CHECK YOUR LEARNING SOLUTIONS

Chapter 1

1. The average grade for your statistics class would be a descriptive statistic because it’s being used only to describe the tendency of people in your class with respect to a statistics grade.

2. In this case, the average grade would be an inferential statistic because it is being used to estimate the results of a population of students taking statistics.

1. 1500 Americans

2. All Americans

3. The 1500 Americans in the sample know 600 people, on average.

4. The entire population of Americans has many acquaintances, on average. The sample mean, 600, is an estimate of the unknown population mean.

1. These data are continuous because they can take on a full range of values.

2. The variable is a ratio observation because there is a true zero point.

3. On an ordinal scale, Lorna’s score would be 2 (or 2nd).

1. The nominal variable is gender. This is a nominal variable because the levels of gender, male and female, have no numerical meaning even if they are arbitrarily labeled 1 and 2.

2. The ordinal variable is hair length. This is an ordinal variable because the three levels of hair length (short, mid-length, and very long) are arranged in order, but we do not know the magnitude of the differences in length.

3. The scale variable is probability score. This is a scale variable because the distances between probability scores are assumed to be equal.

1. There are two independent variables: beverage and subject to be remembered. The dependent variable is memory.

2. Beverage has two levels: caffeine and no caffeine. The subject to be remembered has three levels: numbers, word lists, and aspects of a story.

1. Whether or not trays were available

2. Trays were available; trays were not available.

3. Food waste; food waste could have been measured via volume or weight as it was thrown away.

4. The measure of food waste would be consistent over time.

5. The measure of food waste was actually measuring how much food was wasted.

1. Researchers could randomly assign a certain number of women to be told about a gender difference on the test and randomly assign a certain number of other women to be told that no gender difference existed on this test.

2. If researchers did not use random assignment, any gender differences might be due to confounding variables. The women in the two groups might be different in some way (e.g., in math ability or belief in stereotypes) to begin with.

3. There are many possible confounds. Women who already believed the stereotype might do so because they had always performed poorly in mathematics, whereas those who did not believe the stereotype might be those who always did particularly well in math. Women who believed the stereotype might be those who were discouraged from studying math because “girls can’t do math,” whereas those who did not believe the stereotype might be those who were encouraged to study math because “girls are just as good as boys in math.”

4. Math performance is operationalized as scores on a math test.

5. Researchers could have two math tests that are similar in difficulty. All women would take the first test after being told that women tend not to do as well as men on this test. After taking that test, they would be given the second test after being told that women tend to do as well as men on this test.

Chapter 2

1. Interval	Frequency
   100–104	4
   95–99	14
   90–94	6
   85–89	4
   80–84	2
   75–79	6
   70–74	6
   65–69	3
   60–64	5

2. 

3. 

1. We can now get a sense of the overall pattern of the data.

2. Schools in different academic systems may place different emphases on faculty conducting independent research and getting published, as well as have varying ranges of funding or financial resources to assist faculty research and publication.

1. Early-onset Alzheimer’s disease would create negative skew in the distribution for age of onset.

2. Because all humans eventually die, there is a sort of ceiling effect.

Chapter 3

1. A scatterplot is the best graph choice to depict the relation between two scale variables such as depression and stress.

2. A time plot, or time series plot, is the best graph choice to depict the change in a scale variable, such as the rise or decline in the number of facilities over time.

   D-3

3. For one scale variable, such as number of siblings, the best graph choice would be either a frequency histogram or frequency polygon.

4. In this case, there is a nominal variable (region of the United States) and a scale variable (years of education). The best choice would be a bar graph, with one bar depicting the mean years of education for each region. In a Pareto chart, the bars would be arranged from highest to lowest, allowing for easier comparisons.

5. Calories and hours are both scale variables, and the question is about prediction rather than relation. In this case, we would calculate and graph a line of best fit.

1. Scatterplot or line graph

2. Bar graph

3. Scatterplot or line graph

Chapter 4

1. 

   = [10 + 8 + 22 + 5 + 6 + 1 + 19 + 8 + 13 + 12 + 8]/11

   = 112/11 = 10.18

   The median is found by arranging the scores in numeric order—1, 5, 6, 8, 8, 8, 10, 12, 13, 19, 22—then dividing the total number of scores, 11, by − and adding 1/2 to get 6. The 6th score in our ordered list of scores is the median, and in this case the 6th score is the number 8.

   The mode is the most common score. In these data, the score 8 occurs most often (three times), so 8 is our mode.

2. 

   = (122.5 + 123.8 + 121.2 + 125.8 + 120.2 + 123.8 + 120.5 + 119.8 + 126.3 + 123.6)/10

   = 122.7/10 = 122.75

   The data ordered are: 119.8, 120.2, 120.5, 121.2, 122.5, 123.6, 123.8, 123.8, 125.8, 126.3. Again, we find the median by ordering the data and then dividing the number of scores (here there are 10 scores) by − and adding 1/2. In this case, we get 5.5, so the mean of the 5th and 6th data points is the median. The median is (122.5 1 123.6)/2 5 123.05.

   The mode is 123.8, which occurs twice in these data.

3. 

   = (0.100 + 0.866 + 0.781 + 0.555 + 0.222 + 0.245 + 0.234)/7

   = 3.003/7 = 0.429.

   Note that three decimal places are included here (rather than the standard two places used throughout this book) because the data are carried out to three decimal places.

   The median is found by first ordering the data: 0.100, 0.222, 0.234, 0.245, 0.555, 0.781, 0.866. Then the total number of scores, 7, is divided by − to get 3.5, to which 1/2 is added to get 4. So, the 4th score, 0.245, is the median. There is no mode in these data. All scores occur once.

1. 

   = (1 + 0 + 1 + − + 5 + ...4 + 6)/20 = 50/20 = 2.50

2. In this case, the scores would comprise a sample taken from the whole population, and this mean would be a statistic. The symbol, therefore, would be either M or X.

3. In this case, the scores would constitute the entire population of interest, and the mean would be a parameter. Thus, the symbol would be m.

