Exam 2 HW Questions
For each of the following, assume that the two samples are obtained from populations with the same mean, and calculate how much difference should be expected, on average, between the two sample means. A) Each sample has n = 4 scores with s2 = 68 for the first sample and s2 = 76 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 = 16 scores with s2 = 68 for the first sample and s2 = 76 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?
A) The estimated standard error for the sample mean difference is 6 points. B) The estimated standard error for the sample mean difference is 3 points. C) Larger samples produce a smaller standard error.
Childhood participation in sports, cultural groups, and youth groups appears to be related to improved self-esteem for adolescents. In a represent study, a sample of n=100 adolescents with a history of group participation is given a standardized self0esteem questionnaire. For the general population of adolescents, scores on this questionnaire form a normal distribution with a mean of mu=50 and a standard deviation of sigma=15. The sample of group participation adolescents had an average of M=53.8 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 alpha=0.05. b) Compute Conhen'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
A) The null hypothesis states that participation in sports, cultural groups, and youth groups has no effect on self esteem on hypothesis testing Step 1: H0: μ = 50 (even with participation, the mean is still 50 so did not change). H1: μ ≠ 50 (the mean has changed with the participation) Step 2: The critical region consists of z-scores beyond ±1.96. This test is two-tailed with an alpha-level of 0.05 Step 3: For these data of n = 100, the standard error is: 15 / sqrt(100) = 1.5 and z = (53.8-50)/1.5 = +2.53. This is beyond the critical value of 1.96, so we conclude that there is a significant effect. Step 4: Reject the null hypothesis. There is a significant effect of group participation on self-esteem. z = 2.53, p < .05, d = 0.253 B). Cohen's d = 3.8/15 = 0.253. C). The results indicate that group participation has a significant effect on self-esteem. z = 2.53, p < .05, d = 0.253
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 then null hypothesis that the true mean is μ = 3.1 using a two-tailed test with α = .05. B) Is the estimate from the participants significantly higher than the actual number (μ = 3.1)? Use a one-tailed test with α = .05.
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. Fail to reject H0. 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.
Ackerman and Goldsmith (2011) found that students who studied text from printed hardcopy had better test scores than students who studied from text presented on a screen. In a related study, a professor noticed that several students in a large class had purchased the e-book version of the course textbook. For the final exam, the overall average for the entire class was μ = 81.7 but the n = 9 students who used e-books had a mean of M = 77.2 with a standard deviation of s = 5.7. B) Construct the 90% confidence interval to estimate the mean exam score if the entire population used e-books
B) For 90% confidence, use t = ±1.860. The interval is 77.2 ± (1.860)1.9 and extends from 73.666 to 80.734.
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 μ = 15.1. A sample of n = 16 of today's children produces a mean score of M = 23.3 with SS = 240. B) Make a 90% confidence interval estimate of today's population mean 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 = 23.3 ± (1.753)*(1)
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 = 34.5 minutes on the plain chamber with SS = 210 for a sample of n = 15 birds. (Note: If the eye spots have no effect, then the birds should spend an average of μ = 30 minutes in each chamber.) C) Construct the 90% confidence interval to estimate the mean amount of time spent on the plain side for the population of birds.
C) With df = 14, the t values for 90% confidence are ±1.761, and the interval extends from 32.739 min to 36.261 minutes = 34.5 ± (1.761)*(1)
Researchers at a National Weather Center in the northeastern US recorded the number of 90 degree days each year since records first started in 1875. The numbers for a normal shaped distribution with a mean of mu=9.6 and a standard deviation of sigma=1.9 To see if the data showed any evidence of global warming, they also computed the mean number of 90 degree days for the most recent n=4 years and obtained M=12.25. Do the data indicate that the past four years have had significantly more 90 degree days than would be expected for a random sample from this population? Use a one-tailed test with alpha=0.05
Step 1: H0: μ <= 9.6 H1: μ > 9.6 Step 2: The critical region consists of z-scores beyond +1.65. This test is one-tailed with an alpha-level of 0.05 Step 3: For these data of n = 4, the standard error is: 1.9 / sqrt (4) = 0.95 and the sample mean corresponds to z = (12.25 - 9.6) / 0.95 = +2.79. This is well beyond the critical boundary of 1.65, so we conclude that there is a significant increase. Step 4: Reject the null hypothesis. There is a significant increase in the number of 90 degree days over the past 4 years. z = 2.79, p < .05, d = (12.25 - 9.6) / 1.9 = 1.39
The distribution of sample means is not always a normal distribution. Under what circumstances is the distribution of sample means not be normal?
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. However, the distribution of sample means is normal for large samples (equal or greater than 30) even if your original data is not normal. If your original data is normal, then the distribution of sample means will always be normal.
Describe the homogeneity of variance assumption and explain why it is important for the independent measures t test
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.
