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The tensile strength of a fiber used in manufacturing cloth is of interest to the purchaser. Previous experience indicates that the standard deviation of tensile strength is 2 psi. A random sample of eight fiber specimens is selected, and the average tensile strength is found to be 127 psi. (a) Test the hypothesis that the mean tensile strength equals 125 psi versus the alternative that the mean exceeds 125 psi. Use (b) What is the P-value for this test

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Answer:

The P-value for this test is 0.0023.

Explanation:

The null hypothesis is:


H_(0) = 125

The alternate hypotesis is:


H_(1) > 125

The test statistic is:


z = (X - \mu)/((\sigma)/(√(n)))

In which X is the sample mean,
\mu is the value tested at the null hypothesis,
\sigma is the standard deviation and n is the size of the sample.

Previous experience indicates that the standard deviation of tensile strength is 2 psi.

This means that
\sigma = 2

A random sample of eight fiber specimens is selected, and the average tensile strength is found to be 127 psi.

This means, respectively, that
n = 8, X = 127

Test the hypothesis that the mean tensile strength equals 125 psi versus the alternative that the mean exceeds 125 psi.

This means that
\mu = 125

What is the P-value for this test

First we find the test statistic. So


z = (X - \mu)/((\sigma)/(√(n)))


z = (127 - 125)/((2)/(√(8)))


z = 2.83

The pvalue is 1 subtracted by the pvalue of Z = 2.83.

Looking at the z-table, we have that:

Z = 2.83 has a pvalue of 0.9977

1 - 0.9977 = 0.0023

The P-value for this test is 0.0023.

User Charlie Elverson
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