Mathematics, 29.07.2020 01:01 hey840

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?

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Answer from: xyzendre

(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.

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In a study of the accuracy of fast food drive-through orders, Restaurant A had 302accurate orders an...