed test with a = .05. b. Compute ?, the percentage of variance accounted for by the gender difference, to measure effect size for this study. c. Write a sentence demonstrating how the result of the hypothesis test and the measure of effect size would appear in a research report.
14. a. The pooled variance is 90, the estimated standard error is 4, and t = 9/4 = 2.25. With df = 22 the critical value is 2.074. Reject the null hypothesis and conclude that there is a significant difference in attitude between males and females. b. r2 = 5.06/27.06 = 0.187 or 18.7% c. The results show a significant difference between males and females in their attitude toward food, t(22) = 2.25, p < .05, r2 = 0.187.
14. The range is completely determined by the two extreme scores in a distribution. The standard deviation, on the other hand, uses every score. a. Compute the range (choose either definition) and the standard deviation for the following sample of n = 5 scores. Note that there are three scores clustered around the mean in the center of the distribution, and two extreme values. Scores: 0, 6, 7, 8, 14. b. Now we break up the cluster in the center of the distribution by moving two of the central scores out to the extremes. Once again compute the range and the standard deviation. New scores: 0, 0, 7, 14, 14. c. According to the range, how do the two distributions compare in variability? How do they compare according to the standard deviation? (10.03.2014 Stats ch 4 textbook extraction/notes)
14. a. The range is either 14 or 15, and the standard deviation is s = 5. b. After spreading out the two scores in the middle, the range is still 14 or 15 but the standard deviation is now s = 7. c. The two distributions are the same according to the range. The range is completely determined by the two extreme scores and is insensitive to the variability of the rest of the scores. The second distribution has more variability according to the standard deviation, which measures variability for the complete set.
14. IQ test scores are standardized to produce a normal distribution with a mean of p = 100 and a standard deviation of cr = 15. Find the proportion of the population in each of the following IQ categories. a. Genius or near genius: IQ greater than 140 b. Very superior intelligence: IQ between 120 and 140 c. Average or normal intelligence: IQ between 90 and 109
14. a. z = 2.67, p = 0.0038 b. p(1.33 < z < 2.67) = 0.0880 c. p(-0.67 < z < 0.60) = 0.4743
14. The population of IQ scores forms a normal distribution with a mean of p = 100 and a standard deviation of a = 15. What is the probability of obtaining a sample mean greater than M = 97, a. for a random sample of n = 9 people? b. for a random sample of n = 25 people?
14. a. σM = 5, z = 0.60, and p = 0.7257 b. σM = 3, z = 1.00, and p = 0.8413
14. For a population with a standard deviation of o- = 8, a score of X = 44 corresponds to z = —0.50. What is the population mean? (10.13.2014 stats ch 6, zscores)
14. μ = 48
15. A sample of n = 7 scores has a mean of M = 9. One score in the sample is changed from X = 19 to X = 5. What is the value for the new sample mean?
15. After the score is changed, n = 7, ΣX= 49, and M= 7.
15. For a sample with a standard deviation of s = 10, a score of X = 65 corresponds to z = 1.50. What is the sample mean? (10.13.2014 stats ch 6, zscores)
15. M = 50
15. In a study examining overweight and obese college football players, Mathews and Wagner (2008) found that on average both offensive and defensive linemen exceeded the at-risk criterion for body mass index (BMI). BMI is a ratio of body weight to height squared and is commonly used to classify people as overweight or obese. Any value greater than 30 kg/m2 is considered to be at risk. In the study, a sample of n = 17 offensive linemen averaged M = 34.4 with a 348 CHAPTER 10 THE t TEST FOR TWO INDEPENDENT SAMPLES standard deviation ofs = 4.0. A sample ofn = 19 defensive linemen averaged M = 31.9 with s = 3.5. a. Use a single-sample t test to determine whether the offensive linemen are significantly above the at-risk criterion for BMI. Use a one-tailed test with a = .01. b. Use a single-sample t test to determine whether the defensive linemen are significantly above the at-risk criterion for BMI. Use a one-tailed test with a = .01. c. Use an independent-measures t test to determine whether there is a significant difference between the offensive linemen and the defensive linemen. Use a two-tailed test with a = .01.
15. a. For the offensive linemen, the standard error is 0.97 and t = 4.54. For a one-tailed test with df = 16, the critical value is 2.583. Reject the null hypothesis. The offensive linemen are significantly above the criterion for BMI. b. For the defensive linemen, the standard error is 0.80 and t = 2.375. For a one-tailed test with df = 18, the critical value is 2.552. Fail to reject the null hypothesis. The defensive linemen are not significantly above the criterion for BMI. c. For the independent-measures t, the pooled variance is 14.01, the estimated standard error is 1.25, and t(34) = 2.00. For a two-tailed test using df = 30 (because 34 is not listed), the critical value is 2.750. Fail to reject the null hypothesis. There is no significant difference between the two groups.
15. The following data are from a repeated-measures study examining the effect of a treatment by measuring a group of n = 4 participants before and after they receive the treatment. a. Calculate the difference scores and MD. b. Compute SS, sample variance, and estimated standard error. c. Is there a significant treatment effect? Use a = .05, two tails. Participant Before Treatment After Treatment A 7 10 B 6 13 C 9 12 D 5 8
15. a. The difference scores are 3, 7, 3, and 3. MD = 4. b. SS = 12, sample variance is 4, and the estimated standard error is 1. c. With df = 3 and α = .05, the critical values are t = ±3.182. For these data, t = 4.00. Reject H0. There is a significant treatment effect.
15. For the data in the following sample: 8, 1, 5, 1, 5 a. Find the mean and the standard deviation. b. Now change the score of X = 8 to X = 18, and find the new mean and standard deviation. c. Describe how one extreme score influences the mean and standard deviation. (10.03.2014 Stats ch 4 textbook extraction/notes)
15. a. The mean is M = 4 and the standard deviation is s = √9 = 3. b. The new mean is M = 6 and the new standard deviation is √49 = 7. c. Changing one score changes both the mean and the standard deviation.
15. For several years researchers have noticed that there appears to be a regular, year-by-year increase in the average IQ for the general population. This phenomenon is called the Flynn effect after the researcher who first reported it (Flynn, 1984, 1999), and it means that psychologists must continuously update IQ tests to keep the population mean at = 100. To evaluate the size of the effect, a researcher obtained a 10-year-old IQ test that was standardized to produce a mean IQ of p, = 100 for the population 10 years ago. The test was then given to a sample of n = 64 of today's 20-year-old adults. The average score for the sample was M = 107 with a standard deviation of s = 12. a. Based on the sample, is the average IQ for today's population significantly different from the average 10 years ago, when the test would have produces a mean ofµ = 100? Use a two-tailed test with a = .01. b. Make an 80% confidence interval estimate of today's population mean IQ for the 10-year-old test.
15. a. With df = 63, the critical values are ±2.660 (using df = 60 in the table). For these data, the estimated standard error is 1.50, and t = 7/1.50= 4.67. Reject the null hypothesis and conclude that there has been a significant change in the average IQ score. b. Using df = 60, the t values for 80% confidence are 1.296, and the interval extends from 105.056 to 108.944.
15. The scores on a standardized mathematics test for 8th-grade children in New York State form a normal distribution with a mean of p = 70 and a standard deviation of a = 10. a. What proportion of the students in the state have scores less than X = 75? b. If samples of n = 4 are selected from the population, what proportion of the samples will have means less than M = 75? c. If samples of n = 25 are selected from the population, what proportion of the samples will have means less than M = 75?
15. a. z = 0.50 and p = 0.6915 b. σM = 5, z = 1.00 and p = 0.8413 c. σM = 2, z = 2.50 and p = 0.9938
16. Montarello and Martins (2005) found that fifth-grade students completed more mathematics problems correctly when simple problems were mixed in with their regular math assignments. To further explore this phenomenon, suppose that a researcher selects a standardized mathematics achievement test that produces a normal distribution of scores with a mean of p. = 100 and a standard deviation of a = 18. The researcher modifies the test by inserting a set of very easy problems among the standardized questions, and gives the modified test to a sample of n = 36 students. If the average test score for the sample is M = 104, is this result sufficient to conclude that inserting the easy questions improves student performance? Use a one-tailed test with a = .01.
16. H0: μ < 100 (performance is not increased). The critical region consists of z-scores beyond z = +2.33. For these data, σM = 3 and z = 1.33. Fail to reject H0 and conclude that performance is not significantly higher with the easy questions added.
16. Calculate SS, variance, and standard deviation for the following sample of n = 4 scores: 7, 4, 2, 1. (Note: The computational formula works well with these scores.) (10.03.2014 Stats ch 4 textbook extraction/notes)
16. SS = 21, the sample variance is 7 and the standard deviation is = √7 = 2.65.
16. A sample of n = 7 scores has a mean of M = 5. After one new score is added to the sample, the new mean is found to be M = 6. What is the value of the new score? (Hint: Compare the values for /X before and after the score was added.)
16. The original sample has n= 7 and ΣX= 35. The new sample has n= 8 and ΣX= 48. The new score must be X= 13.
16. Functional foods are those containing nutritional supplements in addition to natural nutrients. Examples include orange juice with calcium and eggs with omega-3. Kolodinsky, et al. (2008) examined attitudes toward functional foods for college students. For American students, the results indicated that females had a more positive attitude toward functional foods and were more likely to purchase them compared to males. In a similar study, a researcher asked students to rate their general attitude toward functional foods on a 7-point scale (higher score is more positive). The results are as follows: Females Male --= 8 n = 12 M = 4.69 M = 4.43 SS = 1.60 SS = 2.72 a. Do the data indicate a significant difference in attitude for males and females? Use a two-tailed test with a = .05. b. Compute r2, the amount of variance accounted for by the gender difference, to measure effect size. c. Write a sentence demonstrating how the results of the hypothesis test and the measure of effect size would appear in a research report
16. a. The pooled variance is 0.24, the estimated standard error is 0.22, and t = 1.18. For a two-tailed test with df = 18 the critical value is 2.101. Fail to reject the null hypothesis. There is no significant difference between the two groups. b. For these data, r2 = 1.39/19.39 = 0.072 or 7.2%. c. The data showed no significant difference in attitude toward functional foods for males compared with females, t(18) = 1.18, p > .05, r2 = 0.072.
