91Ó°ÊÓ

Among 362 McDonald’s orders,33 orders were not accurate.

The null hypothesis is written as follows:

The proportion of inaccurate orders is equal to 10%.

$H_0:p=0.10$

The alternative hypothesis is written as follows:

The proportion of inaccurate orders is not equal to 10%.

$H_1:p≠0.10$

The test is two-tailed.

The sample proportion of inaccurate orders is equal to the following:

$$\begin{gathered} \hat{p}=\frac{\text{Number}\text{of}\text{inaccurate}\text{orders}}{\text{Total}\text{number}\text{of}\text{orders}} \\ =\frac{33}{362} \\ =0.091 \end{gathered}$$

The population proportion of inaccurate orders is equal to p=0.10.

The sample size (n) is equal to 362.

The value of the test statistic is computed below:

$$\begin{gathered} z=\frac{\hat{p}-p}{\sqrt{\frac{pq}{n}}} \\ =\frac{0.0912-0.10}{\sqrt{\frac{0.10\left(1-0.10\right)}{362}}} \\ =-0.56 \end{gathered}$$

Thus, z=-0.56.

Referring to the standard normal distribution table, the critical value of z at $\mathrm{Î\pm }=0.05$ for a two-tailed test is equal to 1.96.

Referring to the standard normal distribution table, the p-value for the test statistic value of -0.56 is equal to 0.5754.

Since the p-value is greater than 0.05, the null hypothesis is failed to reject.

There is not enough evidence to reject the claim that the proportion of inaccurate orders is equal to 0.10.

As the percentage of inaccurate ordersis 10%, the accuracy rate appears unacceptable,and McDonald’s shouldlowerthe rate.