Question

In a study of the accuracy of fast food drive-through orders, Restaurant A had 298 accurate orders and 51 that were not accurate. a. Construct a 90 %

Confidence intervals
In a study of the accuracy of fast food​ drive-through orders, Restaurant A had 298 accurate orders and 51 that were not accurate. a. Construct a 90​% confidence interval estimate of the percentage of orders that are not accurate. b. Compare the results from part​ (a) to this 90​% confidence interval for the percentage of orders that are not accurate at Restaurant​ B: $$\displaystyle{0.127}{<}{p}{<}{0.191}$$</span>. What do you​ conclude? a. Construct a 90​% confidence interval. Express the percentages in decimal form. ___<p<___ ​(Round to three decimal places as​ needed.) (This is used directly in response to B).

Step 1 It is given that 51 out of 298 orders are not accurate. The sample proportion is $$\displaystyle\frac{{51}}{{298}}={0.17}$$. The sample size is 298. For the confidence interval 90%, the two-tailed z value at 0.05 level of significance is 1.645. The 90% confidence interval for the population proportion is calculated as follows: $$\displaystyle{C}{I}=\hat{{{p}}}\pm{z}_{{\frac{\alpha}{{2}}}}\sqrt{{{\frac{{\hat{{{p}}}{\left({1}-\hat{{{p}}}\right)}}}{{{n}}}}}}$$
$$\displaystyle={0.171}\pm{\left({1.645}\right)}\sqrt{{{\frac{{{0.17}{\left({1}-{0.17}\right)}}}{{{298}}}}}}$$
$$\displaystyle={0.171}\pm{0.036}$$
$$\displaystyle={\left({0.135},{0.207}\right)}$$ Thus, the percentage of orders that is not accurate of restaurant A lies between 0.135 and 0.207. Step 2 The percentage of orders that are not accurate of restaurant B lies between 0.127 and 0.191. From the intervals it can be observed that neither restaurant appears to have a significant different percentage of orders that are not accurate because these two confidence intervals are overlapped.