Explore Westonci.ca, the leading Q&A site where experts provide accurate and helpful answers to all your questions. Experience the ease of finding accurate answers to your questions from a knowledgeable community of professionals. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
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.
Thank you for your visit. We are dedicated to helping you find the information you need, whenever you need it. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.