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In testing the accuracy of drive-through orders, for Burger King, 264 orders were accurate, and 54 were not accurate, while for McDonald鈥檚, 329 orders were accurate and 33 were not accurate.

The level of significance to test the hypothesis is 0.05.

Null hypothesis:Burger King and McDonald鈥檚 have the same accuracy rates.

$H_0:p_1=p_2$

Alternate Hypothesis:Burger King and McDonald鈥檚 do not have the same accuracy rates.

$H_1:p_1鈮�p_2$

The sample size $\left(n_1\right)$ is computed below:

$$\begin{gathered} n_1=264+54 \\ =318 \end{gathered}$$

The sample size $\left(n_2\right)$ is computed below:

$$\begin{gathered} n_2=329+33 \\ =362 \end{gathered}$$

Assume that$x_1$ and $x_2$ are the number of accurate orders for Burger King and McDonald鈥檚 respectively.

Let${\hat{p}}_1$ be the sample accuracy rate of Burger King.

Thus,

$$\begin{gathered} {\hat{p}}_1=\frac{x_1}{n_1} \\ =\frac{264}{318} \\ =0.83 \end{gathered}$$

$$\begin{gathered} {\hat{q}}_1=1-{\hat{p}}_1 \\ =0.17 \end{gathered}$$

Let ${\hat{p}}_2$ be the sample accuracy rate of McDonald鈥檚.

Thus,

$$\begin{gathered} {\hat{p}}_2=\frac{x_2}{n_2} \\ =\frac{329}{362} \\ =0.91 \end{gathered}$$

$$\begin{gathered} {\hat{q}}_2=1-{\hat{p}}_2 \\ =0.09 \end{gathered}$$

The value of the pooled sample proportion is equal to:

$$\begin{gathered} \stackrel{炉}{p}=\frac{x_1+x_2}{n_1+n_2} \\ =\frac{264+329}{318+362} \\ =0.872 \end{gathered}$$

Hence,

$$\begin{gathered} \stackrel{炉}{q}=1-\stackrel{炉}{p} \\ =1-0.872 \\ =0.128 \end{gathered}$$

The test statistic is equal to:

$$\begin{gathered} z=\frac{\left({\hat{p}}_1-{\hat{p}}_2\right)-\left(p_1-p_2\right)}{\sqrt{\frac{\stackrel{炉}{p}\stackrel{炉}{q}}{n_1}+\frac{\stackrel{炉}{p}\stackrel{炉}{q}}{n_2}}} \\ =\frac{\left(0.83-0.91\right)-0}{\sqrt{\frac{\left(0.872\right)\left(0.128\right)}{318}+\frac{\left(0.872\right)\left(0.128\right)}{362}}} \\ =-3.064 \end{gathered}$$

Thus, z=-3.064.

Referring to the standard normal distribution table, the critical values of z corresponding to $伪=0.05$for a two-tailed test are equal to -1.96 and 1.96.

Referring to the standard normal distribution table, the corresponding p-value is equal to 0.0022.

Here, the value of the test statistic does not lie between the two critical values.

Therefore, reject the null hypothesis under 0.05 significance level.

a.

There is sufficient evidence to reject the claim thatBurger King and McDonald鈥檚 have the same accuracy rates.

If the level of significance for a two-tailed test is equal to 0.05, then the corresponding confidence level to construct the confidence interval is equal to 95%.

The confidence interval estimate has the following formula:

$\left({\hat{p}}_1-{\hat{p}}_2\right)-E<p_1-p_2<\left({\hat{p}}_1-{\hat{p}}_2\right)+E$

Here, E is the margin of error.

E is the margin of error and has the following formula:

$$\begin{gathered} E=z_{\frac{伪}{2}}\sqrt{\frac{{\hat{p}}_1{\hat{q}}_1}{n_1}+\frac{{\hat{p}}_2{\hat{q}}_2}{n_2}} \\ =1.96脳\sqrt{\frac{\left(0.83\right)\left(0.17\right)}{318}+\frac{\left(0.91\right)\left(0.09\right)}{362}} \\ =0.051 \end{gathered}$$

b.

Substituting the required values, the following confidence interval is obtained:

$$\begin{gathered} \left({\hat{p}}_1-{\hat{p}}_2\right)-E<p_1-p_2<\left({\hat{p}}_1-{\hat{p}}_2\right)+E \\ (0.83-0.91)-0.051<p_1-p_2<(0.83-0.91)+0.051 \\ -0.129<p_1-p_2<-0.028 \end{gathered}$$

Thus, the 95% confidence interval is equal to (-0.129, -0.028).

This confidence interval does not contain zero that means there is a significant difference between the two proportions of accurate orders.

Therefore, the confidence interval suggests that there is sufficient evidence to reject the claim thatBurger King and McDonald鈥檚 have the same accuracy rates.

c.

There is a significant difference between the proportions of accurate orders of Burger King and McDonald鈥檚, and the confidence interval contains only negative values.

Therefore, the accuracy rate of Burger King is less than that of McDonald鈥檚, and McDonald鈥檚 can be considered better.