4. To find the median, we would arrange the scores in order: 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 4, 5, 6, 7. We would then divide the total number of scores, 20, by 2 and add 1/2, which is 10.5. The median, therefore, is the mean of the 10th and 11th scores. Both of these scores are 2; therefore, the median is 2.

5. The mode is the most common score—in this case, there are six 2’s, so the mode is 2.

6. The mean is a little higher than the median. This indicates that there are potential outliers pulling the mean higher; outliers would not affect the median.

1. The range is: X_highest − X_lowest = 22 − 1 = 21

   The variance is:

   

   We start by calculating the mean, which is 10.182. We then calculate the deviation of each score from the mean and the square of that deviation.

   X	X − M	(X − M )2
   10	−0.182	0.033
   8	−2.182	4.761
   22	11.818	139.665
   5	−5.182	26.853
   6	−4.182	17.489
   1	−9.182	84.309
   19	8.818	77.757
   8	−2.182	4.761
   13	2.818	7.941
   12	1.818	3.305
   8	−2.182	4.761

   

   The standard deviation is: or

   

2. The range is: X_highest − X_lowest = 126.3 − 119.8 = 6.5

   The variance is:

   

   We start by calculating the mean, which is 122.750. We then calculate the deviation of each score from the mean and the square of that deviation.

   X	X − M	(X − M )2
   122.500	−0.250	0.063
   123.800	1.050	1.103
   121.200	−1.550	2.403
   125.800	3.050	9.303
   120.200	−2.550	6.503
   123.800	1.050	1.103
   120.500	−2.250	5.063
   119.800	−2.950	8.703
   126.300	3.550	12.603
   123.600	0.850	0.723

   

   The standard deviation is:

   

3. The range is: X_highest − X_lowest = 0.866 − 0.100 = 0.766

   The variance is:

   

   We start by calculating the mean, which is 0.429. We then calculate the deviation of each score from the mean and the square of that deviation.

   X	X − M	(X − M)2
   0.100	−0.329	0.108
   0.866	0.437	0.191
   0.781	0.352	0.124
   0.555	0.126	0.016
   0.222	−0.207	0.043
   0.245	−0.184	0.034
   0.234	−0.195	0.038

   

   The standard deviation is: or

   

1. range = X_highest − X_lowest = 1460 − 450 = 1010

2. We do not know whether scores cluster at some point in the distribution—for example, near one end of the distribution—or whether the scores are more evenly spread out.

   D-5

3. The formula for variance is

   

   The first step is to calculate the mean, which is 927.500. We then create three columns: one for the scores, one for the deviations of the scores from the mean, and one for the squares of the deviations.

   X	X − M	(X − M )2
   450	−477.50	228,006.25
   670	−257.50	66,306.25
   1130	202.50	41,006.25
   1460	532.50	283,556.25

   We can now calculate variance:

   

   = (228,006.25 + 66,306.25 + 41,006.25 + 283,556.25)/4

   = 618,875/4 = 154,718.75.

4. We calculate standard deviation the same way we calculate variance, but we then take the square root.

   

5. If the researcher were interested only in these four students, these scores would represent the entire population of interest, and the variance and standard deviation would be parameters. Therefore, the symbols would be s² and s, respectively.

6. If the researcher hoped to generalize from these four students to all students at the university, these scores would represent a sample, and the variance and standard deviation would be statistics. Therefore, the symbols would be SD², s², or MS for variance and SD or s for standard deviation.

Chapter 5

1. The likely population is all patients who will undergo surgery; the researcher would not be able to access this population, and therefore random selection could not be used. Random assignment, however, could be used. The psychologist could randomly assign half of the patients to counseling and half to a control group.

2. The population is all children in this school system; the psychologist could identify all of these children and thus could use random selection. The psychologist could also use random assignment. She could randomly assign half the children to the interactive online textbook and half to the printed textbook.

3. The population is patients in therapy; because the whole population could not be identified, random selection could not be used. Moreover, random assignment could not be used. It is not possible to assign people to either have or not have a diagnosed personality disorder.

1. probability = successes/trials = 5/100 = 0.05

2. 8/50 = 0.16

3. 130/1044 = 0.12

1. In the short run, we might see a wide range of numbers of successes. It would not be surprising to have several in a row or none in a row. In the short run, the observations seem almost like chaos.

2. Given the assumptions listed for this problem, in the long run, we’d expect 0.50, or 50%, to be women, although there would likely be strings of men and of women along the way.

1. The null hypothesis is that a decrease in temperature does not affect mean academic performance (or does not decrease mean academic performance).

2. The research hypothesis is that a decrease in temperature does affect mean academic performance (or decreases mean academic performance).

3. The researchers would reject the null hypothesis.

4. The researchers would fail to reject the null hypothesis.

1. If the virtual-reality glasses really don’t have any effect, this is a Type I error, which is made when the null hypothesis is rejected but is really true.

2. If the virtual-reality glasses really do have an effect, this is a Type II error, which is made when the researchers fail to reject the null hypothesis but the null hypothesis is not true.

Chapter 6

1. X = z(σ) + µ = 2(2.5) + 14 = 19

2. X = z(σ) + µ = − 1.4(2.5) + 14 = 10.5

1. approximately 16% of students have a CFC score of 2.5 or less.

2. this score is at approximately the 98th percentile.

3. This student has a z score of 1.

4. X = z(σ) + µ = 1(0.70) + 3.20 = 3.90; this answer makes sense because 3.90 is above the mean of 3.20, as a z score of 1 would indicate.

1. Nicole is in better health because her score is above the mean for her measure, whereas Samantha’s score is below the mean.

2. Samantha’s z score is

   Nicole’s z score is

   Nicole is in better health, being 1 standard deviation above the mean, whereas Samantha is − standard deviations below the mean.

3. We can conclude that approximately 98% of the population is in better health than Samantha, who is − standard deviations below the mean. We can conclude that approximately 16% of the population is in better health than Nicole, who is 1 standard deviation above the mean.

1. The scores range from 2.0 to 4.5, which gives us a range of 4.5 − 2.0 = 2.5.

   D-7

2. The means are 3.4 for the first row, 3.4 for the second row, and 3.15 for the third row [e.g., for the first row, M = (3.5 + 3.5 + 3.0 + 4.0 + 2.0 + 4.0 + 2.0 + 4.0 + 3.5 + 4.5)/10 = 3.4]. These three means range from 3.15 to 3.40, which gives us a range of 3.40 − 3.15 = 0.25.

