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?

in Statistics Answers by

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
To avoid this verification in future, please log in or register.

1 Answer

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 (646k points)

Related questions

Welcome to MathHomeworkAnswers.org, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!
82,946 questions
87,612 answers
1,966 comments
4,297 users