A simple random sample of 50 stainless steel metal screws is obtained. The screws have a mean length of 0.73 inches.  Assume the population standard deviation is 0.012 inches and the confidence level is a=0.05. Find the p-value you would use to test the claim that the screws have a mean length equal to 0.75 inches as indicated on the package label. I'm struggling to find the p-value, could someone help?

We need to find the Z score for the difference between the sample mean and the expected population mean.

The difference is 0.02 inches, which is 0.02/0.012=20/12=5/3=1.667 standard deviations, and this is the Z score. From tables the Z score corresponds to a probability of 0.4522 of the left side of the normal distribution so it leaves a tail of 0.5000-0.4522=0.0478. This is the p-value associated with the screw length of 0.73 inches. If the significance level is 0.05, the confidence level is 95%, but that extends across the whole distribution. Half the distribution is to the left of the mean, which is 95/2=47.5% or 0.475, leaving 0.500-0.475=0.025 in the left tail. The p-value is greater than this, which means we fail to reject the claim that the screws have a mean length of 0.75 inches. And so the p-value supports the claim within the significance level specified.

by Top Rated User (680k points)