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In: Statistics and Probability

Test the following claim. Identify the null​ hypothesis, alternative​ hypothesis, test​ statistic, P-value, conclusion about the...

Test the following claim. Identify the null​ hypothesis, alternative​ hypothesis, test​ statistic, P-value, conclusion about the null​ hypothesis, and final conclusion that addresses the original claim. A simple random sample of 60 screws supplied by a certain manufacturer is​ obtained, and the length of each screw is measured. The sample mean is found to be 0.622 in. Assume that the standard deviation of all such lengths is 0.016 ​in, and use a 0.01 significance level to test the claim that the screws have a mean length equal to 5 divided by 85/8 in​ (or 0.625 ​in), as indicated on the package labels. Do the screw lengths appear to be consistent with the package​ label?

Solutions

Expert Solution

2 Tailed Z test, Single Mean

Given: = 0.625 inches, = 0.622 inches meters, = 0.016 meters, n = 60, = 0.01

The Hypothesis:

H0: = 0.625: The mean length of the screws is equal to 0.625 inches meters. (Claim)

Ha: 0.625: The mean length of the screws is not equal to 0.625 inches.

This is a 2 tailed test

The Test Statistic: Since the population standard deviation is known and n > 30, we use the z test.

The test statistic is given by the equation:

t observed = -1.45

The p Value: The p value (2 tailed) for z = -1.45 is; p value = 0.1470

The Decision Rule: If P value is < , Then Reject H0.

The Decision: Since P value (0.1470) is > (0.01) , We Fail to Reject H0.

The Conclusion: Fail to reject H0.There is not sufficient evidence at the 99% significance level to warrant rejection of the claim that the mean length of the screws is equal to 0.625 inches

orchestra answered 2 years ago
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