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Sagot :
Answer:
The P-value for this test is 0.0023.
Step-by-step explanation:
The null hypothesis is:
[tex]H_{0} = 125[/tex]
The alternate hypotesis is:
[tex]H_{1} > 125[/tex]
The test statistic is:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, [tex]\sigma[/tex] 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 [tex]\sigma = 2[/tex]
A random sample of eight fiber specimens is selected, and the average tensile strength is found to be 127 psi.
This means, respectively, that [tex]n = 8, X = 127[/tex]
Test the hypothesis that the mean tensile strength equals 125 psi versus the alternative that the mean exceeds 125 psi.
This means that [tex]\mu = 125[/tex]
What is the P-value for this test
First we find the test statistic. So
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{127 - 125}{\frac{2}{\sqrt{8}}}[/tex]
[tex]z = 2.83[/tex]
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.
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