Answer to Question #243500 in Statistics and Probability for Goldie

Question

A researcher claims that the monthly consumption of coffee per person is more than 19 cups. In a sample of 60 randomly selected people, the mean monthly consumption was 20. The standard deviation of the sample was 4 cups. Find the p-value of the test. On the basis of the p-value, is the researchers claim valid at 0.01 level of significance?

Answer 1

$H_0: \mu≤ 19 \\ H_1: \mu>19 \\ \bar{x} = 20 \\ n=60 \\ s=4$

Test-statistic:

$t = \frac{\bar{x} -\mu}{ s / \sqrt{n}} \\ t = \frac{20-19}{4 / \sqrt{60}} =1.937$

The EXCEL formula to find the p-value for a two-tailed t-test and df=60-1=59 is

=tdist(1.937, 59, 1)

p-value = 0.0287

p>0.01

At significance level of 1%, the researcher can NOT claim that the monthly consumption of coffee per person is more than 19 cups.