Question

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 25 psi. Use ? = 0.05.
(c) Discuss why a one-sided alternative was chosen in part (a).

Answer 1

Answer:

a) $z=\frac{127-125}{\frac{2}{\sqrt{8}}}=2.828$        

Since is a one-side upper test the p value would be:        

$p_v =P(z>2.828)=1-P(z<2.828) =1-0.9977=0.0023$    

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v<\alpha$ so we can conclude that we reject the null hypothesis, and we can conclude that the true mean is higher than 125

b) We select a one sided interval since the interest is in analyze if the true mean is higher than 125, and then the alternative hypothesis needs to be the alternative hypothesis.

Step-by-step explanation:

Data given and notation        

$\bar X=127$ represent the mean for the sample    

$\sigma=2$ represent the standard deviation for the population        

$n=8$ sample size        

$\mu_o =125$ represent the value that we want to test      

$\alpha$ represent the significance level for the hypothesis test.      

z would represent the statistic (variable of interest)        

$p_v$ represent the p value for the test (variable of interest)    

Part a

State the null and alternative hypotheses.        

We need to conduct a hypothesis in order to determine if the true mean is higher than 125:      

Null hypothesis:$\mu = 125$        

Alternative hypothesis:$\mu > 125$        

We know the population deviation, so for this case is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:        

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)        

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".    

Calculate the statistic        

We can replace in formula (1) the info given like this:        

$z=\frac{127-125}{\frac{2}{\sqrt{8}}}=2.828$        

Calculate the P-value        

Since is a one-side upper test the p value would be:        

$p_v =P(z>2.828)=1-P(z<2.828) =1-0.9977=0.0023$    

Conclusion        

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v<\alpha$ so we can conclude that we reject the null hypothesis, and we can conclude that the true mean is higher than 125

Part b

We select a one sided interval since the interest is in analyze if the true mean is higher than 125, and then the alternative hypothesis needs to be the alternative hypothesis.