3. The range is smaller for the means of samples of 10 scores than for the individual scores because the more extreme scores are balanced by lower scores when samples of 10 are taken. Individual scores are not attenuated in that way.

4. The mean of the distribution of means will be the same as the mean of the individual scores: µ_M = µ = 3.32. The standard error will be smaller than the standard deviation; we must divide by the square root of the sample size of 10:

   

Chapter 7

1. µ_M = µ = 156.8

   

   

   50% below the mean; 49.87% between the mean and this score; 50 + 49.87 = 99.87th percentile

2. 100 − 99.87 = 0.13% of samples of this size scored higher than the students at Baylor.

3. At the 99.87th percentile, these 36 students from Baylor are truly outstanding. If these students are representative of their majors, clearly these results reflect positively on Baylor’s psychology and neuroscience department.

1. 0.15

2. 0.03

3. 0.055

1. (1) The dependent variable—diagnosis (correct versus incorrect)—is nominal, not scale, so this assumption is not met. Based only on this, we should not proceed with a hypothesis test based on a z distribution. (2) The samples include only outpatients seen over 2 specific months and only those at one community mental health center. The sample is not randomly selected, so we must be cautious about generalizing from it. (3) The populations are not normally distributed because the dependent variable is nominal.

2. (1) The dependent variable, health score, is likely scale. (2) The participants were randomly selected; all wild cats in zoos in North America had an equal chance of being selected for this study. (3) The data are not normally distributed; we are told that a few animals had very high scores, so the data are likely positively skewed. Moreover, there are fewer than 30 participants in this study. It is probably not a good idea to proceed with a hypothesis test based on a z distribution.

Chapter 8

1. First, we draw a normal curve with the sample mean, 3.20, in the center. Then we put the bounds of the 95% confidence interval on either end, writing the appropriate percentages under the segments of the curve: 2.5% beyond the cutoffs on either end and 47.5% between the mean and each cutoff. Now we look up the z statistics for these cutoffs; the z statistic associated with 47.5%, the percentage between the mean and the z statistic, is 1.96. Thus, the cutoffs are −1.96 and 1.96. Next, we calculate standard error so that we can convert these z statistics to raw means:

   

   M_lower = −z(σ_M) + M_sample = −1.96(0.104) + 3.45 = 3.25

   M_upper = z(σ_M ) + M_sample = 1.96(0.104) + 3.45 = 3.65

   Finally, we check to be sure the answer makes sense by demonstrating that each end of the confidence interval is the same distance from the mean: 3.25 − 3.45 = −0.20 and 3.65 − 3.45 = 0.20. The confidence interval is [3.25, 3.65].

2. If we were to conduct this study over and over, with the same sample size, we would expect the population mean to fall in that interval 95% of the time. Thus, it provides a range of plausible values for the population mean. Because the null-hypothesized population mean of 3.20 is not a plausible value, we can conclude that those who attended the discussion group seem to have higher mean CFC scores than those who did not. This conclusion matches that of the hypothesis test, in which we rejected the null hypothesis.

3. The confidence interval is superior to the hypothesis test because not only does it lead to the same conclusion but it also gives us an interval estimate, rather than a point estimate, of the population mean.

1. We calculate Cohen’s d, the effect size appropriate for data analyzed with a z test. We use standard deviation in the denominator, rather than standard error, because effect sizes are for distributions of scores rather than distributions of means.

   Cohen’s

2. Cohen’s conventions indicate that 0.2 is a small effect and 0.5 is a medium effect. This effect size, therefore, would be considered a small-to-medium effect.

3. If the career discussion group is easily implemented in terms of time and money, the small-to-medium effect might be worth the effort. For university students, a higher mean level of Consideration of Future Consequences might translate into a higher mean level of readiness for life after graduation, a premise that we could study.

1. The statistical power calculation means that, if population 1 really does exist, we have a 77.64% chance of observing a sample mean, based on 45 observations, that will allow us to reject the null hypothesis. We fall just short of the desired 80% statistical power.

2. We can increase statistical power by increasing the sample size, extending or enhancing the career discussion group such that we create a bigger effect, or by changing alpha.

Chapter 9

1. We would use a distribution of means, specifically a t distribution. It is a distribution of means because we have a sample consisting of more than one individual. It is a t distribution because we are comparing one sample to a population, but we know only the population mean, not its standard deviation.

2. The appropriate mean: µ_M = µ = 25. The calculations for the appropriate standard deviation (in this case, standard error, s_M):

   

   X	X − M	(X − M )2
   20	−1.6	2.56
   19	−2.6	6.76
   27	5.4	29.16
   24	2.4	5.76
   18	−3.6	12.96

   Numerator: ∑(X − M)2 = (2.56 + 6.76 + 29.16 + 5.76 + 12.96)

   = 57.2

   

   

3. 

1. df = N − 1 = 35 − 1 = 34

2. df = N − 1 = 14 − 1 = 13

1. ±2.201

2. Either −2.584 or +2.584, depending on the tail of interest

1. We first find the t values associated with a two-tailed hypothesis test and alpha of 0.05. These are ±2.776. We then calculate s_M by dividing s by the square root of the sample size, which results in s_M = 0.548.

   M_lower = −t(s_M) + M_sample = −2.776(0.548) + 1.0 = −0.52

   M_upper = t(s_M) + M_sample = 2.776(0.548) + 1.0 = 2.52

   The confidence interval can be written as [−0.52, 2.52]. Because this confidence interval includes 0, we would fail to reject the null hypothesis. Zero is one of the likely mean differences we would get when repeatedly sampling from a population with a mean difference score of 1.

2. We calculate Cohen’s d as:

   This is a large effect size.

1. Step 1: Population 1 is students for whom we’re measuring energy levels before lunch. Population 2 is students for whom we’re measuring energy levels after lunch.

   The comparison distribution is a distribution of mean difference scores. We use the paired-samples t test because each participant contributes a score to each of the two samples we are comparing.

