# 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 %

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: $0.127. What do you​ conclude? a. Construct a 90​% confidence interval. Express the percentages in decimal form. ___

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Derrick

Step 1 It is given that 51 out of 298 orders are not accurate. The sample proportion is $\frac{51}{298}=0.17$. 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: $CI=\stackrel{^}{p}±{z}_{\frac{\alpha }{2}}\sqrt{\frac{\stackrel{^}{p}\left(1-\stackrel{^}{p}\right)}{n}}$
$=0.171±\left(1.645\right)\sqrt{\frac{0.17\left(1-0.17\right)}{298}}$
$=0.171±0.036$
$=\left(0.135,0.207\right)$ 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.