17. Researchers have often noted increases in violent crimes when it is very hot. In fact, Reifman, Larrick, and Fein (1991) noted that this relationship even extends to baseball. That is, there is a much greater chance of a batter being hit by a pitch when the temperature increases. Consider the following hypothetical data. Suppose that over the past 30 years, during any given week of the major-league season, an average of p. = 12 players are hit by wild pitches. Assume that the distribution is nearly normal with a = 3. For a sample of n = 4 weeks in which the daily temperature was extremely hot, the weekly average of hit-by-pitch players was M = 15.5. Are players more likely to get hit by pitches during hot weeks? Set alpha to .05 for a one-tailed test.
17. H0: μ < 12 (no increase during hot weather). H1: μ > 12 (there is an increase). The critical region consists of z score values greater than +1.65. For these data, the standard error is 1.50, and z = 2.33 which is in the critical region so we reject the null hypothesis and conclude that there is a significant increase in the average number of hit players during hot weather.
17. Calculate SS, variance, and standard deviation for the following population of N = 8 scores: 0, 0, 5, 0, 3, 0, 0, 4. (Note: The computational formula works well with these scores.) (10.03.2014 Stats ch 4 textbook extraction/notes)
17. SS = 32, the population variance is 4, and the standard deviation is 2.
17. A variety of research results suggest that visual images interfere with visual perception. In one study, Segal and Fusella (1970) had participants watch a screen, looking for brief presentations of a small blue arrow. On some trials, the participants were also asked to form a mental image (for example, imagine a volcano). The results for a sample ofn = 6, show that participants made an average of MD = 4.3 more errors while forming images than while not forming images. The difference scores had SS = 63. Do the data indicate a significant difference between the two conditions? Use a two-tailed test with a = .05
17. The null hypothesis states that the images have no effect on performance. For these data, the sample variance is 12.6, the estimated standard error is 1.45, and t(5) = 2.97. With df = 5 and α = .05, the critical values are t = ±2.571. Reject the null hypothesis, the images have a significant effect.
17. Belsky, Weinraub, Owen, and Kelly (2001) reported on the effects of preschool childcare on the development of young children. One result suggests that children who spend more time away from their mothers are more likely to show behavioral problems in kindergarten. Using a standardized scale, the average rating of behavioral problems for kindergarten children is p = 35. A sample of n = 16 kindergarten children who had spent at least 20 hours per week in childcare during the previous year produced a mean score of M = 42.7 with a standard deviation ofs = 6. a. Are the data sufficient to conclude that children with a history of childcare show significantly more behavioral problems than the average kindergarten child? Use a one-tailed test with a = .01. b. Compute r2, the percentage of variance accounted for, to measure the size of the preschool effect. c. Write a sentence showing how the outcome of the hypothesis test and the measure of effect size would appear in a research report
17. a. The estimated standard error is 1.50, and t = 7.7/1.50 = 5.13. For a one-tailed test, the critical value is 2.602. Reject the null hypothesis, children with a history of day care have significantly more behavioral problems. b. The percentage of variance accounted for is r2 = 26.32/41.32 = 0.637 or 63.7%. c. The results show that kindergarten children with a history of day care have significantly more behavioral problems than other kindergarten children, t(15) = 5.13, p < .01, r2 = 0.637.
17. A recent newspaper article reported the results of a survey of well-educated suburban parents. The responses to one question indicated that by age 2, children were watching an average of |i = 60 minutes of television each day. Assuming that the distribution of television-watching times is normal with a standard deviation of ct = 20 minutes, find each of the following proportions. a. What proportion of 2-year-old children watch more than 90 minutes of television each day? b. What proportion of 2-year-old children watch less than 20 minutes a day?
17. a. p(z > 1.50) = 0.0668 b. p(z < -2.00) = 0.0228
17. A population of scores forms a normal distribution with a mean of p = 80 and a standard deviation of if = 10. a. What proportion of the scores have values between 75 and 85? b. For samples of n = 4, what proportion of the samples will have means between 75 and 85? c. For samples of n = 16, what proportion of the samples will have means between 75 and 85?
17. a. z = ±0.50 and p = 0.3830 b. σM = 5, z = ±1.00 and p = 0.6826 c. σM = 2.5, z = ±2.00 and p = 0.9544
17. For a population with a mean of p. = 70, a score of X = 62 corresponds to z = —2.00. What is the population standard deviation? (10.13.2014 stats ch 6, zscores)
17. σ = 4
18. Calculate SS, variance, and standard deviation for the following population of N = 7 scores: 8, 1, 4, 3, 5, 3, 4. (Note: The definitional formula works well with these scores.)| (10.03.2014 Stats ch 4 textbook extraction/notes)
18. SS = 28, the population variance is 4, and the standard deviation is 2.
18. One sample has a mean of M = 4 and a second sample has a mean of M = 8. The two samples are combined into a single set of scores. a. What is the mean for the combined set if both of the original samples have n = 7 scores? b. What is the mean for the combined set if the first sample has n = 3 and the second sample has n = 7? c. What is the mean for the combined set if the first sample has n = 7 and the second sample has n = 3?
18. a. The new mean is M = 6. b. The new mean is (12 + 56)/10 = 6.8. c. The new mean is (28 + 24)/10 = 5.2.
18. Other research examining the effects of preschool childcare has found that children who spent time in day care, especially high-quality day care, perform better on math and language tests than children who stay home with their mothers (Broberg, Wessels, Lamb, & Hwang, 1997). Typical results, for example, show that a sample of n = 25 children who attended day care before starting school had an average score of M = 87 with SS = 1536 on a standardized math test for which the population mean is p = 81. a. Is this sample sufficient to conclude that the children with a history of preschool day care are significantly different from the general population? Use a two-tailed test with a = .01. b. Compute Cohen's d to measure the size of the preschool effect. c. Write a sentence showing how the outcome of the hypothesis test and the measure of effect size would appear in a research report
18. a. The sample variance is 64, the estimated standard error is 1.60, and t = 6/1.60 = 3.75. For a two tailed test, the critical value is 2.797. Reject the null hypothesis and conclude that children with a history of day care have significantly different cognitive skills. b. Cohen's d = 6/8 = 0.75 c. The results show that standardized math test scores are significantly different for children with a history of day care than for other children, t(24) = 3.75, p < .01, d = 0.75.
. Assume that the data are from an independentmeasures study using two separate samples, each with n = 6 participants. Compute the pooled variance and the estimated standard error for the mean difference. b. Now assume that the data are from a repeatedmeasures study using the same sample of n = 6 participants in both treatment conditions. Compute the variance for the sample of difference scores and the estimated standard error for the mean difference. (You should find that the repeatedmeasures design substantially reduces the variance and the standard error.) Treatment 1 Treatment 2 Difference 10 13 3 12 12 0 8 10 2 6 10 4 5 6 1 7 9 2 M = 8 SS = 34 M = 10 SS = 30 MD = 2 SS = 10
19. a. The pooled variance is 6.4 and the estimated standard error is 1.46. b. For the difference scores the variance is 12.8, the estimated standard error is 1.46.
18. A researcher plans to conduct an experiment testing the effect of caffeine on reaction time during a driving simulation task. A sample of n = 9 participants is selected and each person receives a standard dose of caffeine before being tested on the simulator. The caffeine is expected to lower reaction time by an average of 30 msec. Scores on the simulator task for the regular population (without caffeine) form a normal distribution with p. = 240 msec. and a = 30. a. If the researcher uses a two-tailed test with a = .05, what is the power of the hypothesis test? b. Again assuming a two-tailed test with a = .05, what is the power of the hypothesis test if the sample size is increased to n = 25?
18. a. With no treatment effect the distribution of sample means is centered at = 240 with a standard error of 10 points, and the critical boundary of z = 1.96 corresponds to a sample mean of M = 220.4. With a 30-point treatment effect, the distribution of sample means is centered at = 210. In this distribution a mean of M = 220.4 corresponds to z = 1.04. The power for the test is the probability of obtaining a z-score less than 1.04, which is p = 0.8508. b. With a sample of n = 25, the standard error is 6 points. In this case, the critical boundary of z = 1.96 corresponds to a sample mean of M = 228.24. With a 30-point treatment effect, the distribution of sample means is centered at = 210. In this distribution a mean of M = 228.24 corresponds to z = 3.04. The power for the test is the probability of obtaining a z-score less than 3.04, which is p = 0.9988.
18. Information from the Department of Motor Vehicles indicates that the average age of licensed drivers is p = 45.7 years with a standard deviation of cr = 12.5 years. Assuming that the distribution of drivers' ages is approximately normal, a. What proportion of licensed drivers are older than 50 years old? b. What proportion of licensed drivers are younger than 30 years old?
