Question

In a study of the accuracy of fast food​ drive-through orders, Restaurant A had 270 accurate orders and 51 that were not accurate.

a. Construct a 90 ​% confidence interval estimate of the percentage of orders that are not accurate. Express the percentages in decimal form.

______<p<_______

b. Compare the results from part​ (a) to this 90 ​% confidence interval for the percentage of orders that are not accurate at Restaurant​ B: 0.137: 0.137 <p<0.205. What do you conclude?

Answer 1

Solution:

a)

For Restaurant A

No. of accurate orders = 270

No. of not accurate orders = 51

n = 270 + 51 = 321

Let be the sample proportion of f not accurate orders

= 51/321= 0.159

Our aim is to construct 90% confidence interval for the proportion of not accurate orders.

c = 0.90

= 1- c = 1- 0.90 = 0.10

  /2 = 0.10 2 = 0.05 and 1- /2 = 0.950

= 1.645 (use z table)

Now , the margin of error is given by

E = /2 *  

= 1.645 * [0.159*(1 - 0.159)/321]

= 0.0336

Now the confidence interval is given by

( - E)   ( + E)

(0.159 - 0.0336)   (0.159 + 0.0336)

0.1254     0.1926

In percentage form , 12.54% 19.26%

b)

Compare this result with Restaurant​ B: 0.137: 0.137 < p <0.205

i.e. 13.7% < p < 20.5 %

Lower limit of interval for A is less than lower limit interval for B

Upper limit of interval for A is less than upper limit of interval for B

So , we can say that ,

Restaurant A has less number of not accurate orders than that of Restaurant B.