Answer to Question #282532 in Statistics and Probability for Ufgigg

Question

Rick Douglas, the new manager of Food Barn, is

interested in the percentage of customers who are

totally satisfied with the store. The previous manager

had 86 percent of the customers totally satisfied, and

Rick claims the same is true today. Rick sampled 187

customers and found 157 were totally satisfied. At the

1 percent significance level, is there evidence that

Rick's claim is valid?

Answer 1

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

$H_0:p=0.86$

$H_1:p\not=0.86$

This corresponds to a two-tailed test, for which a z-test for one population proportion will be used.

Based on the information provided, the significance level is $\alpha = 0.01,$ and the critical value for a two-tailed test is $z_c = 2.5758.$

The rejection region for this two-tailed test is $R = \{z: |z| > 2.5758\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}=\dfrac{157/187-0.86}{\sqrt{\dfrac{0.86(1-0.86)}{187}}}$$

$$=-0.80506$$

Since it is observed that $|z| = 0.80506 \le 2.5758=z_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value is $p = 2P(Z<-0.80506)=0.420785,$ and since $p = 0.420785>0.01=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population proportion $p$ is different than $0.86,$ at the $\alpha = 0.01$ significance level.