   We meet the assumption that the dependent variable is a scale measurement. However, we do not know whether the participants were randomly selected or if the population is normally distributed, and the sample is less than 30.

   Step 2: The null hypothesis is that there is no difference in mean energy levels before and after lunch—H₀: µ₁ = µ₂.

   The research hypothesis is that there is a mean difference in energy levels—H₁: µ₁ ≠ µ₂.

   Step 3:

   Difference Scores	Difference − Mean Difference	Squared Deviation
   −3	−1.6	2.56
   −3	−1.6	2.56
   2	3.4	11.56
   −1	0.4	0.16
   −2	−0.6	0.36

   M_difference = −1.4

   

   

   µ_M = 0, s_M = 0.928

   Step 4: The degrees of freedom are 5 − 1 = 4, and the cutoffs, based on a two-tailed test and a p level of 0.05, are ±2.776.

   Step 5:

   Step 6: Because the test statistic, −1.51, did not exceed the critical value of −2.776, we fail to reject the null hypothesis.

1. M_lower = −t(s_M) + M_sample = −2.776(0.928) + (−1.4) = −3.98

   M_upper = t(s_M) + M_sample = 2.776(0.928) + (−1.4) = 1.18

   The confidence interval can be written as [ − 3.98, 1.18]. Notice that the confidence interval spans 0, the null-hypothesized difference between mean energy levels before and after lunch. Because the null value is within the confidence interval, we fail to reject the null hypothesis.

2. 

   This is a medium-to-large effect size, according to Cohen’s guidelines.

Chapter 10

1. Group 1 is treated as the X variable; its mean is 3.0.

   X	X − M	(X − M)2
   3	0	0
   2	−1	1
   4	1	1
   6	3	9
   1	−2	4
   2	−1	1

   

   Group 2 is treated as the Y variable; its mean is 4.6.

   Y	Y − M	(Y − M )2
   5	0.4	0.16
   4	−0.6	0.36
   6	1.4	1.96
   2	−2.6	6.76
   6	1.4	1.96

   

   = 2.8

2. df_X = N − 1 = 6 − 1 = 5

   df_y = N − 1 = 5 − 1 = 4

   

3. The variance version of standard error is calculated for each sample as:

   

4. The variance of the distribution of differences between means is:

   s²difference = s²_MX + s²_MY = 0.504 + 0.604 = 1.108

   This can be converted to standard deviation units by taking the square root:

   

5. 

1. The null hypothesis asserts that there are no average between-group differences; employees with low trust in their leader show the same mean level of agreement with decisions as those with high trust in their leader. Symbolically, this would be written H₀: µ₁ = µ₂.

   The research hypothesis asserts that mean level of agreement is different between the two groups—H₁: µ₁ ≠ µ₂.

2. The critical values, based on a two-tailed test, a p level of 0.05, and df_total of 9, are −2.262 and 2.262.

   The t value we calculated, −1.519, does not exceed the cutoff of −2.262, so we fail to reject the null hypothesis.

3. Based on these results, we did not find evidence that mean level of agreement with a decision is different across the two levels of trust, t(9) = −1.519, p > 0.05.

4. Despite having similar means for the two groups, we failed to reject the null hypothesis, whereas the original researchers rejected the null hypothesis. The failure to reject the null hypothesis is likely due to the low statistical power from the small samples we used.

1. The upper and lower bounds of the confidence interval are calculated as:

   (M_X − M_Y)_lower = −t(s_difference) + (M_X − M_Y)_sample

   = −2.262(1.053) + (−1.6) = −3.98

   (M_X − M_Y)_upper = t(s_difference) + (M_X − M_Y)_sample

   = 2.262(1.053) + (−1.6) = 0.78

   The confidence interval is [−3.98, 0.78].

2. To calculate Cohen’s d, we need to calculate the pooled standard deviation for the data:

   

Chapter 11

1. We would use an F distribution because there are more than two groups.

2. We would determine the variance among the three sample means—the means for those in the control group, for those in the 2-hour communication ban, and for those in the 4-hour communication ban.

3. We would determine the variance within each of the three samples, and we would take a weighted average of the three variances.

1. df_between = N_groups − 1 = 3 − 1 = 2

2. df_within = df₁ + df₂ + . . . + df_last = (4 − 1) + (4 − 1) + (3 − 1)

   = 3 + 3 + 2 = 8

3. df_total = df_between + df_within = 2 + 8 = 10

1. Total sum of squares is calculated here as SS_total = ∑(X − GM)²:

   Sample	X	(X −GM)	(X −GM)2
   Group 1	37	−1.727	2.983
   M1 = 29.5	30	−8.727	76.161
   	22	−16.727	279.793
   	29	−9.727	94.615
   Group 2	49	10.273	105.535
   M2 = 45.25	52	13.273	176.173
   	41	2.273	5.167
   	39	0.273	0.075
   Group 3	36	−2.727	7.437
   M3 = 42.333	49	10.273	105.535
   	42	3.273	10.713

   GM = 38.73    SS_total = 864.187

2. Within-groups sum of squares is calculated here as SS_within = ∑(X − M)².

   Sample	X	(X −M)	(X −M)2
   Group 1	37	7.500	56.250
   M1 = 29.5	30	0.500	0.250
   	22	−7.500	56.250
   	29	−0.500	0.250
   Group 2	49	3.750	14.063
   M2 = 45.25	52	6.750	45.563
   	41	−4.250	18.063
   	39	−6.250	39.063
   Group 3	36	−6.333	40.107
   M3 = 42.333	49	6.667	44.449
   	42	−0.333	0.111

   GM = 38.73    SS_within = 314.419

3. Between-groups sum of squares is calculated here as SS_between = ∑(M − GM)²:

   Sample	X	(M −GM )	(M −GM )2
   Group 1	37	−9.227	85.138
   M1 = 29.5	30	−9.227	85.138
   	22	−9.227	85.138
   	29	−9.227	85.138
   Group 2	49	6.523	42.550
   M2 = 45.25	52	6.523	42.550
   	41	6.523	42.550
   	39	6.523	42.550
   Group 3	36	3.606	13.003
   M3 = 42.333	49	3.606	13.003
   	42	3.606	13.003

   GM = 38.73    SS_between = 549.761

1. According to the null hypothesis, there are no mean differences in efficacy among these three treatment conditions; they would all come from one underlying distribution. The research hypothesis states that there are mean differences in efficacy across some or all of these treatment conditions.

