Question

In a study of the accuracy of fast food​ drive-through orders, Restaurant A had 206 accurate orders and 58 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: 0.205 < p < 0.283. What do you​ conclude?

Answer 1

Solution :

Given that,

n = 206

x = 58

Point estimate = sample proportion = = x / n = 58/206=0.282

1 -   = 1- 0.282 =0.718

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 *((0.282*0.718) /206 )

E = 0.052

A 90% confidence interval proportion p is ,

- E < p < + E

0.282-0.052 < p < 0.282+0.052

0.230< p < 0.334

B.

0.205 < p < 0.283.

THIS IS LOWER BOUND 0.205

UPPER BOUND 0.283

and sample proportion lies between them