Question

In a study of the accuracy of fast food​ drive-through orders, Restaurant A had 302accurate orders and 59that 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: 0.143less thanpless than0.219.What do you​ conclude?

Answer 1

(a) A 95​% confidence interval estimate of the percentage of orders that are not accurate is [0.125, 0.201].

(b) We can conclude that both restaurants can have the same inaccuracy rate due to the overlap of interval areas.

Explanation:

We are given that in a study of the accuracy of fast food​ drive-through orders, Restaurant A had 302 accurate orders and 59 orders that were not accurate.

Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;

P.Q. =
`\frac{\hat p-p}{\sqrt{(\hat p(1-\hat p))/(n) } }` ~ N(0,1)

where,
`\hat p` = sample proportion of orders that were not accurate =
`(59)/(361)` = 0.163

n = sample of total orders = 302 + 59 = 361

p = population proportion of orders that are not accurate

Here for constructing a 95% confidence interval we have used a One-sample z-test for proportions.

So, 95% confidence interval for the population proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 <
`\frac{\hat p-p}{\sqrt{(\hat p(1-\hat p))/(n) } }` < 1.96) = 0.95

P(
`-1.96 * {\sqrt{(\hat p(1-\hat p))/(n) } }` <
`{\hat p-p}` <
`1.96 * {\sqrt{(\hat p(1-\hat p))/(n) } }` ) = 0.95

P(
`\hat p-1.96 * {\sqrt{(\hat p(1-\hat p))/(n) } }` < p <
`\hat p+1.96 * {\sqrt{(\hat p(1-\hat p))/(n) } }` ) = 0.95

95% confidence interval for p = [
`\hat p-1.96 * {\sqrt{(\hat p(1-\hat p))/(n) } }` ,
`\hat p+1.96 * {\sqrt{(\hat p(1-\hat p))/(n) } }` ]

= [
`0.163 -1.96 * {\sqrt{(0.163(1-0.163))/(361) } }` ,
`0.163 +1.96 * {\sqrt{(0.163(1-0.163))/(361) } }` ]

= [0.125, 0.201]

(a) Therefore, a 95​% confidence interval estimate of the percentage of orders that are not accurate is [0.125, 0.201].

(b) We are given that the 95​% confidence interval for the percentage of orders that are not accurate at Restaurant​ B is [0.143 < p < 0.219].

Here we can observe that there is a common area of inaccurate order of 0.058 or 5.85% for both the restaurants.

So, we can conclude that both restaurants can have the same inaccuracy rate due to the overlap of interval areas.