18. a. z = 0.34, p = 0.3669 b. z = -1.26, p = 0.1038
18. In a population of exam scores, a score of X = 48 corresponds to z = +1.00 and a score of X = 36 corresponds to z = —0.50. Find the mean and standard deviation for the population. (Hint: Sketch the distribution and locate the two scores on your sketch.) (10.13.2014 stats ch 6, zscores)
18. μ = 40 and σ = 8. The distance between the two scores is 12 points which is equal to 1.5
2. If the sample data are sufficient to reject the null hypothesis for a one-tailed test, then the same data would also reject Ho for a two-tailed test. (True or false?)
2. False. Because a two-tailed test requires a larger mean difference, it is possible for a sample to be significant for a one-tailed test but not for a two-tailed test
19. A sample of n = 40 is selected from a normal population with µ = 75 msec. and a = 12, and a treatment is administered to the sample. The treatment is expected to increase scores by an average of 4 points. a. If the treatment effect is evaluated with a two-tailed hypothesis test using a = .05, what is the power of the test? b. What is the power of the test if the researcher uses a one-tailed test with a = .05?
19. a. With no treatment effect the distribution of sample means is centered at = 75 with a standard error of 1.90 points. The critical boundary of z = 1.96 corresponds to a sample mean of M = 78.72. With a 4-point treatment effect, the distribution of sample means is centered at = 79. In this distribution a mean of M = 78.72 corresponds to z = 0.15. The power for the test is the probability of obtaining a z-score greater than 0.15, which is p = 0.5596. b. With a one-tailed test, critical boundary of z = 1.65 corresponds to a sample mean of M = 78.14. With a 4-point treatment effect, the distribution of sample means is centered at = 79. In this distribution a mean of M = 78.14 corresponds to z = 0.45. The power for the test is the probability of obtaining a z-score greater than 0.45, which is p = 0.6736.
19. A consumer survey indicates that the average household spends p = $185 on groceries each week. The distribution of spending amounts is approximately normal with a standard deviation of cr = $25. Based on this distribution, a. What proportion of the population spends more than $200 per week on groceries? b. What is the probability of randomly selecting a family that spends less than $150 per week on groceries? c. How much money do you need to spend on groceries each week to be in the top 20% of the distribution?
19. a. z = 0.60, p = 0.2743 b. z = -1.40, p = 0.0808 c. z = 0.84, X = $206 or more
19. In a distribution of scores, X = 64 corresponds to z = 1.00, and X = 67 corresponds to z = 2.00. Find the mean and standard deviation for the distribution (10.13.2014 stats ch 6, zscores)
19. μ = 61 and σ = 3. The distance between the two scores is 3 points which is equal to 1.0 standard deviation.
2. False. The expected value does not depend on sample size.
2. As sample size increases, the value of expected value also increases. (True or false?)
2. Describe the distribution of sample means (shape, expected value, and standard error) for samples of n = 36 selected from a population with a mean of p = 100 and a standard deviation of cr = 12
2. The distribution of sample means will be normal (because n > 30), have an expected value of μ = 100, and a standard error of σM = 12/√36 = 2.
2. Changing the value of a score in a distribution always changes the mean. (True or false?)
2. True.
2. For a distribution with p, = 40 and o- = 12, find the X value corresponding to each of the following z-scores. a. z= 1.50 b. z = -1.25 c. z=1/3 (10.13.2014 stats ch 6, zscores) (10.13.2014 stats ch 6, zscores)
2. a) x=58, b) x=25 c) x=44
2. For a normal distribution, find the z-score location that divides the distribution as follows: a. Separate the top 20% from the rest. b. Separate the top 60% from the rest. c. Separate the middle 70% from the rest
2. a. c = 0.84 b. z = -0 .2 5 c. z = — 1.04 and + 1.04
2. Scores on the Mathematics section of the SAT Reasoning Test form a normal distribution with a mean of |i = 500 and a standard deviation of ct = 100. a. If the state college only accepts students who score in the top 60% on this test, what is the minimum score needed for admission? b. What is the minimum score necessary to be in the top 10% of the distribution? c. What scores form the boundaries for the middle 50% of the distribution?
2. a. z = -0.25; X = 475 b. z = 1.28; X = 628 c. z = ±0.67; X =433 and X =567
2. A distribution has a standard deviation of et = 12. Find the z-score for each of the following locations in the distribution. a. Above the mean by 3 points. b. Above the mean by 12 points. c. Below the mean by 24 points. d. Below the mean by 18 points. (10.13.2014 stats ch 6, zscores)
2. a. z = 0.25 b. z = 1.00 c. z = -2.00 d. z = -1.50
20. When people learn a new task, their performance usually improves when they are tested the next day, but only if they get at least 6 hours of sleep (Stickgold, Whidbee, Schirmer, Patel, & Hobson, 2000). The following data demonstrate this phenomenon. The participants learned a visual discrimination task on one day, and then were tested on the task the following day. Half of the participants were allowed to have at least 6 hours of sleep and the other half were kept awake all night. Is there a significant difference between the two conditions? Use a two-tailed test with a = .05. Performance Scores 6 Hours of Sleep No Sleep n = 14 n = 14 M = 72 M = 65 SS = 932 SS = 706
20. The pooled variance is 63, the estimated standard error is 3.00, and t = 7/3 = 2.33. With df = 26 the critical value is 2.056. Reject the null hypothesis and conclude that there is a significant difference between the two sleep conditions.
20. The average age for licensed drivers in the county is 1.1 = 40.3 years with a standard deviation of a = 13.2 years. a. A researcher obtained a random sample of n = 16 parking tickets and computed an average age of M = 38.9 years for the drivers. Compute the z-score for the sample mean and find the probability of obtaining an average age this young or younger for a random sample of licensed drivers. Is it reasonable to conclude that this set of n = 16 people is a representative sample of licensed drivers? b. The same researcher obtained a random sample of n = 36 speeding tickets and computed an average age of M = 36.2 years for the drivers. Compute the z-score for the sample mean and find the probability of obtaining an average age this young or younger for a random sample of licensed drivers. Is it reasonable to conclude that this set of n = 36 people is a representative sample of licensed drivers?
20. a. With a standard error of σM = 3.3, M = 38.9 corresponds to z = -0.42 and p = 0.3372. This is not an unusual sample. It is representative of the population. b. With a standard error of σM = 2.2, M = 36.2 corresponds to z = -1.86 and p = 0.0314. The sample mean is unusually small and not representative.
20. A random sample is obtained from a population with a mean of p = 70. A treatment is administered to the individuals in the sample and, after treatment, the sample mean is M = 78 with a standard deviation of s = 20. a. Assuming that the sample consists ofn = 25 scores, compute r2 and the estimated Cohen's d to measure the size of treatment effect. b. Assuming that the sample consists of n = 16 scores, compute r2 and the estimated Cohen's d to measure the size of treatment effect. c. Comparing your answers from parts a and b, how does the number of scores in the sample influence the measures of effect size?
20. a. With n = 25 the estimated standard error is 4 and t = 8/4 = 2. r2 = 4/28 = 0.143. Cohen's d = 8/20 = 0.40. b. With n = 16 the estimated standard error is 5 and t = 8/5 = 1.60. r2 = 2.56/17.56 = 0.146. Cohen's d = 8/20 = 0.40. c. The sample size does not have any influence on Cohen's d and has only a minor effect on r2.
20. For each of the following populations, would a score of X = 50 be considered a central score (near the middle of the distribution) or an extreme score (far out in the tail of the distribution)? a. p, = 45 and cr = 10 b. p, = 45 and o- = 2 c. μ =90anda=20 d. = 60 and o- = 20 (10.13.2014 stats ch 6, zscores)
20. a. central (z = 0.50) b. extreme (z = 2.50) c. extreme (z = -2.00) d. central (z = -0.50)
20. Over the past 10 years, the local school district has measured physical fitness for all high school freshmen. During that time, the average score on a treadmill endurance task has been p = 19.8 minutes with a standard deviation of cr = 7.2 minutes. Assuming that the distribution is approximately normal, find each of the following probabilities. a. What is the probability of randomly selecting a student with a treadmill time greater than 25 minutes? In symbols. p(X > '25) = ? b. What is the probability of randomly selecting a student with a time greater than 30 minutes? In symbols, p(X > 30) = ? c. If the school required a minimum time of 10 minutes for students to pass the physical education course, what proportion of the freshmen would fail?
20. a. p(z > 0.72) = 0.2358 b. p(z > 1.42) = 0.0778 c. p(z < -1.36) = 0.0869
21. Steven Schmidt (1994) conducted a series of experiments examining the effects of humor on memory. In one study, participants were given a mix of humorous and nonhumorous sentences and significantly more humorous sentences were recalled. However, Schmidt argued that the humorous sentences were not necessarily easier to remember, they were simply preferred when participants had a choice between the two types of sentence. To test this argument, he switched to an independent-measures design in which one group got a set of exclusively humorous sentences and another group got a set of exclusively nonhumorous sentences. The following data are similar to the results from the independentmeasures study. Humorous Sentences Non humorous Sentence 4 5 2 4 6 3 5 3 6 7 6 6 3 4 2 6 2 5 4 3 4 3 4 4 3 3 5 3 5 2 6 4 Do the results indicate a significant difference in the recall of humorous versus nonhumorous sentences? Use a two-tailed test with a = .05.
21. The humorous sentences produced a mean of M = 4.25 with SS = 35, and the non- humorous sentences had M = 4.00 with SS = 26. The pooled variance is 2.03, the estimated standard error is 0.504, and t = 0.496. With df = 30, the critical value is 2.042. Fail to reject the null hypothesis and conclude that there is no significant difference in memory for the two types of sentences
21. Identify the circumstances in which the median rather than the mean is the preferred measure of central tendency.
21. The median is used instead of the mean when there is a skewed distribution (a few extreme scores), an open ended distribution, undetermined scores, or an ordinal scale.
