Question

1. a supplier submits lots of fabric to a textile manufacturer. the customer wants to know if the lot average breaking strength exceeds 200 psi. if so, she wants to accept the lot. past experience indicates that a reasonable value for the variance of breaking strength is 100(psi)². four specimens are randomly selected, and the average breaking strength observed is y=214 psi. is the customer accepting this lot at alpha=0.05?

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=200$

$H_1:\mu>200$

This corresponds to a right-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

Based on the information provided, the significance level is $\alpha=0.05,$ and the critical value for a right-tailed test is $z_c=1.6449.$

The rejection region for this right-tailed test is $R=\{z: z>1.6449\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}$$

$$=\dfrac{214-200}{10/\sqrt{4}}=2.8$$

Since it is observed that $z=2.8>1.6449=z_c,$ it is then concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$  is greater than 200, at the $\alpha=0.05$ significance level.

Using the P-value approach: The p-value is $p=P(z>2.8)=0.002555,$ and since $p=0.002555<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is greater than 200, at the $\alpha=0.05$ significance level.

 Therefore, there is enough evidence to claim that the customer is accepting this lot at $\alpha=0.05.$