company wants to test the claim that their batteries last more than 40 hours. random sample of 15 batteries yielded a mean of 44.9 hours, with a standard deviation of 8.9 hours. alpha 0.05.

Question

i calculated the t-test and it's 2.13.

Can someone explain the following and how we got 0.026?

p-value = P(tdf=14 > 2.13) = 0.026

because p=0.026 <alpha=0.05 reject H0

Answer 1

Basics. We are assuming normal distribution of lifetimes for the batteries. We do not know the population variance. We have a sample X which has a mean of 44.9 and a variance of 8.9² * (n-1/n) or 77.6258 (this is an assumption as the question is incomplete and does not indicate whether the standard deviation given is a sample standard deviation or a population standard deviation, we assume a sample (estimate) of the standard deviation) Under these assumptions the sample standard deviation s is 8.9

Based on standard formula for standard error, (as we are testing the mean of a sample) we get a standard error (se) as follows

standard deviation of mean (se) = s / √15. or 2.297970

test mean (given) = 40

Conducting the hypothesis test we use a student's - t distribution (population parameters are estimated)

t = (x-mean)/se or (44.9-40)/2.297970 which is 2.132316 with a d.f. of n-1 or 14

Using on line assets to calculate P(t > 2.1323) d.f. = 14: P = 0.025586 (rounded 0.0256)

( see https://stattrek.com/online-calculator/t-distribution and remember to take the t<x results and subtract from one to get t>x)

"because p=0.026 <alpha=0.05 reject H0". should be self explanatory

So: you see, your t value (rounded to 2 decimal) is correct. A p value of .0256 agrees with what you got to 2 decimal places. You are correct. The generation of this p value using the t distribution was done by standard algorithms (available on line) for a d.f. of 14 an a t value of 2.1323