21. Explain how the power of a hypothesis test is influenced by each of the following. Assume that all other factors are held constant. a. Increasing the alpha level from .01 to .05. b. Changing from a one-tailed test to a two-tailed test.
21. a. Increasing alpha increases power. b. Changing from one- to two-tailed decreases power
21. There is some evidence suggesting that you are likely to improve your test score if you rethink and change answers on a multiple-choice exam (Johnston, 1975). To examine this phenomenon, a teacher gave the same final exam to two sections of a psychology course. The students in one section were told to turn in their exams immediately after finishing, without changing any of their answers. In the other section, students were encouraged to reconsider each question and to change answers whenever they felt it was appropriate. Before the final exam, the teacher had matched 9 students in the first section with 9 students in the second section based on their midterm grades. For example, a student in the no-change section with an 89 on the midterm exam was matched with student in the change section who also had an 89 on the midterm. The final exam grades for the 9 matched pairs of students are presented in the following table. a. Do the data indicate a significant difference between the two conditions? Use a two-tailed test with a = .05. b. Construct a 95% confidence interval to estimate the size of the population mean difference. c. Write a sentence demonstrating how the results of the hypothesis test and the confidence interval would appear in a research report. Matched Pair No-Change Section Change Section #1 71 86 #2 68 80 #3 91 88 #4 65 74 #5 73 82 #6 81 89 #7 85 85 #8 86 88 #9 65 76
21. a. The null hypothesis says that changing answers has no effect, H0: μD = 0. With df = 8 and α = .05, the critical values are t = ±2.306. For these data, MD = 7, SS = 288, the standard error is 2, and t(8) = 3.50. Reject H0 and conclude that changing answers has a significant effect on exam performance. b. For 95% confidence use t = ±2.306. The interval extends from 2.388 to 11.612. c. Changing answers resulted in significantly higher exam scores, t(8) = 3.50, p < .05, 95% CI [2.388, 11.612].
21. People are selected to serve on juries by randomly picking names from the list of registered voters. The average age for registered voters in the county is 1.1 = 44.3 years with a standard deviation of a = 12.4. A statistician computes the average age for a group of n = 12 people currently serving on a jury and obtains a mean of M = 48.9 years. a. How likely is it to obtain a random sample of n = 12 jurors with an average age equal to or greater than 48.9? b. Is it reasonable to conclude that this set of n = 12 people is not a representative random sample of registered voters?
21. a. With a standard error of 3.58 this sample mean corresponds to a z score of z = 1.28. A z score this large (or larger) has a probability of p = 0.1003. b. A sample mean this large should occur only 1 out of 10 times. This is not a very representative sample.
21. A distribution of exam scores has a mean of p. = 80. a. If your score is X = 86, which standard deviation would give you a better grade: a = 4 Cr = 8? b. If your score is X = 74, which standard deviation would give you a better grade: Cr = 4 or Cr = 8? (10.13.2014 stats ch 6, zscores)
21. a. σ = 4 b. σ = 8
22. Welsh, Davis, Burke, and Williams (2002) con ducted a study to evaluate the effectiveness of a carbohydrate-electrolyte drink on sports performance and endurance. Experienced athletes were given either a carbohydrate-electrolyte drink or a placebo while they were tested on a series of high-intensity exercises. One measure was how much time it took for the athletes to run to fatigue. Data similar to the results obtained in the study are shown in the following table. Time to Run Fatigue (In Minutes) Mean SE Placebo 21.7 2.2 Carb-Elec 28.6 2.7 a. Construct a bar graph that incorporates all of the information in the table. b. Looking at your graph, do you think that the carbohydrate-electrolyte drink helps performance?
22. a. 40 │ │ Mean 30 │ Running │ Time 20 │ │ 10 │ │ └─────────────────────────── Placebo Carb. Drink b. Even considering the standard error for each mean, there is no overlap between the two groups. The carbohydrate-electrolyte does seem to have increased endurance even allowing for the standard error.
22. In an extensive study involving thousands of British children, Arden and Plomin (2006) found significantly higher variance in the intelligence scores for males than for females. Following are hypothetical data, similar to the results obtained in the study. Note that the scores are not regular IQ scores but have been standardized so that the entire sample has a mean of M = 10 and a standard deviation of s = 2. a. Calculate the mean and the standard deviation for the sample of n = 8 females and for the sample of n = 8 males. b. Based on the means and the standard deviations, describe the differences in intelligence scores for males and females. Female Male 9 8 11 10 10 11 13 12 8 6 9 10 11 14 9 9 (10.03.2014 Stats ch 4 textbook extraction/notes)
22. a. For the females, M = 10 and s = 1.60. For the males, M = 8 and s = 2.45. b. The males and females have the same mean IQ score but the male's scores are more variable.
2. In studies examining the effect of humor on interpersonal attractions, McGee and Shevlin (2009) found that an individual's sense of humor had a significant effect on how the individual was perceived by others. In one part of the study, female college students were given brief descriptions of a potential romantic partner. The fictitious male was described positively as being single, ambitious, and having good job prospects. For one group of participants, the description also said that he had a great sense of humor. For another group, it said that he had no sense of humor. After reading the description, each participant was asked to rate the attractiveness of the man on a seven-point scale from 1 (very attractive) to 7 (very unattractive). A score of 4 indicates a neutral rating. a. The females who read the "great sense of humor" description gave the potential partner an average attractiveness score of M = 4.53 with a standard deviation of s = 1.04. If the sample consisted of n = 16 participants, is the average rating significantly higher than neutral (p = 4)? Use a one-tailed test with a = .05. b. The females who read the description saying "no sense of humor" gave the potential partner an average attractiveness score of M = 3.30 with a standard deviation of s = 1.18. If the sample consisted of n = 16 participants, is the average rating significantly lower than neutral (p = 4)? Use a one-tailed test with a = .05.
22. a. H0: µ ≤ 4 (not greater than neutral). The estimated standard error is 0.26 and t = 2.04. With a critical value of 1.753, reject H0 and conclude that the males with a great sense of humor were rated significantly higher than neutral. b. H0: µ ≥ 4 (not lower than neutral). The estimated standard error is 0.295 and t = -2.37. With a critical value of -1.753, reject H0 and conclude that the males with a no sense of humor were rated significantly lower than neutral.
22. For each of the following situations, identify the measure of central tendency (mean, median, or mode) that would provide the best description of the average score: a. A news reporter interviewed people shopping in a local mall and asked how much they spent on summer vacations. Most people traveled locally and reported modest amounts but one couple had flown to Paris for a month and paid a small fortune. b. A marketing researcher asked consumers to select their favorite from a set of four designs for a new product logo. c. A driving instructor recorded the number of orange cones that each student ran over during the first attempt at parallel parking.
22. a. With one extreme score, the median would be better than the mean. b. With a nominal scale, the mode is the only option. c. With numerical scores, the mean is usually best.
23. At the Olympic level of competition, even the smallest factors can make the difference between winning and losing. For example, Pelton (1983) has shown that Olympic marksmen shoot much better if they fire between heartbeats, rather than squeezing the trigger during a heartbeat. The small vibration caused by a heartbeat seems to be sufficient to affect the marksman's aim. The following hypothetical data demonstrate this phenomenon. A sample of n = 8 Olympic marksmen fires a series of rounds while a researcher records heartbeats. For each marksman, a score is recorded for shots fired during heartbeats and for shots fired between heartbeats. Do these data indicate a significant difference? Test with a = .05. Participant During Heartbeats Between Heartbeats A 93 98 B 90 94 C 95 96 D 92 91 E 95 97 F 91 97 G 92 95 H 93 9
23. The null hypothesis says that there is no difference between shots fired during versus between heart beats, H0: μD = 0. With α = .05, the critical region consists of t values beyond ±2.365. For these data, MD = 3, SS = 36, s2 = 5.14, the standard error is 0.80, and t(7) = 3.75. Reject H0 and conclude that the timing of the shot has a significant effect on the marksmen's scores.
23. In the Preview section at the beginning of this chapter we reported a study by Wegesin and Stern (2004) that found greater consistency (less variability) in the memory performance scores for younger women than for older women. The following data represent memory scores obtained for two women, one older and one younger, over a series of memory trials. a. Calculate the variance of the scores for each woman. b. Are the scores for the younger woman more consistent (less variable)? Younger Older 8 7 6 5 6 8 7 5 8 7 7 6 8 8 8 5 (10.03.2014 Stats ch 4 textbook extraction/notes)
23. a. For the younger woman, the variance is s2 = 0.786. For the older woman, the variance is s2 = 1.696. b. The variance for the younger woman is only half as large as for the older woman. The younger woman's scores are much more consistent.
23. A psychologist would like to determine whether there is a relationship between depression and aging. It is known that the general population averages p = 40 on a standardized depression test. The psychologist obtains a sample of n = 9 individuals who are all more than 70 years old. The depression scores for this sample are as follows: 37, 50, 43, 41, 39, 45, 49, 44, 48. a. On the basis of this sample, is depression for elderly people significantly different from depression in the general population? Use a two-tailed test with a = .05. b. Compute the estimated Cohen's d to measure the size of the difference. c. Write a sentence showing how the outcome of the hypothesis test and the measure of effect size would appear in a research report.
23. a. H0: µ = 40. With df = 8 the critical values are t = ±2.306. For these data, M = 44, SS = 162, s2 = 20.25, the standard error is 1.50, and t = 2.67. Reject H0 and conclude that depression for the elderly is significantly different from depression for the general population. b. Cohen's d = 4/4.5 = 0.889. c. The results indicate that depression scores for the elderly are significantly different from scores for the general population, t(8) = 2.67, p < .05, d = 0.889.
