esting Claims About Proportions. In Exercises 7–22, test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim.

Accuracy of Fast Food Drive-Through Orders In a study of Burger King drive-through orders, it was found that 264 orders were accurate and 54 were not accurate. For McDonald’s, 329 orders were found to be accurate while 33 orders were not accurate (based on data from QSR magazine). Use a 0.05 significance level to test the claim that Burger King and McDonald’s have the same accuracy rates.

a. Test the claim using a hypothesis test.

b. Test the claim by constructing an appropriate confidence interval.

c. Relative to accuracy of orders, does either restaurant chain appear to be better?

In testing the accuracy of drive-through orders, for Burger King, 264 orders were accurate, and 54 were not accurate, while for McDonald’s, 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’s have the same accuracy rates.

\({H_0}:{p_1} = {p_2}\)

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

\({H_1}:{p_1} \ne {p_2}\)

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

\(\begin{array}{c}{n_1} = 264 + 54\\ = 318\end{array}\)

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

\(\begin{array}{c}{n_2} = 329 + 33\\ = 362\end{array}\)

Assume that\({x_1}\)and\({x_2}\)are the number of accurate orders for Burger King and McDonald’s respectively.

Let \({\hat p_1}\)be the sample accuracy rate of Burger King.

Thus,

\(\begin{array}{c}{{\hat p}_1} = \frac{{{x_1}}}{{{n_1}}}{\rm{ }}\\ = \frac{{264}}{{318}}\\ = 0.83\end{array}\)

\(\begin{array}{c}{{\hat q}_1} = 1 - {{\hat p}_1}\\ = 0.17\end{array}\)

Let \({\hat p_2}\)be the sample accuracy rate of McDonald’s.

Thus,

\(\begin{array}{c}{{\hat p}_2} = \frac{{{x_2}}}{{{n_2}}}{\rm{ }}\\ = \frac{{329}}{{362}}\\ = 0.91\end{array}\)

\(\begin{array}{c}{{\hat q}_2} = 1 - {{\hat p}_2}\\ = 0.09\end{array}\)

The value of the pooled sample proportion is equal to:

\(\begin{array}{c}\bar p = \frac{{{x_1} + {x_2}}}{{{n_1} + {n_2}}}\\ = \frac{{264 + 329}}{{318 + 362}}\\ = 0.872\end{array}\)

Hence,

\(\begin{array}{c}\bar q = 1 - \bar p\\ = 1 - 0.872\\ = 0.128\end{array}\)

The test statistic is equal to:

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

Thus, z=-3.064.

Referring to the standard normal distribution table, the critical values of z corresponding to\(\alpha = 0.05\)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’s 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:

\(\begin{array}{c}E = {z_{\frac{\alpha }{2}}}\sqrt {\frac{{{{\hat p}_1}{{\hat q}_1}}}{{{n_1}}} + \frac{{{{\hat p}_2}{{\hat q}_2}}}{{{n_2}}}} \\ = 1.96 \times \sqrt {\frac{{\left( {0.83} \right)\left( {0.17} \right)}}{{318}} + \frac{{\left( {0.91} \right)\left( {0.09} \right)}}{{362}}} \\ = 0.051\end{array}\)

b.

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

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

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’s have the same accuracy rates.

c.

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

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