Question

In a study of the accuracy of fast food​ drive-through orders, Restaurant A had 224 accurate orders and 60 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: 0.189 < p < 0.289. What do you​ conclude?

Answer 1

p : sample proportion of orders that were not accurate at restaurant A

p= order that were not accurate / total number of orders at restaurant A

p=60/284 = 0.2113

n = 284

100(1-​​​​​​) % confidence interval is given by,

( p - Z/2*(p*(1-p)/n) , p + Z/2*(p*(1-p)/n) ​​​​​​)

Z is critical value taken from N(0,1) probability table.

95 % confidence interval is,

=(0.2113 - Z0.025 *0.0242 , 0.2213 + Z0.025 * 0.0242)

= (0.164 , 0.259)

This means that 95% times, order that are not accurate at restaurant A varies between 16.4% and 25.9%

The 95% confidence interval is given by,

(0.189 , 0.289)

This means that 95% times, order that are not accurate at restaurant B varies between 18.9% and 28.9%

This tells us that Restaurant B has higher chance of orders that are not accurate than those of Restaurant A.

This means Restaurant A is better at accurate orders than Restaurant B.