23. Research has shown that people are more likely to show dishonest and self-interested behaviors in darkness than in a well-lit environment (Zhong, Bohns, & Gino, 2010). In one experiment, participants were given a set of 20 puzzles and were paid $0.50 for each one solved in a 5-minute period. However, the participants reported their own performance and there was no obvious method for checking their honesty. Thus, the task provided a clear opportunity to cheat and receive undeserved money. One group of participants was tested in a room with dimmed lighting and a second group was tested in a well-lit room. The reported number of solved puzzles was recorded for each individual. The following data represent results similar to those obtained in the study. Well-Lit Room Dimly Lit Room 7 9 8 II 10 13 6 10 8 II 5 7 IS 12 14 5 10 a. Is there a significant difference in reported performance between the two conditions? Use a two-tailed test with a = .01. b. Compute Cohen's d to estimate the size of the treatment effect.
23. a. The null hypothesis states that the lighting in the room does not affect behavior. For the well-lit room the mean is M = 7.55 with SS = 42.22. For the dimly-lit room, M = 11.33 with SS = 38. The pooled variance is 5.01, the standard error is 1.06, and t(16) = 3.57. With df = 16 the critical values are 2.921. Reject the null hypothesis and conclude that the lighting did have an effect on behavior. b. d = 3.78/2.24 = 1.69
23. Use interpolation to find the requested percentiles and percentile ranks requested for the following distribution of scores. X f fc f c% 14-15 3 50 100 12-13 6 47 94 10-11 8 41 82 8-9 18 33 66 6-7 10 15 30 4-5 4 5 10 2-3 1 1 2 a. What is the percentile rank for X = 5? b. What is the percentile rank for X = 12? c. What is the 25th percentile? d. What is the 70th percentile?
23. a. The percentile rank for X = 5 is 8%. b. The percentile rank for X = 12 is 85%. c. The 25th percentile is X = 7. d. The 70th percentile is X = 10.
23. A distribution with a mean of p, = 62 and a standard deviation of o• = 8 is transformed into a standardized distribution with p, = 100 and = 20. Find the new, standardized score for each of the following values from the original population. a. X = 60 b. X = 54 c. X = 72 d. X = 66 (10.13.2014 stats ch 6, zscores)
23. a. X = 95 (z = -0.25) b. X = 80 (z = -1.00) c. X = 125 (z = 1.25) d. X = 110 (z = 0.50)
24. The Preview section of this chapter presented a repeated-measures research study demonstrating that swearing can help reduce pain (Stephens, Atkins, & Kingston, 2009). In the study, each participant was asked to plunge a hand into icy water and keep it there as long as the pain would allow. In one condition, the participants repeated their favorite curse words while their hands were in the water. In the other condition, the participants repeated a neutral word. Data similar to the results obtained in the study are shown in the following table. a. Do these data indicate a significant difference in pain tolerance between the two conditions? Use a two-tailed test with a = .05. b. Compute ,-2, the percentage of variance accounted for, to measure the size of the treatment effect. c. Write a sentence demonstrating how the results of the hypothesis test and the measure of effect size would appear in a research report. Amount of Time (in Seconds) Participant Swear Words Neutral Words 1 94 59 2 70 61 3 52 47 4 83 60 5 46 35 6 117 92 7 69 53 8 39 30 9 51 56 10 73 61
24. a. For these data, MD = 14, s2 = 128, the standard error is 3.58, and t(9) = -3.91. For a two- tailed test with α = .05 the critical boundary is t = 2.262. Reject H0 and conclude that swearing significantly increases pain tolerance. b. r2 = 15.288/24.288 = 0.629. c. Swearing significantly increased the amount of time that participants could tolerate the icy water, t(9) = -3.91, p < .05, r2 = 0.629.
24. The following frequency distribution presents a set of exam scores for a class of N = 20 students. X f cf c% 90-99 4 20 100 80-89 7 16 80 70-79 4 9 45 60-69 3 5 25 50-59 2 2 10 a. Find the 30th percentile. b. Find the 88th percentile. c. What is the percentile rank for X = 77? d. What is the percentile rank for X = 90?
24. a. The 30th percentile is X = 72. b. The 88th percentile is X = 93.5. c. The percentile rank for X = 77 is 40%. d. The percentile rank for X = 90 is 81%.
24. A distribution with a mean of p, = 56 and a standard deviation of o- = 20 is transformed into a standardized distribution with p„ = 50 and a = 10. Find the new, standardized score for each of the following values from the original population. a. X = 46 b. X = 76 c. X = 40 d. X = 80 (10.13.2014 stats ch 6, zscores)
24. a. X = 45 (z = -0.50) b. X = 60 (z = 1.00) c. X = 42 (z = -0.80) d. X = 62 (z = 1.20)
24. A roulette wheel has alternating red and black numbered slots into one of which the ball finally stops to determine the winner. If a gambler always bets on black to win, what is the probability of winning at least 24 times in a series of 36 spins? (Note that at least 24 wins means 24 or more.)
24. p = q = 1/2, and with n = 36 the normal approximation has μ = 18 and σ = 3. Using the lower real limit of 23.5, p(X > 23.5) = p(z > 1.83) = 0.0336.
25. Does it ever seem to you that the weather is nice during the work week, but lousy on the weekend? Cerveny and Balling (1998) have confirmed that this is not your imagination—pollution accumulating during the work week most likely spoils the weekend weather for people on the Atlantic coast. Consider the following hypothetical data showing the daily amount of rainfall for 10 weeks during the summer. Week Average Drain (weekday) Average DRain (weekend) 1 1.2 1.5 2 0.6 2.0 3 0.0 1.8 4 1.6 1.5 5 0.8 2.2 6 2.1 2.4 7 0.2 0.8 8 0.9 1.6 9 1.1 1.2 10 1 .4 1.7 a. Calculate the average daily rainfall (the mean) during the week, and the average daily rainfall for weekends. b. Based on the two means, does there appear to be a pattern in the data?
25. a. For weekdays M = 0.99 inches and for weekend days is M = 1.67 inches. b. There does appear to be more rain on weekend days than there is on weekdays.
25. One test for ESP involves using Zener cards. Each card shows one of five different symbols (square, circle, star, cross, wavy lines), and the person being tested has to predict the shape on each card before it is selected. Find each of the probabilities requested for a person who has no ESP and is just guessing. a. What is the probability of correctly predicting 20 cards in a series of 100 trials? b. What is the probability of correctly predicting more than 30 cards in a series of 100 trials? c. What is the probability of correctly predicting 50 or more cards in a series of 200 trials?
25. a. With five options, p = 1/5 for each trial. μ = 20 and σ = 4 For X = 20, z = ±0.13 and p = 0.1034. b. For X = 30.5, z = 2.63 and p = 0.0043. c. μ = 40 and σ = 5.66 For X = 49.5, z = 1.68 and p = 0.0465
27. For a balanced coin: a. What is the probability of getting more than 30 heads in 50 tosses? b. What is the probability of getting more than 60 heads in 100 tosses? c. Parts a and b both asked for the probability of getting more than 60% heads in a series of coin tosses ( ^ = = 60%). Why do you think the two probabilities are different?
27. a. With n = 50 and p = q = 1/2, you may use the normal approximation with μ = 25 and σ = 3.54. Using the upper real limit of 30.5, p(X > 30.5) = p(z > 1.55) = 0.0606. b. The normal approximation has μ = 50 and σ = 5. Using the upper real limit of 60.5, p(X > 60.5) = p(z > 2.10) = 0.0179. c. Getting 60% heads with a balanced coin is an unusual event for a large sample. Although you might get 60% heads with a small sample, you should get very close to a 50-50 distribution as the sample gets larger. With a larger sample, it becomes very unlikely to get 60% heads.
3. In a perfectly symmetrical distribution. the mean, the median, and the mode will all have the same value. (True or false?)
3. False, if the distribution is bimodal.
3. In a research report, the results of a hypothesis test include the phrase "z = 3.15, p < .01." This means that the test failed to reject the null hypothesis. (True or false?)
3. False. The probability is less than .01, which means it is very unlikely that the result occurred without any treatment effect. In this case, the data are in the critical region, and H0 is rejected.
3. As sample size increases, the value of the standard error also increases. (True or false?)
3. False. The standard error decreases as sample size increases.
3. A distribution of English exam scores has p. = 70 and a = 4. A distribution of history exam scores has p. = 60 and a = 20. For which exam would a score of X = 78 have a higher standing? Explain your answer. (10.13.2014 stats ch 6, zscores)
3. For the English exam, X = 78 corresponds to z = 2.00, which is a higher standing than z = 0.90 for the history exam.
3. Is it possible to obtain a negative value for the variance or the standard deviation? (10.03.2014 Stats ch 4 textbook extraction/notes)
3. Standard deviation and variance are measures of distance and are always greater than or equal to zero.
3. W hat are the two requirem ents that must be satisfied for a random sample?
3. The two requirements for a random sample are: (1) each individual has an equal chance of being selected, and (2) if more than one individual is selected, the probabilities must stay constant for all selections.
4. What is sampling with replacement, and why is it used?
3. The two requirements for a random sample are: (1) each individual has an equal chance of being selected, and (2) if more than one individual is selected, the probabilities must stay constant for all selections.
3. How is a matched-subjects design similar to a repeated-measures design? How do they differ?
3. They are similar in that the role of individual differences in the experiment is reduced. They differ in that there are two samples in a matched-subjects design and only one in a repeatedmeasures study.