2. There are three assumptions: that the participants were selected randomly, that the underlying populations are normally distributed, and that the underlying populations have similar variances. Although we can’t say much about the first two assumptions, we can assess the last one using the sample data.

   Sample	Group 1	Group 2	Group 3
   Squared deviations of scores from sample means	56.25	14.063	40.107
   	0.25	45.563	44.449
   	56.25	18.063	0.111
   	0.25	39.063
   Sum of squares	113	116.752	84.667
   N – 1	3	3	2
   Variance	37.67	38.92	42.33

   Because these variances are all close together, with the biggest being no more than twice as large as the smallest, we can conclude that we meet the third assumption of homoscedastic samples.

3. The critical value for F with a p value of 0.05, two between-groups degrees of freedom, and 8 within-groups degrees of freedom, is 4.46. The F statistic exceeds this cutoff, so we can reject the null hypothesis. There are mean differences between these three groups, but we do not know where.

1. 

2. Children’s grade level accounted for 64% of the variability in reaction time. This is a large effect.

1. df_between = N_groups − 1 = 3 − 1 = 2

2. df_subjects = n − 1 = 3 − 1 = 2

3. df_within = (df_between)(df_subjects) = (2)(2) = 4

4. df_total = df_between + df_subjects + df_within = 2 + 2 + 4 = 8; or we can calculate it as df_total = N_total − 1 = 9 − 1 = 8

1. SS_total = ∑(X − GM)² = 24.886

   Group	Rating (X)	X − GM	(X − GM )2
   1	7	0.111	0.012
   1	9	2.111	4.456
   1	8	1.111	1.234
   2	5	−1.889	3.568
   2	8	1.111	1.234
   2	9	2.111	4.456
   3	6	−0.889	0.790
   3	4	−2.889	8.346
   3	6	−0.889	0.790

   GM = 6.889    ∑(X − GM )² = 24.886

2. SS_between = ∑(M − GM)² = 11.556

   GROUP	Rating (X)	Group Mean (M)	M − GM	(M − GM)2
   1	7	8	1.111	1.234
   1	9	8	1.111	1.234
   1	8	8	1.111	1.234
   2	5	7.333	0.444	0.197
   2	8	7.333	0.444	0.197
   2	9	7.333	0.444	0.197
   3	6	5.333	−1.556	2.421
   3	4	5.333	−1.556	2.421
   3	6	5.333	−1.556	2.421

   GM = 6.889    ∑(M − GM )² = 11.556

3. SS_subject = ∑(M_participant − GM )² = 4.221

   Participant	Group	Rating (X )	Participant Mean (MPARTICIPANT)	MPARTICIPANT − GM	(MPARTICIPANT − GM )2
   1	1	7	6	−0.889	0.790
   2	1	9	7	0.111	0.012
   3	1	8	7.667	0.778	0.605
   1	2	5	6	−0.889	0.790
   2	2	8	7	0.111	0.012
   3	2	9	7.667	0.778	0.605
   1	3	6	6	−0.889	0.790
   2	3	4	7	0.111	0.012
   3	3	6	7.667	0.778	0.605

   GM = 6.889    ∑(M_participant − GM )² = 4.221

4. 

1. Null hypothesis: People rate the driving experience of these three cars the same, on average—H₀: µ₁ = µ₂ = µ₃. Research hypothesis: People do not rate the driving experience of these three cars the same, on average.

2. Order effects are addressed by counterbalancing. We could create a list of random orders of the three cars to be driven. Then, as a new customer arrives, we would assign him or her to the next random order on the list. With a large enough sample size (much larger than the three participants we used in this example), we could feel confident that this assumption would be met using this approach.

   D-15

3. The critical value for the F statistic for a p level of 0.05 and 2 and 4 degrees of freedom is 6.95. The between-groups F statistic of 2.54 does not exceed this critical value. We cannot reject the null hypothesis, so we cannot conclude that there are differences among mean ratings of cars.

1. First, we calculate s_M:

   

   Next, we calculate HSD for each pair of means.

   For time 1 versus time 2:

   

   For time 1 versus time 3:

   

   For time − versus time 3:

   

2. We have an independent variable with three levels and df_within = 10, so the q cutoff value at a p level of 0.05 is 3.88. Because we are performing a two-tailed test, the cutoff values are 3.88 and −3.88.

3. We reject the null hypothesis for all three of the mean comparisons because all of the HSD calculations exceed the critical value of −3.88. This tells us that all three of the group means are statistically significantly different from one another.

1. 

   This is a large effect size.

2. Because the F statistic did not exceed the critical value, we failed to reject the null hypothesis. As a result, Tukey HSD tests are not necessary.

Chapter 12

1. There are two factors: diet programs and exercise programs.

2. There are three factors: diet programs, exercise programs, and metabolism type.

3. There is one factor: gift certificate value.

4. There are two factors: gift certificate value and store quality.

1. The participants are the stocks themselves.

2. One independent variable is the type of ticker-code name, with two levels: pronounceable and unpronounceable. The second independent variable is time lapsed since the stock was initially offered, with four levels: 1 day, 1 week, 6 months, and 1 year.

3. The dependent variable is the stocks’ selling price.

4. This would be a two-way mixed-design ANOVA.

5. This would be a 2 × 4 mixed-design ANOVA.

6. This study would have eight cells: 2 × 4 = 8. We multiplied the numbers of levels of each of the two independent variables.

1. There are four cells.

   		IV 2
   		Level A	Level B
   IV I	Level A
   	Level B

   D-16

2. 		IV 2
   		Level A	Level B
   IV I	Level A	M = (2 + 1 + 1 + 3)/4 = 1.75	M = (2 + 3 + 3 + 3)/4 = 2.75
   	Level B	M = (5 + 4 + 3 + 4)/4 = 4	M = (3 + 2 + 2 + 3)/4 = 2.5

3. Because the sample size is the same for each cell, we can compute marginal means as simply the average between cell means.

