Question

The average gasoline price of one of the major oil companies has been hovering around $2.20 per gallon. Because of cost reduction measures, it is announced that there will be a significant reduction in the average price over the next month. In order to test this belief, we wait one month, then randomly select a sample of 36 of the company's gas stations. We find that the average price for the stations in the sample was $2.15. The standard deviation of the prices for the selected gas stations is $.10. Given that the test statistic for this sample is t = –3, determine the p-value.

Answer 1

Answer: 0.0025

Step-by-step explanation:

Let $\mu$ be the population mean for the gasoline price of one of the major oil companies .

As per given , we hav

$H_0: \mu=2.20\\\\ H_a: \mu$

Since the alternative hypothesis is left-tailed , so the test is a left-tailed test.

Given : n= 36

Degree of freedom : 35    (df= n-1)

The given test statistic : t=-3

Using the p-value table or p-value calculator for a student's t -value left-test having degree of freedom 35, we have

P-value : $P(t$

Thus , the required p-value = 0.0025