Although there is a popular belief that herbal remedies such as Ginkgo biloba and Ginsen may improve learning and memory in healthy adults, these effects are usually not supported by well controlled research. In a typical study, a researcher obtains a sample of n=16 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 form a normal distribution with a mean of mu=50 and a standard deviation of sigma=12. The sample of research participants had an average of M=54. a) Assuming a two-tailed test, state the null hypothesis in a sentence that includes the two variables being examined. b)Using the standard 4-step procedure, conduct a two tailed hypothesis test with alpha=0.05 to evaluate the effect of the supplements
a) The null hypothesis states that the herb has no effect on memory scores b) step 1: Ho: mu= 50 (even with the herbs, the mean is still 50 so did not change) Hi: mu cannot equal 50 (the mean has changed with the herb Step 2: The critical region consists of z-scores beyond +- 1.96. This test is two-taield with an alpha level of 0.05 Step 3: For these data, the standard error is: 12/sqrt(16) = 3 and z= (54-50)/3 = +1.33 Step 4: -Fail to reject the null hypothesis -The herbal supplements do not have a significant effect on memory scores - z=+1.33, p<0.05, d=(4/12) 0.33
If the population standard deviation is sigma = 10, how large a sample is necessary to have a standard error that is a. less than 5 points? b. less than 2 points? c. less than 1 point?
a. 5 = 10/squar root n ; n > 4 b. 2 = 10/squar root n ; n > 25 c. 1 = 10/squar root n ; n > 100 ** Note: as sample size (n) goes up, the standard error (of the means) decreases, assuming the variance does not change.
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 0.01 to 0.05 b) Changing from a one-tailed test to a two tailed test
a. Increasing alpha will increase power. The critical value is moved towards the center of the distribution. b. Changing from one- to two-tailed decreases power (if the effect is in the predicted direction). The critical value is moved toward the tail of the distribution.
Briefly define each of the following: a) Distribution of sample means b) Expected value of M c) Standard error of M
a. The distribution of sample means consist 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 distribution of sample mean (mu) c. The standard error of M is the standard deviation of distribution of sample means ** Note: Expected value is the (weighted) average or mean (μ) of all the Ms (or X bar). It is what you would expect to get if one sample is selected.
One sample has SS = 60 and a second sample has SS = 48. A)If n = 7 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 = 7 for the first sample and n = 5 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.
a. The first sample has s2 = 10 and the second has s2 = 8. The pooled variance is 108/12 = 9 (halfway between). b. The first sample has s2 = 10 and the second has s2 = 12. The pooled variance is 108/10 = 10.8 (closer to the variance for the larger sample).
One sample has SS = 36 and a second sample has SS = 18. A)If n = 4 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 = 4 for the first sample and n = 7 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.
a. The first sample has s2 = 12 and the second has s2 = 6. The pooled variance is 54/6 = 9(halfway between). b. The first sample has s2 = 12 and the second has s2 = 3. The pooled variance is 54/9 = 6(closer to the variance for the larger sample).
Find the z-score location of a vertical line that separates a normal distribution as described in each of the following: a) 5% in the tail on the left b) 30% in the tail on the right c) 65% in the body on the left d) 80% in the body on the right
a. z = -1.64 or -1.65 b. z = 0.52 c. z = 0.39 d. z = -0.84 **z-scores represent the closest scores to the percentages/proportions
In the Chapter Preview we presented a study showing that handling money reduces the perception pain (Zhou, Vohs, & Baumeister, 2009). In the experiment, a group of college students was told that they were participating in a manual dexterity study. Half of the students were given a stack of money to count and the other half got a stack of blank pieces of paper. After the counting task, the participants were asked to dip their hands into bowls of very hot water (122°F) and rate how uncomfortable it was. The following data show ratings of pain similar to the results obtained in the study. Counting Money Counting Paper 7 9 8 11 10 13 6 10 8 11 5 9 7 . 15 12 14 5 10 A) Is there a significant difference in reported pain between the two conditions? Use a two-tailed test with α = .01. B) Compute Cohen's d to estimate the size of the treatment effect.
a. The null hypothesis states that counting money versus counting paper does not affect the perception of pain. For the money group the mean is M = 7.56 with SS = 42.22. For the paper group, 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 there is a significant difference in the amount of pain experienced after counting money versus counting paper. b.d = 3.77/2.24 = 1.68
Two separate samples receive different treatments. After treatment, the first sample has n = 9 with SS = 462, and the second has n = 7 with SS = 420. 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 two-tailed test with α = .05?
a. The pooled variance is 63. b. The estimated standard error is 4. c. A mean difference of 10 points produces t = 2.50. With df = 14 the critical boundaries are ±2.145. Reject H0
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. Smashed into HIT n=15 n= 15 M=40.8 M= 34.0 SS= 510 SS=414 A) .Do the results indicate a significantly higher estimated speed for the "smashed into" group? Use a one-tailed test with α = .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 would appear in a research report.
