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

Answer:

(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.

Step-by-step 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.  =  $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$  ~ N(0,1)

where, $\hat p$ = sample proportion of orders that were not accurate = $\frac{59}{361}$ = 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 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.96) = 0.95

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

P( $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

95% confidence interval for p = [ $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

  = [ $0.163 -1.96 \times {\sqrt{\frac{0.163(1-0.163)}{361} } }$ , $0.163 +1.96 \times {\sqrt{\frac{0.163(1-0.163)}{361} } }$ ]

  = [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.