Question

1.

A group of 73 college students from a certain liberal arts college were randomly sampled and asked about the number of alcoholic drinks they have in a typical week. The purpose of this study was to compare the drinking habits of the students at the college to the drinking habits of college students in general. In particular, the dean of students, who initiated this study, would like to check whether the mean number of alcoholic drinks that students at his college in a typical week differs from the mean of U.S. college students in general, which is estimated to be 4.73.

The group of 73 students in the study reported an average of 4.23 drinks per with a standard deviation of 3.75 drinks.

Find the p-value for the hypothesis test.  

The p-value should be rounded to 4-decimal places.

2.

Commute times in the U.S. are heavily skewed to the right. We select a random sample of 65 people from the 2000 U.S. Census who reported a non-zero commute time.

In this sample the mean commute time is 28.3 minutes with a standard deviation of 18.6 minutes. Can we conclude from this data that the mean commute time in the U.S. is less than half an hour? Conduct a hypothesis test at the 5% level of significance.

What is the p-value for this hypothesis test?  

(Your answer should be rounded to 4 decimal places.)

3.

Find the p-value for the hypothesis test. A random sample of size 55 is taken. The sample has a mean of 422 and a standard deviation of 85.

H0: µ = 400

Ha: µ > 400

The p-value for the hypothesis test is  .

Your answer should be rounded to 4 decimal places.

Answer 1

HUU NO. Date: 1 solution) Given that x = 4.23 6 = 3.75 ha 73 l- 4.73 Hoiu=4.73 Hall +4073 This is two-tailed test. Test statistic, 6/ h 4.23-4.73 3.75 / V73 z = -1.14 The test statistic is -1.14 I DaValue = 2xP(Z <-1-14) - 2x0.1271 30-2542 P-value = 0.2542

Dete: 1 x=28.3 6=18.6 n=65 u- 30 L= 0.05 Hotl=30 Halu <30 this is Left a tailed test Test statistic, 61 Jn 28. 3- 30 18-67 165 2 -0.74 The test statistic is no. 74 p-value 2PC25-0.74) = 0.2296 P-value = 0.2296 0.2296 do 05 P-value 72 fail to reject Ho.

X= 422 6=85 n= 5s l=400 Ho: el = 400 Hai l> 400 This is right -tailed test Test statistic, z = 422-400 85/ 5 z = 1.92 The test statistic is 1.92 p-value = l-P(Z > 1.92) - -P(Z <lic2 = -0.9726 = 0.0274 P-value = 0.0274