Question

7.4.13-T Question Help In a study of the accuracy of fast-food drive through orders, a restaurant had 38 orders that were not accurate among 381 orders observed Use the bootstrap method to construct a 95% confidence interval estimate of the proportion of orders that are not accurate. Use the 200 accompanying bootstrap samples. How does the result compare to the 95% confidence interval 0.070<p <0.130 constructed using the sample proportion? B Click the icon to view the bootstrap samples The 95% confidence interval is <p (Round to three decimal places as needed.) Incorre Enter your answer in the edit fields and then click Check Answer Clear All part remaining o

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Answer 1

ANSWER:

Given

Total orders = n = 381

not accurate orders = x = 38

sample proportion = p=  x / n = 0.0997

At 95% confidence interval (C.I) = 1 - 0.95 = 0.05

Z_0.05 = 1.96 = Z (using critical values of z)

$\small C.I ={p}\pm Z\sqrt{\frac{{p}(1-{p})}{n}}$

$\small C.I =0.0997\pm 1.96\sqrt{\frac{0.0997*0.9003}{381}}$

C.I = 0.0997 $\small \pm$ 0.0301

C.I = (0.0696 , 0.1298) $\small \simeq$ (0.070 ,0.130)

The 95% confidence interval is 0.070 < p < 0.130

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