Question

In a study of the accuracy of fast food​ drive-through orders, Restaurant A had 211 accurate orders and 56 that were not accurate.

a. Construct a 90​% confidence interval estimate of the percentage of orders that are not accurate.

b. Compare the results from part​ (a) to this 90​% confidence interval for the percentage of orders that are not accurate at Restaurant​ B: 0.188<p<0.271

Answer 1

Solution:

a) Given that,

n = 211

x = 56

Point estimate = sample proportion = = x / n = 56/211=0.2654

1 -    = 1-0.307 =0.7346

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

Margin of error = E Z/2 *(( * (1 - )) / n)

= 1.645 *((0.2654*0.7346) /211 )

E = 0.05

A 90% confidence interval is ,

- E < p < + E

0.2654-0.05< p < 0.2654+0.05

0.2154< p < 0.3154

(0.2154 , 0.3154)

b) Restaurant​ B:

0.188<p<0.271

The range of the both confidence interval Restaurant A and B are almost equal.