Psy201 Ch 8 INTRODUCTION TO HYPOTHESIS TESTING
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
2. In a research report, the term significant is used when the null hypothesis is rejected. (True or false?)
2. True
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
3. How does increasing sample size influence the power of a hypothesis test?
3. Increasing sample size increases the power of a test
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. 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.
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.
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. 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
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. 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. 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.
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 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%
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.
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.
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. 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. 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 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.
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.
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.
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.
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%.
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. 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.
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.
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.
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
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.
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
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.
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.
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.
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. 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.
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.
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.
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
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. 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.
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%.
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.
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.
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. 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. 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.
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.