Westonci.ca is the premier destination for reliable answers to your questions, brought to you by a community of experts. Discover the answers you need from a community of experts ready to help you with their knowledge and experience in various fields. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.
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.
Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.