   		IV 2		Marginal Means
   		Level A	Level B
   IV I	Level A	1.75	2.75	2.25
   	Level B	4	2.5	3.25
   	Marginal Means	2.875	2.625

4. 

1. 1. Independent variables: student (Caroline, Mira); class (philosophy, psychology)

   2. Dependent variable: performance in class

   3. 	Caroline	Mira
      Philosophy class
      Psychology class

   4. This describes a qualitative interaction because the direction of the effect reverses. Caroline does worse in philosophy class than in psychology class, whereas Mira does better.

2. 1. Independent variables: game location (home, away); team (own conference, other conference)

   2. Dependent variable: number of wins

   3. 	Home	Away
      Own conference
      Other conference

   4. This describes a qualitative interaction because the direction of the effect reverses. The team does worse at home against teams in the other conference, but does well against those teams while away; the team does better at home against teams in its own conference, but performs poorly against teams in its own conference when away.

3. 1. Independent variables: amount of caffeine (caffeine, none); exercise (worked out, did not work out)

   2. Dependent variable: amount of sleep

   3. 	Caffeine	No Caffeine
      Working out
      Not working out

   4. This describes a quantitative interaction because the effect of working out is particularly strong in the presence of caffeine versus no caffeine (and the presence of caffeine is particularly strong in the presence of working out versus not). The direction of the effect of either independent variable, however, does not change depending on the level of the other independent variable.

1. Population 1 is students who received an initial grade of C who received e-mail messages aimed at bolstering self-esteem. Population 2 is students who received an initial grade of C who received e-mail messages aimed at bolstering their sense of control over their grades. Population 3 is students who received an initial grade of C who received e-mails with just review questions. Population 4 is students who received an initial grade of D or F who received e-mail messages aimed at bolstering self-esteem. Population 5 is students who received an initial grade of D or F who received e-mail messages aimed at bolstering their sense of control over their grades. Population 6 is students who received an initial grade of D or F who received emails with just review questions.

2. Step 2: Main effect of first independent variable—initial grade:

   Null hypothesis: The mean final exam grade of students with an initial grade of C is the same as that of students with an initial grade of D or F. H₀: µ_C = µ_D/F.

   Research hypothesis: The mean final exam grade of students with an initial grade of C is not the same as that of students with an initial grade of D or F. H₁: µ_C ≠ µ_D/F.

   Main effect of second independent variable—type of email:

   Null hypothesis: On average, the mean exam grades among those receiving different types of e-mails are the same—H₀: µ_SE = µ_CG = µ_TR.

   Research hypothesis: On average, the mean exam grades among those receiving different types of e-mails are not the same.

   Interaction: Initial grade × type of e-mail:

   Null hypothesis: The effect of type of e-mail is not dependent on the levels of initial grade.

   Research hypothesis: The effect of type of e-mail depends on the levels of initial grade.

3. Step 3: df_{between/grade} + N_groups − 1 = 2 − 1 = 1

   df_{between/e-mail} = N_groups − 1 = 3 − 1 = 2

   df_interaction = (df_{between/grade})(df_{between/e-mail}) = (1)(2) = 2

   df_C,SE = N − 1 = 14 − 1 = 13

   df_C,C = N − 1 = 14 − 1 = 13

   df_C,TR = N − 1 = 14 − 1 = 13

   df_D/F,SE = N − 1 = 14 − 1 = 13

   df_D/F,C = N − 1 = 14 − 1 = 13

   df_D/F,TR = N − 1 =14 − 1 = 13

   df_within = df_C,SE + df_C,C + df_C,TR + df_D/F,SE + df_D/F,C + df_D/F,TR = 13 + 13 + 13 + 13 + 13 + 13 = 78

   Main effect of initial grade: F distribution with 1 and 78 degrees of freedom

   Main effect of type of e-mail: F distribution with − and 78 degrees of freedom

   Main effect of interaction of initial grade and type of e-mail: F distribution with − and 78 degrees of freedom

4. Step 4: Note that when the specific degrees of freedom is not in the F table, you should choose the more conservative—that is, larger—cutoff. In this case, go with the cutoffs for a within-groups degrees of freedom of 75 rather than 80. The three cutoffs are:

   Main effect of initial grade: 3.97

   Main effect of type of e-mail: 3.12

   Interaction of initial grade and type of e-mail: 3.12

5. Step 6: There is a significant main effect of initial grade because the F statistic, 20.84, is larger than the critical value of 3.97. The marginal means, seen in the accompanying table, tell us that students who earned a C on the initial exam have higher scores on the final exam, on average, than do students who earned a D or an F on the initial exam. There is no statistically significant main effect of type of e-mail, however. The F statistic of 1.69 is not larger than the critical value of 3.12. Had this main effect been significant, we would have conducted a post hoc test to determine where the differences were. There also is not a significant interaction. The F statistic of 3.02 is not larger than the critical value of 3.12. (Had we used a cutoff based on a p level of 0.10, we would have rejected the null hypothesis for the interaction. The cutoff for a p level of 0.10 is 2.38.) If we had rejected the null hypothesis for the interaction, we would have examined the cell means in tabular and graph form.

   	Self-Esteem	Take Responsibility	Control Group	Marginal Means
   C	67.31	69.83	71.12	69.42
   D/F	47.83	60.98	62.13	56.98
   Marginal means	57.57	65.41	66.63

Chapter 13

1. According to Cohen, this is a large (strong) correlation. Note that the sign (negative in this case) is not related to the assessment of strength.

   D-18

2. This is just above a medium correlation.

3. This is lower than the guideline for a small correlation, 0.10.

1. A scatterplot for a correlation coefficient of –0.60 might look like this:

   

2. A scatterplot for a correlation coefficient of 0.35 might look like this:

   

3. A scatterplot for a correlation coefficient of 0.04 might look like this:

   

1. The test does not have sufficient reliability to be used as a diagnostic tool. As stated in the chapter, when using tests to diagnose or make statements about the ability of individuals, a reliability of at least 0.90 is necessary.

2. We do not have enough information to determine whether the test is valid. The coefficient alpha tells us only about the reliability of the test.

