Question

3. In a study of the accuracy of fast food​drive-through orders, Restaurant A had 233 accurate orders and 58 that were not accurate.

a. Construct a 95​% confidence interval estimate of the percentage of orders that are not accurate.

b. Compare the results from part​ (a) to this 95​% confidence interval for the percentage of orders that are not accurate at Restaurant​ B:

0.171 <p< 0.262. 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

3)

Solution :

Given that,

n = 233

x = 58

Point estimate = sample proportion = = x / n = 58 / 233 = 0.249

1 - = 1 - 0.249 = 0.751

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

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

= 1.96 * (((0.249 * 0.751) / 233)

= 0.056

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.249 - 0.056< p < 00.249 + 0.056

0.193 < p < 0.305

b) Compare to the given part 0.171 <p< 0.262 part A is the wider because the margin of error is part A > part B