Question

In a study of fast food accuracy on drive-thru orders, Burger King had 264 accurate orders and 54 that were not accurate.

a) Construct a 95% CI for the percentage of orders that are not accurate.

b) A similar survey at Wendy's yielded a 95% CI of inaccurate orders of 6.2% < p < 15.9%. Comparing the two results, what do you find?

Answer 1

Solution:

a)

Given,

n = 264 ....... Sample size

x = 54 .......no. of successes in the sample

Let denotes the sample proportion.

     = x/n   = 54/264 = 0.2045

Our aim is to construct 95% confidence interval.

c = 0.95

= 1- c = 1- 0.95 = 0.05

  /2 = 0.025 and 1- /2 = 0.975

Search the probability 0.975 in the Z table and see corresponding z value

= 1.96

Now , the margin of error is given by

E = /2 *  

= 1.96 * [0.2045 *(1 - 0.2045)/264]

= 0.04865425826

Now the confidence interval is given by

( - E)   ( + E)

(0.2045 - 0.04865425826)   (0.2045 + 0.04865425826)

0.156 0.253

15.6% < p < 25.3%

b)

Compare above interval with given interval 6.2% < p < 15.9%.

Two intervals are very different . because each confidence interval do not contain the mean of other confidence interval.