3. To assess the validity of the test, we would need the correlation between this measure and other measures of reading ability or between this measure and students’ grades in the nonremedial class (i.e., do students who perform very poorly in the nonremedial reading class also perform poorly on this measure?).

1. The psychometrician could assess test–retest reliability by administering the quiz to 100 readers who have boyfriends, and then one week later readministering the test to the same 100 readers. If their scores at the two times are highly correlated, the test would have high test–retest reliability. She also could calculate a coefficient alpha using computer software. The computer would essentially calculate correlations for every possible two groups of five items and then calculate the average of all of these split-half correlations.

2. The psychometrician could assess validity by choosing criteria that she believed assessed the underlying construct of interest, a boyfriend’s devotion to his partner. There are many possible criteria. For example, she could correlate the amount of money each participant’s boyfriend spent on his partner’s most recent birthday with the number of minutes the participant spent on the phone with his or her boyfriend today or with the number of months the relationship ends up lasting.

   D-20

3. Of course, we assume that these other measures actually assess the underlying construct of a boyfriend’s devotion, which may or may not be true! For example, the amount of money that the boyfriend spent on the participant’s last birthday might be a measure of his income, not his devotion.

Chapter 14

1. 

2. z_ŷ = (r_XY)(z_X)

   = (0.28)(1.5) = 0.42

3. Ŷ = z_ŷ(SD_Y) + M_Y = (0.42)(15) + 155 = 161.3 pounds

1. Ŷ = 12 + 0.67 [X] = 12 + 0.67 [78] = 64.26

2. Ŷ = 12 + 0.67 [−14 ] = 2.62

3. Ŷ = 12 + 0.67 [52] = 46.84

1. The y intercept, 2.586, is the GPA we might expect if someone played no minutes. The slope of 0.016 is the increase in GPA that we would expect for each 1-minute increase in playing time. Because the correlation is positive, it makes sense that the slope is also positive.

2. The standardized regression coefficient is equal to the correlation coefficient for simple linear regression, 0.344. We can also check that this is correct by computing ß:

   X	(X −MX)	(X −MX)2	Y	(Y −MY)	(Y −MY)2
   29.70	13.159	173.159	3.20	0.343	0.118
   32.14	15.599	243.329	2.88	0.023	0.001
   32.72	16.179	261.760	2.78	−0.077	0.006
   21.76	5.219	27.238	3.18	0.323	0.104
   18.56	2.019	4.076	3.46	0.603	0.364
   16.23	−0.311	0.097	2.12	−0.737	0.543
   11.80	−4.741	22.477	2.36	−0.497	0.247
   6.88	−9.661	93.335	2.89	0.033	0.001
   6.38	−10.161	103.246	2.24	−0.617	0.381
   15.83	−0.711	0.506	3.35	0.493	0.243
   2.50	−14.041	197.150	3.00	0.143	0.020
   4.17	−12.371	153.042	2.18	−0.677	0.458
   16.36	−0.181	0.033	3.50	0.643	0.413

   Σ(X − M_X )² = 1279.448    Σ(Y − M_Y )² = 2.899

   

   There is some difference due to rounding decisions, but in both cases, these numbers would be expressed as 0.34.

3. Strong correlations indicate strong linear relations between variables. Because of this, when we have a strong correlation, we have a more useful regression line, resulting in more accurate predictions.

4. According to the hypothesis test for the correlation, this r value, 0.34, fails to reach statistical significance. The critical values for r with 11 degrees of freedom at a p level of 0.05 are −0.553 and 0.553. In this chapter, we learned that the outcome of hypothesis testing is the same for simple linear regression as it is for correlation, so we know we also do not have a statistically significant regression.

1. X	Y	(MY) Mean For Y	Error (Y −MY)	Squared Error (Y −MY)2
   5	6	5.333	0.667	0.445
   6	5	5.333	−0.333	0.111
   4	6	5.333	0.667	0.445
   5	6	5.333	0.667	0.445
   7	4	5.333	−1.333	1.777
   8	5	5.333	−0.333	0.111

   SS_total = Σ(Y − M_Y)² = 3.334

2. X	Y	Regression Equation	Ŷ	Error (Y −Ŷ)	Squared Error
   5	6	Ŷ = 7.846 − 0.431 (5) =	5.691	0.309	0.095
   6	5	Ŷ = 7.846 − 0.431 (6) =	5.260	−0.260	0.068
   4	6	Ŷ = 7.846 − 0.431 (4) =	6.122	−0.122	0.015
   5	6	Ŷ = 7.846 − 0.431 (5) =	5.691	0.309	0.095
   7	4	Ŷ = 7.846 − 0.431 (7) =	4.829	−0.829	0.687
   8	5	Ŷ = 7.846 − 0.431 (8) =	4.398	0.602	0.362

   SS_error = Σ(Y − Ŷ)² = 1.322

3. We have reduced error from 3.334 to 1.322, which is a reduction of 2.012. Now we calculate this reduction as a proportion of the total error:

   

   D-21

This can also be written as:



We have reduced 0.603, or 60.3%, of error using the regression equation as an improvement over the use of the mean as our predictor.

4. r²=(−0.77)(−0.77) = 0.593, which closely matches our calculation of r² above, 0.603. These numbers are slightly different due to rounding decisions.

1. Ŷ = 5.251 + 0.06(40) + 1.105(14) = 23.12

2. Ŷ = 5.251 + 0.06(101) + 1.105(39) = 54.41

3. Ŷ = 5.251 + 0.06(76) + 1.105(20) = 31.91

1. Ŷ = 2.695 + 0.069(X₁) + 0.015(X₂) − 0.072(X₃)

2. Ŷ = 2.695 + 0.069(6) + 0.015(20) − 0.072(4) = 3.12

3. The negative sign in the slope (−0.072) tells us that those with higher levels of admiration for Pamela Anderson tend to have lower GPAs, and those with lower levels of admiration for Pamela Anderson tend to have higher GPAs.

Chapter 15

1. The independent variable is city, a nominal variable. The dependent variable is whether a woman is pretty or not so pretty, an ordinal variable.

