Question

You are interested in estimating the average number of pizza slices a college student eats in one month. In a random sample of 30 college students, you find the average to be 18.924 slices with a sample variance of 120. What is the p-value for a two-sided hypothesis test where the null is that the true population mean is 14 slices per month?

Answer 1

Given:

Null Hypothesis: population mean (μ) = 14

Alternative Hypothesis: population mean (μ) ≠ 14

Sample mean () = 18.924

Sample size (n) = 30

Sample variance (V) = 120

We use t-test since sample size is small and population variance is also unknown

t-statistic (t) = - μ / s/√n

where s =standard deviation = √sample variance = √120 = 10.95445

and s/√n = standard error of the mean = 10.95445/√30 = 10.95445/5.477226 = 2

              now, t = - s/√n

                    = 18.924 – 14 / 2

                      = 4.924 / 2

= 2.462

Now, we find p-value in excel by following formula

= tdist(x,deg_freedom,tails)

Where x = t = 2.462

             deg_freedom = n-1 = 29

             tails = 2

= tdist(2.462,29,2) = 0.02001 = p-value