3. A z-score value in the critical region means that you should reject the null hypothesis. (True or false?)
3. True. A z-score value in the critical region means that the sample is not consistent with th null hypothesis.
3. A positively skewed distribution has p = 60 and a = 8. a. What is the probability of obtaining a sample mean greater than M = 62 for a sample of n = 4 scores? (Be careful. This is a trick question.)
3. a. The distribution of sample means does not satisfy either of the criteria for being normal. Therefore, you cannot use the unit normal table, and it is impossible to find the probability. b. With n = 64, the distribution of sample means is nearly normal. The standard error is 8/A74 = 1, the z-score is +2.00, and the probability is 0.0228.
3. A sample is selected from a population with a mean of p = 40 and a standard deviation of o- = 8. a. If the sample has n = 4 scores, what is the expected value of M and the standard error of M? b. If the sample has n = 16 scores, what is the expected value of M and the standard error of M?
3. a. The expected value is μ = 40 and σM = 8/√4 = 4. b. The expected value is μ = 40 and σM = 8/√16 = 2.
3. If other factors are held constant, explain how each of the following influences the value of the independent-measures t statistic and the likelihood of rejecting the null hypothesis: a. Increasing the number of scores in each sample. b. Increasing the variance for each sample.
3. a. The size of the two samples influences the magnitude of the estimated standard error in the denominator of the t statistic. As sample size increases, the value of t also increases (moves farther from zero), and the likelihood of rejecting H0 also increases. b. The variability of the scores influences the estimated standard error in the denominator. As the variability of the scores increases, the value of t decreases (becomes closer to zero), and the likelihood of rejecting H0 decreases.
3. The SAT scores for the entering freshman class at a local college form a normal distribution with a mean of p = 530 and a standard deviation of a = 80. a. For a random sample of n = 16 students, what range of values for the sample mean would be expected 95% of the time? b. What range of values would be expected 95% of the time if the sample size were n = 100?
3. a. With n = 16 the standard error is am = 20 points. Using z = ± 1.96, the 95% range extends from 490.8 to 569.2. b. With n = 100 the standard error is only 8 points and the range extends from 514.32 to 545.68.
3. A distribution has a standard deviation of a = 6. Describe the location of each of the following z-scores in terms of position relative to the mean. For example, z = +1.00 is a location that is 6 points above the mean. a. z = +2.00 b. z = +0.50 c. z = -2.00 d. z = -0.50 (10.13.2014 stats ch 6, zscores)
3. a. above the mean by 12 points b. above the mean by 3 points c. below the mean by 12 points d. below the mean by 3 points
3. Find each value requested for the distribution of scores in the following table. a.n b.EX c. EX^2 x f 5 2 4 3 3 5 2 1 1 1
3. a. n = 12 b. ΣX = 40 c. ΣX2 = 148
3. Suppose that you are going to select a random sample of /; = 1 score from the distribution in Figure 6.2. Find the following probabilities: a. p(X > 2) b. p(X > 5) c. p(X < 3)
3. a. p=7/10 b- P = 0.1 c- P = m = 0.3
3. For a population with g. = 30 and a = 8, find the z-score for each of the following scores: a. X = 32 b. X = 26 c. X = 42 (10.13.2014 stats ch 6, zscores)
3. a. z = +0.25 b. z = —0.50 c. z = +1.50
4. Describe the homogeneity of variance assumption and explain why it is important for the independentmeasures t test
4. The homogeneity of variance assumption specifies that the variances are equal for the two populations from which the samples are obtained. If this assumption is violated, the t statistic can cause misleading conclusions for a hypothesis test.
4. Explain why t distributions tend to be flatter and more spread out than the normal distribution.
4. The sample variance (s2) in the t formula changes from one sample to another and contributes to the variability of the t statistics. A z-score uses the population variance which is constant from one sample to another.
4. A distribution of English exam scores has p. = 50 and a = 12. A distribution of history exam scores has p. = 58 and a = 4. For which exam would a score of X = 62 have a higher standing? Explain your answer. (10.13.2014 stats ch 6, zscores)
4. The score X = 62 corresponds to z = +1.00 in both distributions. The score has exactly the same standing for both exams.
4. If the alpha level is changed from a = .05 to a = .01, a. What happens to the boundaries for the critical region? b. What happens to the probability of a Type I error?
4. a. Lowering the alpha level causes the boundaries of the critical region to move farther out into the tails of the distribution. b. Lowering α reduces the probability of a Type I error.
4. For a population with p. = 50 and a = 12, find the X value corresponding to each of the following z-scores: a. z = —0.25 b. z = 2.00 c. z = 0.50 (10.13.2014 stats ch 6, zscores)
4. a. X = 47 b. X = 74 c. X = 56
4. A sample of n = 6 scores has a mean of M = 40. One new score is added to the sample and the new mean is found to be M = 35. What can you conclude about the value of the new score? a. It must be greater 40. b. It must be less than 40.
4. b
4. A researcher reports a t statistic with df = 20. How many individuals participated in the study?
4. n = 21
5. Although there is a popular belief that herbal remedies such as ginkgo biloba and ginseng may improve learning and memory in healthy adults, these effects are usually not supported by well-controlled research (Persson, Bringlov, Nilsson, & Nyberg, 2004). In a typical study, a researcher obtains a sample of n = 36 participants and has each person take the herbal supplements every day for 90 days. At the end of the 90 days, each person takes a standardized memory test. For the general population, scores from the test are normally distributed with a mean of IL = 80 and a standard deviation of a = 18. The sample of research participants had an average of M = 84. a. Assuming a two-tailed test, state the null hypothesis in a sentence that includes the two variables being examined. b. Using symbols, state the hypotheses (H0 and HI) for the two-tailed test. c. Sketch the appropriate distribution, and locate the critical region for a = .05. d. Calculate the test statistic (z-score) for the sample. e. What decision should be made about the null hypothesis, and what decision should be made about the effect of the herbal supplements?
5. a. The null hypothesis states that the herb has no effect on memory scores. b. H0: μ = 80 (even with the herbs, the mean is still 80). H1: μ 80 (the mean has changed) c. The critical region consists of z-scores beyond 1.96. d. For these data, the standard error is 3 and z = 4/3 = 1.33. e. Fail to reject the null hypothesis. The herbal supplements do not have a significant effect on memory scores.
5. A sample has a mean of M = 30 and a standard deviation of s = 8. a. Would a score of X = 36 be considered a central score or an extreme score in the sample? b. If the standard deviation were s = 2, would X = 36 be central or extreme? (10.13.2014 stats ch 6, zscores)
5. a. X = 36 is a central score corresponding to z = 0.75. b. X = 36 would be an extreme score corresponding to z = 3.00.
6. A population has a mean of p. = 80 and a standard deviation of if = 20. a. Would a score of X = 70 be considered an extreme value (out in the tail) in this sample? b. If the standard deviation were if = 5, would a score of X = 70 be considered an extreme value? (10.03.2014 Stats ch 4 textbook extraction/notes)
6. a. No. X = 70 is 10 points away from the mean, only ½ of the standard deviation. b. Yes. With s = 5, 10 points is equal to a distance of 2 standard deviations.
6. One sample has SS = 70 and a second sample has SS = 42. a. Ifn = 8 for both samples, find each of the sample variances, and calculate the pooled variance. Because the samples are the same size, you should find that the pooled variance is exactly halfway between the two sample variances. b. Now assume that n = 8 for the first sample and n = 4 for the second. Again, calculate the two sample variances and the pooled variance. You should find that the pooled variance is closer to the variance for the larger sample
6. a. The first sample has a variance of 10, the second sample variance is 6, and the pooled variance is 8 (halfway between). b. The first sample has a variance of 10, the second sample variance is 14, and the pooled variance is 112/10 = 11.2 (closer to the variance for the larger sample).
6. Childhood participation in sports, cultural groups, and youth groups appears to be related to improved self-esteem for adolescents (McGee, Williams, Howden Chapman, Martin, & Kawachi, 2006). In a representativ study, a sample ofn = 100 adolescents with a history of group participation is given a standardized self-esteem questionnaire. For the general population of adolescents scores on this questionnaire form a normal distribution with a mean of p = 40 and a standard deviation of o = 12. The sample of group-participation adolescents had an average of M = 43.84. a. Does this sample provide enough evidence to conclude that self-esteem scores for these adolescents are significantly different from those of the general population? Use a two-tailed test with a = .01. b. Compute Cohen's d to measure the size of the difference. c. Write a sentence describing the outcome of the hypothesis test and the measure of effect size as it would appear in a research report.
6. a. The null hypothesis states that participation in sports, cultural groups, and youth groups has no effect on self-esteem. H0: µ = 40, even with participation. With n = 100, the standard error is 1.2 points and z = 3.84/1.2 = 3.20. This is beyond the critical value of 2.58, so we conclude that there is a significant effect. b. Cohen's d = 3.84/12 = 0.32. c. The results indicate that group participation has a significant effect on self-esteem, z = 3.20, p < .01, d = 0.32.
6. Draw a vertical line through a normal distribution for each of the following z-score locations. Determine whether the body is on the right or left side of the line and find the proportion in the body. a. z = 2.20 b. z = 1.60 c. z = -1.50 d. z = -0.70
6. a. body to the left, p = 0.9861 b. body to the left, p = 0.9452 c. body to the right, p = 0.9332 d. body to the right, p = 0.7580
The third is or called the standard variation which is the average squared distance from the mean. This involves finding the distance of each integer of the mean, square each distance, find the average of the result and then square rooting it. In the situation of a sample you would subtract 1 from the number dividing for the mean of the squared numbers because of a concept called degrees of freedom which mainly is that there is a tendency for deviations to be undervalues and needs to be accounted for. With this method you can achieve what is called one standard deviation from the mean which is ___% variation of the score, while 2 standard deviation involves 95%. (10.02.2014, stats lec 4 variability)
68
7. On an exam with a mean of M = 78, you obtain a score of X = 84. a. Would you prefer a standard deviation of s = 2 or s = 10? (Hint: Sketch each distribution and find the location of your score.) b. If your score were X = 72, would you prefer s = 2 or s = 10? Explain your answer. (10.03.2014 Stats ch 4 textbook extraction/notes)
7. a. s = 2 is better (you are above the mean by 3 standard deviations). b. s = 10 is better (you are below the mean by less than half a standard deviation).