2. The independent variable is city, a nominal variable. The dependent variable is beauty, assessed on a scale of 1–10. This is a scale variable.

3. The independent variable is intelligence, likely a scale variable. The dependent variable is beauty, assessed on a scale of 1–10. This is a scale variable.

4. The independent variable is ranking on intelligence, an ordinal variable. The dependent variable is ranking on beauty, also an ordinal variable.

1. We’d choose a hypothesis test from category III. We’d use a nonparametric test because the dependent variable is not scale and would not meet the primary assumption of a normally distributed dependent variable, even with a large sample.

2. We’d choose a test from category II because the independent variable is nominal and the dependent variable is scale. (In fact, we’d use a one-way between-groups ANOVA because there is only one independent variable and it has more than two levels.)

3. We’d choose a hypothesis test from category I because we have a scale independent variable and a scale dependent variable. (If we were assessing the relation between these variables, we’d use the Pearson correlation coefficient. If we wondered whether intelligence predicted beauty, we’d use simple linear regression.)

4. We’d choose a hypothesis test from category III because both the independent and dependent variables are ordinal. We would not meet the assumption of having a normal distribution of the dependent variable, even if we had a large sample.

1. df_x2 = k − 1 = 2 − 1 = 1

2. Observed:

   Clear Blue Skies	Unclear Skies
   59 days	19 days

   Expected:

   Clear Blue Skies	Unclear Skies
   (78)(0.80) = 62.4	(78)(0.20) = 15.6 days

   D-22

3. Category	Observed (O)	Expected (E)	O − E	(O − E)2
   Clear blue skies	59	62.4	−3.4	11.56	0.185
   Unclear skies	19	15.6	3.4	11.56	0.741

   

1. The participants are the lineups. The independent variable is type of lineup (simultaneous, sequential), and the dependent variable is outcome of the lineup (suspect identification, other identification, no identification).

2. Step 1: Population 1 is police lineups like those we observed. Population − is police lineups for which type of lineup and outcome are independent. The comparison distribution is a chi-square distribution. The hypothesis test is a chi-square test for independence because we have two nominal variables. This study meets three of the four assumptions. The two variables are nominal; every participant (lineup) is in only one cell; and there are more than five times as many participants as cells (eight participants and six cells).

   Step 2: Null hypothesis: Lineup outcome is independent of type of lineup.

   Research hypothesis: Lineup outcome depends on type of lineup.

   Step 3: The comparison distribution is a chi-square distribution with − degrees of freedom:

   df_x2 = (k_row − 1)(k_column − 1) = (2 − 1) (3 − 1) = (1)(2) = 2

   Step 4: The cutoff chi-square statistic, based on a p level of 0.05 and − degrees of freedom, is 5.992. (Note: It is helpful to include a drawing of the chi-square distribution with the cutoff.)

   Step 5:

   Observed
   	Suspect Id	Other Id	No Id
   Simultaneous	191	8	120	319
   Sequential	102	20	107	229
   	293	28	227	548

   We can calculate the expected frequencies in one of two ways. First, we can think about it. Out of the total of 548 lineups, 293 led to identification of the suspect, an identification rate of 293/548 = 0.535, or 53.5%. If identification were independent of type of lineup, we would expect the same rate for each type of lineup. For example, for the 319 simultaneous lineups, we would expect: (0.535)(319) = 170.665. For the 229 sequential lineups, we would expect: (0.535)(229) = 122.515. Or we can use the formula; for these same two cells (the column labeled “Suspect ID”), we calculate:

   

   For the column labeled “Other ID”:

   

   For the column labeled “No ID”:

   

   Expected
   	Suspect Id	Other Id	No Id
   Simultaneous	170.665	16.269	132.066	319
   Sequential	122.515	11.679	94.806	229
   	293	28	227	548

   Category	Observed (O)	Expected (E)	O − E	(O − E )2
   Sim; suspect	191	170.665	20.335	413.512	2.423
   Sim; other	8	16.269	−8.269	68.376	4.203
   Sim; no	120	132.066	−12.066	145.588	1.102
   Seq; suspect	102	122.515	−20.515	420.865	3.435
   Seq; other	20	11.679	8.321	69.239	5.929
   Seq; no	107	94.806	12.194	148.694	1.568

   

   Step 6: Reject the null hypothesis. It appears that the outcome of a lineup depends on the type of lineup. In general, simultaneous lineups tend to lead to a higher-than-expected rate of suspect identification, lower-than-expected rates of identification of other members of the lineup, and lower-than-expected rates of no identification at all. (Note: It is helpful to add the test statistic to the drawing that included the cutoff.)

   D-23

3. χ²(1, N = 548) = 18.66, p < 0.05

4. The findings of this study were the opposite of what had been expected by the investigators; the report of results noted that, prior to this study, police departments believed that the sequential lineup led to more accurate identification of suspects. This situation occurs frequently in behavioral research, a reminder of the importance of conducting two-tailed hypothesis tests. (Of course, the fact that this study produced different results doesn’t end the debate. Future researchers should explore why there are different findings in different contexts in an effort to target the best lineup procedures for specific situations.)

5. 

   This is a small-to-medium effect.

6. To create a graph, we must first calculate the conditional proportions by dividing the observed frequency in each cell by the row total. These conditional proportions appear in the table below and are graphed in the figure.

   	Id Suspect	Id Other	No Id
   SIMULTANEOUS	0.599	0.025	0.376
   SEQUENTIAL	0.445	0.087	0.467

   

7. We must first calculate two conditional probabilities: the conditional probability of obtaining a suspect identification in the simultaneous lineups, which is , and the conditional probability of obtaining a suspect identification in the sequential lineups, which is . We then divide 0.599 by 0.445 to obtain the relative likelihood of 1.35. Suspects are 1.35 times more likely to be identified in simultaneous as opposed to sequential lineups.

1. There is an extreme outlier, 139, suggesting that the underlying population distribution might be skewed. Moreover, the sample size is small.

2. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 (we chose to rank this way, but you could do the reverse, from 10 to 1).

3. The outlier was 25 IQ points (139 − 114 = 25) behind the next-highest score of 114. It now is ranked 10th, compared to the next-highest score’s rank of 9th.