Question

n a study of the accuracy of fast food​ drive-through orders, Restaurant A had 314 accurate orders and 61 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: 0.147less thanpless than0.206. What do you​ conclude?

Answer 1

Answer:

a. 0.1576<p<0.2310

b. The two restaurants likely have similar order rates which are inaccurate.

Step-by-step explanation:

a. We first calculate the proportion, $\hat p$:

$\hat p=\frac{61}{314}\\\\=0.1943$

-We use the z-value alongside the proportion to calculate the margin of error:

$MOE=z\sqrt{\frac{\hat p(1-\hat p)}{n}}\\\\=1.645\times \sqrt{\frac{0.1943(1-0.1943)}{314}}\\\\=0.0367$

The confidence interval at 90% is then calculated as:

$CI=\hat p\pm MOE\\\\=0.1943\pm 0.0367\\\\=[0.1576,0.2310]$

Hence, the confidence interval at 90% is [0.1576,0.2310]

b. From a above, the calculated confidence interval is 0.1576<p<0.2310

-We compare the calculated CI to the stated CI of 0.147<p<0.206

-The two confidence intervals overlap each other and have the same value for 0.1576<p<0.206

-Hence, we conclude that the two restaurants likely have similar order rates which are inaccurate.