7. A local college requires an English composition course for all freshmen. This year they are evaluating a new online version of the course. A random sample of n = 16 freshmen is selected and the students are placed in the online course. At the end of the semester, all freshmen take the same English composition exam. The average score for the sample is M = 76. For the general population of freshmen who took the traditional lecture class, the exam scores form a normal distribution with a mean of p. = 80. a. If the final exam scores for the population have a standard deviation of a = 12, does the sample provide enough evidence to conclude that the new online course is significantly different from the traditional class? Assume a two-tailed test with a = .05. b. If the population standard deviation is Q = 6, is the sample sufficient to demonstrate a significant difference? Again, assume a two-tailed test with a = .05. c. Comparing your answers for parts a and b, explain how the magnitude of the standard deviation influences the outcome of a hypothesis test.
7. a. H0 μ = 80. With σ = 12, the sample mean corresponds to z = 4/3 = 1.33. This is not sufficient to reject the null hypothesis. You cannot conclude that the course has a significant effect. b. H0 μ = 80. With σ = 6, the sample mean corresponds to z = 4/1.5 = 2.67. This is sufficient to reject the null hypothesis and conclude that the course does have a significant effect. c. There is a 4 point difference between the sample and the hypothesis. In part a, the standard error is 3 points and the 4-point difference is not significant. However, in part b, the standard error is only 1.5 points and the 4-point difference is now significantly more than is expected by chance. In general, a larger standard deviation produces a larger standard error, which reduces the likelihood of rejecting the null hypothesis.
7. The following sample was obtained from a population with unknown parameters. Scores: 6, 12, 0, 3, 4 a. Compute the sample mean and standard deviation. (Note that these are descriptive values that summarize the sample data.) b. Compute the estimated standard error for M. (Note that this is an inferential value that describes how accurately the sample mean represents the unknown population mean.)
7. a. M = 5 and s = √20 = 4.47 b. sM = 2.
7. For the following sample a. Assume that the scores are measurements of a continuous variable and find the median by locating the precise midpoint of the distribution. b. Assume that the scores are measurements of a discrete variable and find the median. Scores: 1, 2, 3, 3, 3, 4
7. a. Median = 2.83 (2.5 + 0.33) b. Median = 3
7. a. A repeated-measures study with a sample of n = 9 participants produces a mean difference of MD = 3 with a standard deviation of s = 6. Based on the mean and standard deviation, you should be able to visualize (or sketch) the sample distribution. Use a two-tailed hypothesis test with a = .05 to determine whether it is likely that this sample came from a population with No = 0. b. Now assume that the sample mean difference is MD = 12, and once again visualize the sample distribution. Use a two-tailed hypothesis test with a = .05 to determine whether it is likely that this sample came from a population with 1.tp = 0. c. Explain how the size of the sample mean difference influences the likelihood of finding a significant mean difference.
7. a. The estimated standard error is 2 points and t(8) = 1.50. With a critical boundary of ±2.306, fail to reject the null hypothesis. b. With MD = 12, t(8) = 6.00. With a critical boundary of ±2.306, reject the null hypothesis. c. The larger the mean difference, the greater the likelihood of finding a significant difference.
7. As noted on page 320, when the two population means are equal, the estimated standard error for the independent-measures t test provides a measure of how much difference to expect between two sample means. For each of the following situations, assume that p = i.1.2 and calculate how much difference should be expected between the two sample means. a. One sample has n = 8 scores with SS = 45 and the second sample has n = 4 scores with SS = 15. b. One sample has n = 8 scores with SS = 150 and the second sample has n = 4 scores with SS = 90. c. In part b, the samples have larger variability (bigger SS values) than in part a, but the sample sizes are unchanged. How does larger variability affect the size of the standard error for the sample mean difference?
7. a. The pooled variance is 6 and the estimated standard error is 1.50. b. The pooled variance is 24 and the estimated standard error is 3. c. Larger variability produces a larger standard error.
8. A population has a mean of p. = 30 and a standard deviation of if = 5. a. If 5 points were added to every score in the population, what would be the new values for the mean and standard deviation? b. If every score in the population were multiplied by 3 what would be the new values for the mean and standard deviation? (10.03.2014 Stats ch 4 textbook extraction/notes)
8. a. The mean is μ = 35 and the standard deviation is still σ = 5. b. The new mean is μ = 90 and the new standard deviation is σ = 15.
8. Two separate samples, each with n = 12 individuals, receive two different treatments. After treatment, the first sample has SS = 1740 and the second has SS = 1560. a. Find the pooled variance for the two samples. b. Compute the estimated standard error for the sample mean difference. c. If the sample mean difference is 8 points, is this enough to reject the null hypothesis and conclude that there is a significant difference for a two-tailed test at the .05 level? PROBLEMS 347 d. If the sample mean difference is 12 points, is this enough to indicate a significant difference for a two-tailed test at the .05 level? e. Calculate the percentage of variance accounted for (r2) to measure the effect size for an 8-point mean difference and for a 12-point mean difference.
8. a. The pooled variance is 150. b. The estimated standard error is 5.00. c. A mean difference of 8 would produce t = 8/5 = 1.60. With df = 22 the critical values are ±2.074. Fail to reject H0. d. A mean difference of 12 would produce t = 12/5 = 2.40. With df = 22 the critical values are ±2.074. Reject H0. e. With a mean difference of 8 points, r2 = 0.104. With a difference of 12 points, r2 = 0.207.
8. To evaluate the effect of a treatment, a sample is obtained from a population with a mean of p = 75, and the treatment is administered to the individuals in the sample. After treatment, the sample mean is found to be M = 79.6 with a standard deviation ofs = 12. a. If the sample consists ofn = 16 individuals, are th data sufficient to conclude that the treatment has a significant effect using a two-tailed test with a = .05? b. If the sample consists ofn = 36 individuals, are the data sufficient to conclude that the treatment has a significant effect using a two-tailed test with a = .05? c. Comparing your answer for parts a and b, how does the size of the sample influence the outcome of a hypothesis test?
8. a. With n = 16, sM = 3 and t = 4.6/3 = 1.53. This is not greater than the critical value of 2.131, so there is no significant effect. b. With n = 36, sM = 2 and t = 4.6/2 = 2.30. This value is greater than the critical value of 2.042 (using df = 30), so we reject the null hypothesis and conclude that there is a significant treatment effect. c. As the sample size increases, the likelihood of rejecting the null hypothesis also increases.
8. A random sample is selected from a normal population with a mean of p. = 50 and a standard deviation of a = 12. After a treatment is administered to the individuals in the sample, the sample mean is found to be M = 55. a. If the sample consists of n = 16 scores, is the sample mean sufficient to conclude that the treatment has a significant effect? Use a two-tailed test with a = .05. b. If the sample consists of n = 36 scores, is the sample mean sufficient to conclude that the treatment has a significant effect? Use a two-tailed test with a = .05. c. Comparing your answers for parts a and b, explain how the size of the sample influences the outcome of a hypothesis test.
8. a. With n = 16, the standard error is 3, and z = 5/3 = 1.67. Fail to reject H0. b. With n = 36, the standard error is 2, and z = 5/2 = 2.50. Reject H0. c. A larger sample increases the likelihood of rejecting the null hypothesis.
8. A sample of difference scores from a repeated-measures experiment has a mean ofMD = 4 with a standard deviation of s = 6. a. If n = 4, is this sample sufficient to reject the null hypothesis using a two-tailed test with a = .05? b. Would you reject H0 if n = 16? Again, assume a two-tailed test with a = .05. c. Explain how the size of the sample influences the likelihood of finding a significant mean difference.
8. a. With n = 4, the estimated standard error is 3 and t(3) = 4/3 = 1.33. With critical boundaries of ±3.182, fail to reject H0. b. With n = 16, the estimated standard error is 1.5 and t(15) = 4/1.5 = 2.67. With critical boundaries of ±2.131, reject H0 c. If other factors are held constant, a larger sample increases the likelihood of finding a significant mean difference.
8. If the population standard deviation is a = 8, how large a sample is necessary to have a standard error that is: a. less than 4 points? b. less than 2 points? c. less than 1 point?
8. a. n > 4 b. n > 16 c. n > 64
9. Two separate samples receive two different treatments. The first sample has n = 9 with SS = 710, and the second has n = 6 with SS = 460. a. Compute the pooled variance for the two samples. b. Calculate the estimated standard error for the sample mean difference. c. If the sample mean difference is 10 points, is this enough to reject the null hypothesis using a twotailed test with a = .05? d. If the sample mean difference is 13 points, is this enough to reject the null hypothesis using a two tailed test with a = .05?
