Stats Ch 9 - T test 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

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. 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

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

4. A researcher reports a t statistic with df = 20. How many individuals participated in the study?

4. n = 21

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 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.

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.

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.

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.

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

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].

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.

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.

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

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.

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.

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.

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.

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. 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.

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.

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

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.

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.

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. 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. 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.

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. 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.

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.

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.

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.

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

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

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.