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.
If other factors are held constant, explain how each of the following influences the value of the independent-measures t statistic, the likelihood of rejecting the null hypothesis, and the magnitude of measures of effect size. A) Increasing the number of scores in each sample. B) Increasing the variance for each sample.
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, however sample size has little or no effect on measures of effect size. 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), the likelihood of rejecting H0 decreases, and measures of effect size decrease.
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
a. The z-score (test statistic) increases (farther from zero) as sample size increases. b. Cohen's d (effect size) is not influenced by sample size. c. Power increases as sample size increases.
Scores on a standardized reading test for 4th-grade students form a normal distribution with mu=60 and sigma = 20. What is the probability of obtaining a sample mean greater than M=65 for each of the following: a) a sample of n=16 students b) a sample of n=25 students c) a sample of n=100 students
a. With a standard error of 5, M=65 corresponds to z = 1.00, p(M>65) = 0.1587 b. With a standard error of 4, M=65 corresponds to z= 1.25, p(M>65)= 0.1065 c. With a standard error of 2, M=65 corresponds to z=2.50, p(M>65) = 0.0062
For a population with a mean of mu=40 and a standard deviation of sigma=8, find the z-score corresponding to each of the following samples. a. sigma M = 8 points b. sigma M = 4 points c. sigma M = 1 point
a. You need to find the z-score for an individual score (X=36) so use σ = 8 points. z= (36-40)/8 = 0.50 b. For this problem, you need to find a z-score for a sample mean so you need to calculate the standard error, which comes out to be σM = 4 points. z = (36-40)/4 = 1.00 c. For this problem, you need to find a z-score for a sample mean so you need to calculate the standard error, which comes out to be σM = 2 points. z = (36-40)/2 = 2.00
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 = 2.50 b) z = 0.80 c) z = -0.50 d) z = -0.77
a. body to the left, p = 0.9938 b. body to the left, p = 0.7881 c. body to the right, p = 0.6915 d. body to the right, p = 0.7794
According to a recent report people smile an average of mu=62 times per day. Assuming that the distribution of smiles is approximately normal with a standard deviation of sigma=18, find each of the following values. a) What proportion of people smile more than 80 times a day? b) What proportion of people smile at least 50 times a day?
a. p(z > 1.00) = 0.1587 b. p(z > -0.67) = 0.7486
A population has a mean of mu=30 and a standard deviation of sigma = 8 a) If the population distribution is normal, what is the probability of obtaining a sample mean greater than M=32 for a sample of n=4? b)If the population distribution is positively skewed, what is the probability of obtaining a sample mean greater than M=32 for a sample of n=4 c) If the population distribution is normal , what is the probability of obtaining a sample mean greater than M=32 for a sample of n=64? d) If the population distribution is positively skewed, what is the probability of obtaining a sample mean greater than M=32 for a sample of n=64
a. σM = 8/2 = 4, z = (32-30) / 4 = 0.50 and p = 0.3085 b. Cannot answer because the distribution of sample means is not normal with n = 4 since the original distribution is not normal. The sample size n has to be 30 or above. c. σM = 8/8 = 1, z = (32-30) /1 = 2.00 and p = 0.0228 d. Same as above. With n = 64, the distribution of sample means is normal. σM= 1, z = 2.00 and p = 0.0228
If the alpha level is changed from alpha=.05 to alpha=.01, a) What happens to the boundaries for the critical region b) What happens to the probability of a Type 1 error?
a: Lowering the alpha level causes the boundaries of the critical region to move farther out into the tails of the distribution b: Lowering alpha reduces the probability of a Type 1 error
Callahan conducted a study to evaluate the effectiveness of physical exercise programs for individuals with chronic arthritis. Participants with doctor-diagnosed arthritis either received a Tai Chi course immediately or were placed in a control group to being the course 8 weeks later. At the end of the 8 week period, self-reports of pain were obtained in the study are shown in the following table Mean SE Tai Chi Course 3.7 1.2 No Tai Chi Course . 7.6 1.7 a) Construct a bar graph that incorporates all of the information in the table b) Looking at your graph, do you think that participation in the Tai Chi course reduces arthritis pain?
b. Even considering the standard error for each mean, there is no overlap between the two groups. The Tai Chi course does seem to have decreased self-reported pain even allowing for the standard error.
Rochester, New York, averages mu=21.9 inches of snow for the month of December. The distribution of snowfall amounts is approximately normal with a standard deviation of sigma=6.5 inches. A local jewelry store advertised a refund of 50% off all purchases made in December, if we have more than 3 feet during the month. What is the probability that the jewelry store will have to pay off on its promise?
p(X > 36) = p(z > 2.17) = 0.0150 or 1.50%
