Question

in a study of the accuracy of fast food drive throughorders. Restaurant A had 336 accurate orders and 72 that were not accurate.

a. construct a 95% confidence interval estimate of the percentages of orders that are not accurate.

b. compare rhe results from part (a) to this 95% confidence interval for the percentage of orders that are not accurate at Restaurant B: 0.158 < p < 0.231 what do you conclude?

____________________________________

a. Construct a 95% confidence interval. Express the percentages in decimal form.

__< p < __ (round to three decimal places as needed)

Answer 1

Solution :

Given that,

n = 336

x = 72

Point estimate = sample proportion = = x / n = 72 / 336 = 0.214

1 - = 1 - 0.214 = 0.786

a)

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.10

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z _0.025 = 1.96

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

= 1.96 * (((0.214*0.786) / 336)

= 0.044

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.214 - 0.044 < p < 0.214 + 0.044

0.170 < p < 0.258

b)

The results from part (a) to this 95% confidence interval for the percentage of orders that are not accurate at Restaurant B: 0.158 < p < 0.231

Therefore, the result of (a) is wider than to this 95% confidence interval for the percentage of orders of restaurant B.