Question

10. A random sample of 12 college students were surveyed on their social activities. Students were asked to state how many times during the previous year they had gone to the movies or sporting events.

Student	1	2	3	4	5	6	7	8	9	10	11	12
Movies	12	2	13	17	14	9	24	6	26	22	10	16
Sports	7	4	13	4	9	15	18	4	11	9	5	2
Difference (movies – sports)

1. Once you have completed the table above and entered the data in your calculator, find the mean and standard deviation for the difference in student attendance at movies and sports. Use 3 decimal places. (5pts)
2. What is the standard error for this sample distribution? 3 decimal places (5pts)
3. To compare the difference in mean movie attendance with the mean sports attendance for these students, what statkey test function should we use? Provide the null and alternative hypothesis as well. (5pts)
4. Perform the test on Statkey using function you chose in part c. Then report the test statistic and p-value. 3 decimal places (5pts)
5. At = .05, what is your conclusion of this test to determine whether there is any difference in mean attendance between these two social activities for these students?

Answer 1

The hypothesis being tested is:

H0: µd = 0

Ha: µd ≠ 0

Movies	Sports	Difference
12	7	5
2	4	-2
13	13	0
17	4	13
14	9	5
9	15	-6
24	18	6
6	4	2
26	11	15
22	9	13
10	5	5
16	2	14
	14.250	mean Movies
	8.417	mean Sports
	5.833	mean difference (Movies - Sports)
	6.780	std. dev.
	1.957	std. error
	12	n
	11	df
	2.980	t
	.0125	p-value (two-tailed)

The p-value is 0.0125.

Since the p-value (0.0125) is less than the significance level (0.05), we can reject the null hypothesis.

Therefore, we can conclude that there is a difference in mean attendance between these two social activities for these students.