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AP Statistics 7.10 Skills Focus: Selecting, Implementing, and Communicating Inference Procedures- FRQs - Exam Style Questions

Question

Stefan, a psychologist, conducted a study to investigate the effect of time of day on reading comprehension in children. One hundred children volunteered, with their parents’ consent, to participate in the study. Fifty of the children were randomly assigned to read a story at \(9\) a.m. and then answer \(25\) questions about it. The remaining \(50\) children were assigned to read the same story at \(3\) p.m. and answer the same \(25\) questions. The reading comprehension for each child was measured by a reading score, which was determined by the number of questions that were answered correctly about the story. Stefan is interested in comparing the mean reading scores for the two times of day. Table 1 shows the results of Stefan’s study.
Table 1: Summary Statistics of Reading Scores
 \(n\)MeanStandard Deviation
\(9\) a.m.\(50\)\(15.2\)\(4.12\)
\(3\) p.m.\(50\)\(17.9\)\(4.43\)
Stefan found the conditions for inference were met and conducted a two-sample \(t\)-test for the difference in two population means. Let \(\mu_{AM}\) represent the mean reading score for all children, similar to those in the study, who would read the story at \(9\) a.m. Let \(\mu_{PM}\) represent the mean reading score for all children, similar to those in the study, who would read the story at \(3\) p.m. Stefan’s hypotheses are as shown.
\[H_0: \mu_{AM} = \mu_{PM}\]
\[H_a: \mu_{AM} \neq \mu_{PM}\]
A. The \(p\)-value for Stefan’s hypothesis test was \(0.002\). State an appropriate conclusion, at the \(5\) percent significance level, for Stefan’s test in the context of the investigation. Justify your answer.
B. Explain why it was appropriate for Stefan to conduct a two-sample \(t\)-test for the difference in two population means instead of a paired \(t\)-test for the population mean difference.
C. Researchers are usually interested in the practical importance of their results as well as the statistical significance of the hypothesis test. The practical importance of the results indicates whether the observed results are meaningful in real life. For example, in an investigation of the heights of two groups of students, a difference in the two group means of \(3.8\) inches is much more meaningful, or has more practical importance, than a difference in the two group means of only \(0.2\) inches.
One indicator of practical importance is effect size. A common method for measuring effect size for the difference in two group means is Cohen’s \(d\) coefficient. Cohen’s \(d\) coefficient compares the absolute value of the difference in the means of the two groups to the pooled variability of the observed data values from the two groups.
Cohen’s \(d\) coefficient can be calculated using \(d = \frac{|\overline{x}_1 – \overline{x}_2|}{s_p}\), where \(s_p\) represents the pooled standard deviation, \(\overline{x}_1\) represents the sample mean for the first group, and \(\overline{x}_2\) represents the sample mean for the second group. When the sizes of the groups are equal, \(s_p\) is calculated as \(s_p = \sqrt{\frac{s_1^2 + s_2^2}{2}}\), where \(s_1\) represents the sample standard deviation for the first group and \(s_2\) represents the sample standard deviation for the second group.
Consider the summary statistics from Stefan’s study in Table 1.
i. Calculate Cohen’s \(d\) coefficient for Stefan’s study. Show your work.
ii. Higher values of Cohen’s \(d\) indicate greater practical importance and lower values of Cohen’s \(d\) indicate less practical importance. Typically, we use the intervals listed in Table 2 to help interpret practical importance.
Table 2: Guidelines for Interpreting Cohen’s \(d\) Coefficient
Cohen’s \(d\) CoefficientPractical Importance
\(0 \leq d \leq 0.20\)Not very meaningful in real life
\(0.20 < d < 0.80\)Somewhat meaningful in real life
\(d \geq 0.80\)Very meaningful in real life
Based on your answer to part C (i) and the information in Tables 1 and 2, describe the practical importance of Stefan’s results, in context.
D. Suppose the results of Stefan’s study, summarized in Table 1, instead had a standard deviation for the \(9\) a.m. reading scores, \(s_1\), and a standard deviation for the \(3\) p.m. reading scores, \(s_2\), that were both greater than \(4.43\). Assume the group sample sizes and the means are not changed.
i. Would the Cohen’s \(d\) coefficient in this new situation be smaller than, larger than, or the same as the Cohen’s \(d\) coefficient calculated in part C (i)? Explain your answer.
ii. Does the Cohen’s \(d\) coefficient described in part D (i) indicate that Stefan’s observed difference in the means in the new situation would have more practical importance than, less practical importance than, or the same practical importance as what was originally determined in part C (ii)? Explain your answer.

Most-appropriate topic codes (CED):

TOPIC 7.9: Carrying Out a Test for the Difference of Two Population Means — part (A)
TOPIC 3.2: Introduction to Planning a Study — part (B)
TOPIC 7.10: Skills Focus: Selecting, Implementing, and Communicating Inference Procedures — part (C i)
TOPIC 7.10: Skills Focus: Selecting, Implementing, and Communicating Inference Procedures — part (C ii)
TOPIC 7.10: Skills Focus: Selecting, Implementing, and Communicating Inference Procedures — part (D)
▶️ Answer/Explanation
Detailed solution

A
The \(p\)-value of \(0.002\) is less than the significance level of \(\alpha = 0.05\), so we reject the null hypothesis. There is convincing statistical evidence that there is a difference between the mean reading score for all children who would read the story at \(9\) a.m. and the mean reading score for all children who would read the story at \(3\) p.m.

B
It was appropriate to use a two-sample \(t\)-test because the two groups are independent. The children were randomly assigned to the two groups, and there is no natural pairing between the children in the \(9\) a.m. group and the children in the \(3\) p.m. group.

C i
First, calculate the pooled standard deviation:
\[s_p = \sqrt{\frac{s_1^2 + s_2^2}{2}} = \sqrt{\frac{(4.12)^2 + (4.43)^2}{2}} = \sqrt{\frac{16.9744 + 19.6249}{2}} = \sqrt{\frac{36.5993}{2}} = \sqrt{18.29965} \approx 4.278\]
Now calculate Cohen’s \(d\):
\[d = \frac{|\overline{x}_1 – \overline{x}_2|}{s_p} = \frac{|15.2 – 17.9|}{4.278} = \frac{2.7}{4.278} \approx 0.631\]
\(\boxed{0.63}\)

C ii
Since Cohen’s \(d \approx 0.63\) falls in the interval \(0.20 < d < 0.80\), Stefan’s results are somewhat meaningful in real life. This indicates that the observed difference in mean reading scores between the two times of day has some practical importance.

D i
The Cohen’s \(d\) coefficient would be smaller. With larger standard deviations, the pooled standard deviation \(s_p\) would increase. Since the numerator \(|\overline{x}_1 – \overline{x}_2| = 2.7\) remains unchanged, a larger denominator results in a smaller value of \(d\).

D ii
The smaller Cohen’s \(d\) value would indicate less practical importance than what was originally determined in part C (ii). A smaller effect size suggests that the observed difference in means is less meaningful in real life relative to the increased variability in the data.

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