Question

Critical Thinking. In Exercises 17–28, use the data and confidence level to construct a

confidence interval estimate of p, then address the given question. Fast Food AccuracyIn a study of the accuracy of fast food drive-through orders, Burger King had 264 accurate orders and 54 that were not accurate (based on data from QSRmagazine).

a.Construct a 99% confidence interval estimate of the percentageof orders that are not accurate.

b.Compare the result from part (a) to this 99% confidence interval for the percentage of orders that are not accurate at Wendy’s: 6.2%<p< 15.9%. What do you conclude?

Short answer

a. The 99% confidence interval is between 11.6% and 22.4%.

b. Burger king and Wendy have a similar percentage of orders that are not accurate.

Step-by-step solution

Given information

The number of fast-food orders is recorded.

The number of orders is,

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

Where 264 were accurate and 54 were inaccurate.

The confidence interval is .

Calculate the sample proportion

a.

Here we need to estimate the confidence interval of the order that is not accurate.

Thesample proportion of the order which is not accurate is:

$$\begin{gathered} \hat{p}=\frac{x}{n} \\ =\frac{54}{318} \\ =0.17 \end{gathered}$$

Therefore, the sample proportion is 0.17.

Then,

$$\begin{gathered} \hat{q}=1-\hat{p} \\ =1-0.17 \\ =0.83 \end{gathered}$$

Check the requirements

The requirements for the confidence interval are,

1. The samples are selected randomly.
2. There are two categories of the outcome, either accurate or not accurate.
3. The counts of failures and successes are both greater than 5.

Therefore, all the conditions are satisfied and we can construct a confidence interval for the population proportion.

Compute the critical value

At $99\%$confidence interval, $\alpha =0.01$.

Using standard normal table,

$$\begin{gathered} z_{crit}=z_{\frac{\alpha }{2}} \\ =z_{\frac{0.01}{2}} \\ =2.575 \end{gathered}$$

Compute the margin of error

The margin of error is given by,

$$\begin{gathered} E=z_{crit}\times \sqrt{\frac{\hat{p}\hat{q}}{n}} \\ =2.575\times \sqrt{\frac{0.17\times 0.83}{318}} \\ =0.0542 \end{gathered}$$

Compute the confidence interval

The formula for the confidence interval is given by,

$$\begin{gathered} CI=\hat{p}-E<p<\hat{p}+E \\ =\left(0.17-0.0542<p<0.17+0.0542\right) \\ =(0.1158<p<0.2242) \end{gathered}$$

In percentage,$11.56\%<p<22.40\%$ .

Thus, there is 99% confidence interval for orders that are not accurate and lie between 11.6% and 22.4% .

Compare the results

b.

The 99% confidence interval for not accurate orders for Burger King is between 11.6% and 22.4%.

The 99% confidence interval for the percentage of not accurate orders for Wendy’s is between 6.2% and 15.9%.

Here two confidence interval overlaps.

Hence it can be concluded that Burger King and Wendy have similar percentage of orders that are not accurate.