Question

A travel association claims that the mean daily food cost for two adults traveling together on vacation in San Francisco is $110. A random sample of 14 groups of adults is used to conduct a hypothesis test to show that the mean daily food is greater than 110. Using the mean and sample standard deviation from this sample, the test statistic was calculated to be 1.76. The corresponding p-value of this test is: a. 0.0784 b. 0.05 < p-value < 0.10 c. 0.10 < p-value < 0.20 d. 0.0392

Answer 1

b. 0.05 < p-value < 0.10

Explanation:

We have that the sample size is 14 (n = 14) and the statistical test is 1.76 (t = 1.76)

claim to test, mean daily food is greater than 110

m> 110

Thus,

p-value at t = 1.76 and degree of freedom = n -1 = 14-1 df = 13

p-value = 0.0509, attached excel table

therefore p-value is in the range of 0.05 and 0.10