Help with P-Value

Question

Hi, I am currently working on a project on hypothesis testing. This is the data set of heights of students in inches...

57, 59, 60, 62, 63, 64, 65, 65, 66, 67, 68, 68, 68, 68, 69, 69, 70, 71, 71, 72, 72, 72, 74, 75 .

I am testing whether or not the average height of students is 66.5 inches. H0: μ = 66.5 ; Ha: μ ≠ 66.5

The question asks us to find the p-value, explain what it means in terms of the data set, and sketch a graph. I am having a lot of trouble with understanding the p-value if someone could help me. Thank you!

Answer 1

There is no directionality, as we are testing if the mean is not equal to 66.5, so this is a 2-sided test..

The p-value tells us the probability that we would have as great or greater a value of |x-bar - 66.5|/(s/√n) as with the existing data under the null hypothesis that µ = 66.5 and heights are normally distributed.

Since σ is not given, we calculate s from the data, so |x-bar - 66.5|/(s/√n) will have the t distribution. As n =24, we have 23 df.

x-bar = 67 7/24 =67.2917

s = 4.7133

n=24

(67.2917-66.5)/(4.7133/√24) = .8228

P(|t| >.8228 = in Excel, T.DIST.2T(.8228,23) = .4190

As Excel does not do a 1 sample t test against the mean, you set up a second variable with two values, both equal to the hypothesized mean of 66.5, then do a 2-sample t-test with unequal variances.

If you are just sketching this by hand, draw a bell shape centered at 66.5 and show half of the p-value as the probability to the right of 67.2917.

If you wanted to have Excel graph it for you, let the x-value range from 63 to 70, and let the y-value equal

T.DIST((A18-66.5)/(4.7133/√24),23,0) (a18 is a sample x-value) Then, label half the p-value for x > 67.29.