91Ó°ÊÓ

The number of fast-food orders is recorded.

The number of orders is,

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

Where 264 were accurate and 54 were inaccurate.

The confidence interval is\(99\% \).

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:

\(\begin{array}{c}\hat p = \frac{x}{n}\\ = \frac{{54}}{{318}}\\ = 0.17\end{array}\)

Therefore, the sample proportion is 0.17.

Then,

\(\begin{array}{c}\hat q = 1 - \hat p\\ = 1 - 0.17\\ = 0.83\end{array}\)

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.

At \(99\% \) confidence interval, \(\alpha = 0.01\).

Using standard normal table,

\(\begin{array}{c}{z_{crit}} = {z_{\frac{\alpha }{2}}}\\ = {z_{\frac{{0.01}}{2}}}\\ = 2.575\end{array}\)

The margin of error is given by,

\(\begin{array}{c}E = {z_{crit}} \times \sqrt {\frac{{\hat p\hat q}}{n}} \\ = 2.575 \times \sqrt {\frac{{0.17 \times 0.83}}{{318}}} \\ = 0.0542\end{array}\)

The formula for the confidence interval is given by,

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

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% .

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.