In a study of the accuracy of fast food drive-through orders, Restaurant A had 295 accurate orders and 55 that were not accurate. a. Construct a 95% confidence interval estimate of the percentage of orders that are not accurate. b. Compare the results from part (a) to this 95% confidence interval for the percentage of orders that are not accurate at Restaurant B: 0.138less thanpless than0.211. What do you conclude? a. Construct a 95% confidence interval. Express the percentages in decimal form.
Accepted Solution
A:
Answer: (a) [tex]14.5\%<p<23.5\%[/tex](b) Restaurant B has a significantly lower percentage of orders that are not accurate. Step-by-step explanation:Confidence interval for population proportion is given by :-[tex]\hat{p}\pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]Given: Significance level : [tex]\alpha: 1-0.95=0.05[/tex]Critical value : [tex]z_{\alpha/2}=1.96[/tex]For Restaurant A , The proportion of orders not accurate : [tex]p=\dfrac{55}{295}\approx0.19[/tex]Then , the Confidence interval for population proportion will be :-[tex]0.19\pm (1.96)\sqrt{\dfrac{0.19(1-0.19)}{295}}\\\\=0.19\pm0.045=(0.145,0.235)=(14.5\%,23.5\%)\\\\\text{i.e.}14.5\%<p<23.5\%[/tex]Also, 95% confidence interval for the percentage of orders that are not accurate at Restaurant B:[tex]0.138<p<0.211[/tex]By comparing both the lower confidence limit and upper confidence limit of the interval for Restaurant B is lower than the lower confidence limit of t and the upper confidence limit of the interval for Restaurant A. Therefore, Restaurant B has a significantly lower percentage of orders that are not accurate.