Question

In a study of the accuracy of fast food​ drive-through orders, Restaurant A had 205 accurate orders and 63 that were not accurate. a.

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

Answer 1

Solution :

Given that,

n = 205

x = 63

Point estimate = sample proportion = = x / n = 63/205=0.307

1 -   = 1-0.307 =0.693

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 *((0.307*0.693) /205 )

E = 0.053

A 90% confidence interval is ,

- E < p < + E

0.307-0.053< p < 0.307+0.053

0.254< p < 0.360

(0.254 , 0.360)