1. A facilities manager at a university reads in a research report that the mean amount of time spent in the shower by an adult is 5 minutes. He decides to collect data to see if the mean amount of time that college students spend in the shower is significantly different from 5 minutes. In a sample of 15 students, he found the average time was 4.29 minutes and the standard deviation was 0.75 minutes. Using this sample information, conduct the appropriate hypothesis test at the 0.01 level of significance. Assume normality.

a) What are the appropriate null and alternative hypotheses?

H0: ? = 5 versus Ha: ? ? 5H0: x = 5 versus Ha: x ? 5 H0: ? = 5 versus Ha: ? < 5H0: ? = 5 versus Ha: ? > 5

b) What is the test statistic? Give your answer to four decimal places.

c) What is the P-value for the test? Give your answer to four decimal places.

d) What is the appropriate conclusion?

2.

Reject the claim that the mean time is 5 minutes because the P-value is smaller than 0.01.Fail to reject the claim that the mean time is 5 minutes because the P-value is smaller than 0.01. Reject the claim that the mean time is 5 minutes because the P-value is larger than 0.01.Fail to reject the claim that the mean time is 5 minutes because the P-value is larger than 0.01.

A research report claims that 20% of all individuals use Firefox to browse the web. A software company is trying to determine if the proportion of their users who use Firefox is significantly different from 0.2. In a sample of 200 of their users, 26 users stated that they used Firefox. Using this data, conduct the appropriate hypothesis test using a 0.01 level of significance.

a) What are the appropriate hypotheses?

H0: p = 0.2 versus Ha: p > 0.2H0: p = 0.2 versus Ha: p < 0.2 H0: p = 0.2 versus Ha: p ? 0.2H0: ? = 0.2 versus Ha: ? > 0.2

b) What is the test statistic? Give your answer to four decimal places.

c) What is the P-value for the test? Give your answer to four decimal places.

d) What is the appropriate conclusion?

Conclude that the Firefox proportion is not 0.2 because the P-value is smaller than 0.01.Conclude that the Firefox proportion is not 0.2 because the P-value is larger than 0.01. Fail to reject the claim that the Firefox proportion is 0.2 because the P-value is larger than 0.01.Fail to reject the claim that the Firefox proportion is 0.2 because the P-value is smaller than 0.01.