Question

(Still focusing on the study that found Americans spent an average of 5.1 hours on leisure per day). Without any prior expectations, you want to see if aircraft pilots spend different amounts of time per day on leisure than the national average. You find in a sample of 26 pilots, the average amount of leisure time per day was 290 minutes with a standard deviation of 50 minutes. Use the critical-value approach for a t-test (where α = .001) to determine if there is a significant difference in amount of leisure time per day between the national average and pilots. 5. List your degrees of freedom 6. List your t test statistic 7. List your critical t value(s) 8. Is the difference between the national average and working parents statistically significant? (Yes/No) please answer all questions.

Answer 1

(Still focusing on the study that found Americans spent an average of 5.1 hours on leisure per day). Without any prior expectations, you want to see if aircraft pilots spend different amounts of time per day on leisure than the national average. You find in a sample of 26 pilots, the average amount of leisure time per day was 290 minutes with a standard deviation of 50 minutes. Use the critical-value approach for a t-test (where α = .001) to determine if there is a significant difference in amount of leisure time per day between the national average and pilots. 5. List your degrees of freedom 6. List your t test statistic 7. List your critical t value(s) 8. Is the difference between the national average and working parents statistically significant? (Yes/No) please answer all questions

We are going to test for the population mean. Since the population SD is not given but we assume that the data is normally distributed so we will use t-test. We are testing if the population mean is different from the national average. So it is two tailed test.

= 5.1 hours (5.1*60 = 306 mins)

306 mins

Test Stat =

Where the null mean =

= 0.001

Critical value =   .............using t-dist tables

Reject the null hypothesis if |Test Stat| > C.V.

Sample Mean	290
Sample SD	50
n	26
SE	9.80581
Null mean	306
Test Stat	-1.6317
df	25
Critical value	3.45019
Since |Test Stat| < C.V.
Decision	Do not reject the null hypothesis
Conclusion	There is insufficient evidence to conclude that there is difference between the national average and working parents.

The result is not significant .