Answer to Question #342461 in Statistics and Probability for Ari

Question

A large manufacturing company claims that more than 88% of its customers are fully satisfied with the company’s products. Test this claim at the 5% significance level, if a survey found that 360 out of a random sample of 400 of the company’s customers said that they were fully satisfied with the company’s products.

Answer 1

The following null and alternative hypotheses for the population proportion needs to be tested:

$H_0:p\le0.88$

$H_a:p>0.88$

This corresponds to a right-tailed test, for which a z-test for one population proportion 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 $z_c =1.6469.$

The rejection region for this right-tailed test is $R = \{z: z>1.6469\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}=\dfrac{\dfrac{360}{400}-0.88}{\sqrt{\dfrac{0.88(1-0.88)}{400}}}=1.2309$$

Since it is observed that $z =1.2309<1.6469=z_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value is $p=P(Z>1.2309)= 0.10918,$ and since $p = 0.10918> 0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population proportion $p$ is greater than 0.88, at the $\alpha = 0.05$ significance level.