Chapter 13, Problem 31SE

Sports and Extraversion (Example 5) Are students who participate in sports more extraverted than those who do not? A random sample of students at a small university were asked to indicate whether they participated in sports (yes or no) and to rate their level of extraversion. Extraverts are outgoing, are talkative, and don’t mind being the center of attention. Students were asked whether they agreed with the statement that they were extraverts, using a scale of 1 to 5 with 1 meaning “strongly disagree” and 5 meaning “strongly agree.” There were 51 students who participated in sports and 64 who did not.

The mean extraversion score for those who participated in

sports was 3.618, and the mean for those who did not participate was 3.172, a difference of 0.446 point. To determine whether the mean difference was significant, we performed a randomization test to test whether the mean extraversion level was greater for athletic students.

a. The histogram shows the results of 1000 randomizations of the data. In each randomization, we found the mean difference between two groups that were randomly selected from the combined group of data: the combined data for the sporty and nonsporty. Note that, just as you would expect under the null hypothesis, the distribution is centered at about 0. The red line shows the observed sample mean difference in extraversion for the sporty minus the nonsporty. From the graph, does it look like the observed mean difference is unusual for this data set? Explain.

b. The software output gives us the probability of having an observed difference of 0.446 or more. (See the column labeled “Proportion=7 Observed”). In other words, it gives us the right tail area, which is the $p$ -value for a one-sided alternative that the mean extraversion score is higher for athletes than for nonathletes. State the $p$ -value.

c. Using a significance level of 0.05, can we reject the null hypothesis that the means are equal?

d. If you did not have the computer output, explain how you would use the histogram to get an approximate $p$ -value.