9. a. The pooled variance is 90. b. The estimated standard error is 5. c. A mean difference of 10 points produces t = 2.00. With critical boundaries of ±2.160, fail to reject H0 d. A mean difference of 13 points produces t = 2.60. With critical boundaries of ±2.160, reject H0
9. A random sample of n = 36 scores is selected from a normal population with a mean of p. = 60. After a treatment is administered to the individuals in the sample, the sample mean is found to be M = 52. a. If the population standard deviation is a = 18, is the sample mean sufficient to conclude that the treatment has a significant effect? Use a two-tailed test with a = .05. b. If the population standard deviation is a = 30, is the sample mean sufficient to conclude that the treatment has a significant effect? Use a two-tailed test with a = .05. c. Comparing your answers for parts a and b, explain how the magnitude of the standard deviation influences the outcome of a hypothesis test
9. a. With σ = 18, the standard error is 3, and z = -8/3 = -2.67. Reject H0. b. With σ = 30, the standard error is 5, and z = -8/5 = -1.60. Fail to reject H0. c. Larger variability reduces the likelihood of rejecting H0.
The third is or called the standard variation which is the average squared distance from the mean. This involves finding the distance of each integer of the mean, square each distance, find the average of the result and then square rooting it. In the situation of a sample you would subtract 1 from the number dividing for the mean of the squared numbers because of a concept called degrees of freedom which mainly is that there is a tendency for deviations to be undervalues and needs to be accounted for. With this method you can achieve what is called one standard deviation from the mean which is 68% variation of the score, while 2 standard deviation involves ___%. (10.02.2014, stats lec 4 variability)
95
For hypothesis tests how do you try to find a hypothesized population U?
In the z-score formula plug everything else assumption and if M-U equals 0 then you have found the population mean.
Why isn't a pre-post test considered an experimental study?
In this test a researcher would be measuring changes in two periods of time. There is control for time since the before score is always before the after score. Also nothing controls for other possible variables
There are three forms of verifiability. The first is called the range difference .It is the simplest type but can be influenced by outliers or explain values which make it less useful. The second is the_____________ of 25th to 75th percentile. This is a better representation of the outliers, but may not give the entire picture of the variables. The third is or called the standard variation which is the average squared distance from the mean. This involves finding the distance of each integer of the mean, square each distance, find the average of the result and then square rooting it. (10.02.2014, stats lec 4 variability)
Interquartile range
If you are an RA asked to draft a research paper for the at a about the birth rates of rats and the effect of alcohol in the mother and the data was z=3 thus defeating H0 then how would you state the results
The study with alcohol had a significant effect on the birth weight of the new born rats z=3, p<0.05.
A study concluded that the use of mindfulness helps resolve ADHD though the sample selected was East Asian. What type of statistical error could occur here?
Type 1 error where the null hypothesis is rejected when it is actually the sample that is very extreme.
4. Find each value requested for the distribution of scores in the following table. a. n b. EX c. Ex^2 Frequency table X: 5 4 3 2 1 f: 1 2 3 5 3
a. n=14 b. EX = 35 c. 107
The mean is the arithmetic average. It is computed by adding all of the scores and then dividing by the number of scores. Conceptually, the mean is obtained by dividing the total (EX) equally among the number of individuals (N or n). The mean can also be defined as the balance point for the distribution. The distances above the mean are exactly balanced by the distances below the mean. Although the calculation is the same for a population or a sample mean, a population mean is identified by the symbol __a__, and a sample mean is identified by __b__. In most situations with numerical scores from an ___c___ scale, the mean is the preferred measure of central tendency. (09.28.2014 Stats 3 Ch 3 - Central tendency (notes + questions)
a. u b. M c. interval or a ratio
There are three forms of verifiability. The first is called the range difference .It is the simplest type but can be influenced __________________________. The second is the Interquartile range of 25th to 75th percentile. This is a better representation of the outliers, but may not give the entire picture of the variables. The third is or called the standard variation which is the average squared distance from the mean. This involves finding the distance of each integer of the mean, square each distance, find the average of the result and then square rooting it. (10.02.2014, stats lec 4 variability)
by outliers or explain values which make it less useful
The third is or called the standard variation which is the average squared distance from the mean. This involves finding the distance of each integer of the mean, square each distance, find the average of the result and then square rooting it. In the situation of a sample you would subtract 1 from the number dividing for the mean of the squared numbers because of a concept called ______________ which mainly is that there is a tendency for deviations to be undervalues and needs to be accounted for. With this method you can achieve what is called one standard deviation from the mean which is 68% of the score, while 2 standard deviation involves 95%. (10.02.2014, stats lec 4 variability)
degrees of freedom
The third is or called the standard variation which is the average squared distance from the mean. This involves finding the _____________, square each distance, find the average of the result and then square rooting it. In the situation of a sample you would subtract 1 from the number dividing for the mean of the squared numbers because of a concept called degrees of freedom which mainly is that there is a tendency for deviations to be undervalues and needs to be accounted for. With this method you can achieve what is called one standard deviation from the mean which is 68% of the score, while 2 standard deviation involves 95%. (10.02.2014, stats lec 4 variability)
distance of each integer of the mean
A sampling distribution is a
distribution of statistics obtained by selecting all of the possible samples of a specific size from a population.
There are three forms of verifiability. The first is called the range difference .It is the simplest type but can be influenced by outliers or explain values which make it less useful. The second is the Interquartile range of 25th to 75th percentile. This is a better representation of the outliers, but may not give the _______________. The third is or called the standard variation which is the average squared distance from the mean. This involves finding the distance of each integer of the mean, square each distance, find the average of the result and then square rooting it. (10.02.2014, stats lec 4 variability)
entire picture of the variables
The third is or called the standard variation which is the average squared distance from the mean. This involves finding the distance of each integer of the mean, square each distance, _____________ and then square rooting it. In the situation of a sample you would subtract 1 from the number dividing for the mean of the squared numbers because of a concept called degrees of freedom which mainly is that there is a tendency for deviations to be undervalues and needs to be accounted for. With this method you can achieve what is called one standard deviation from the mean which is 68% of the score, while 2 standard deviation involves 95%. (10.02.2014, stats lec 4 variability)
find the average of the result
In a study involving how eating speed of humans effect the speed at which their household cat eats researchers measure the eating speed of 30 pet owners and the eating speed of each of their cats. They then compute the results to suggest an average for the effect in general. Describe the following in order independent variable, depedent variable, the sample, the statistic.
how quickly the people eat, how quickly the cats eat, the sample is the 30 humans and cats, statistics is the general effect size that was computed.
When you add a constant number to the variables the mean changes but the standard deviation does not change. On the other hand if you multiply by a constant then the standard deviation will ______________. (10.02.2014, stats lec 4 variability)
increase by that change multiple
When you add a constant number to the variables the mean changes but the standard deviation does not change. On the other hand if you _________ constant then the standard deviation will increase by that change multiple. (10.02.2014, stats lec 4 variability)
multiply by a
Sampling error is the
natural discrepancy, or amount of error, between a sample statistic and its corresponding population parameter.
Researchers have observed that those who watch porn tend to have better sex life's at UofT than those who do not watch porn. Is this an experimental study?
no this is quai-experimental since the independent variable is not manipulated.
There are three forms of verifiability. The first is called the ___________ It is the simplest type and can be influenced by outliers or explain values which make it less useful. The second is the Interquartile range of 25th to 75th percentile. This is a better representation of the outliers, but may not give the entire picture of the variables. The third is or called the standard variation which is the average squared distance from the mean. This involves finding the distance of each integer of the mean, square each distance, find the average of the result and then square rooting it. (10.02.2014, stats lec 4 variability)
range difference.
The second is the Interquartile range of 25th to 75th percentile. This is a better representation of the outliers, but may not give the entire picture of the variables. To achieve this you look for the the second quarter of the value. The general idea is (3rd quarter) - (1st quarter). A__________ range is the middle of distribution to the boundary which is defined by the middle 50%. (10.02.2014, stats lec 4 variability)
semi interqualtile
Deviation from the mean is the conventional form of variation. It measures the deviation from a common central tendency. It can tell us how an individual stands in relation to the other scores as well as how accurate this set of scores is to a population. When there is a _____________ then it is a good representation because it means that more scores are contained in an area. (10.02.2014, stats lec 4 variability)
small vulnerability
The third is or called the standard variation which is the average squared distance from the mean. This involves finding the distance of each integer of the mean, _____________, find the average of the result and then square rooting it. In the situation of a sample you would subtract 1 from the number dividing for the mean of the squared numbers because of a concept called degrees of freedom which mainly is that there is a tendency for deviations to be undervalues and needs to be accounted for. With this method you can achieve what is called one standard deviation from the mean which is 68% of the score, while 2 standard deviation involves 95%. (10.02.2014, stats lec 4 variability)
square each distance
The third is or called the standard variation which is the average squared distance from the mean. This involves finding the distance of each integer of the mean, square each distance, find the average of the result and then _____________. In the situation of a sample you would subtract 1 from the number dividing for the mean of the squared numbers because of a concept called degrees of freedom which mainly is that there is a tendency for deviations to be undervalues and needs to be accounted for. With this method you can achieve what is called one standard deviation from the mean which is 68% of the score, while 2 standard deviation involves 95%. (10.02.2014, stats lec 4 variability)
square rooting it
At distribution is
the complete set of t values computed for every possible random sample for a specific sample size (n) or a specific degrees of freedom (df). The t distribution approximates the shape of a normal distribution.
The law of large numbers states that the larger the sample size (n), the more probable it is that
the sample mean is close to the population mean.
Deviation from the mean is the conventional form of variation. It measures the deviation from a common central tendency. It can tell us how an individual stands in relation to the other scores as well as how accurate this set of scores is ____________. When there is a small vulnerability then it is a good representation because it means that more scores are contained in an area. (10.02.2014, stats lec 4 variability)
to a population
Parameter
value that describes a populaiton
Statistic
value that describes a sample
