Question

In a study of the accuracy of fast food​ drive-through orders, Restaurant A had 341accurate orders and 57 that were not accurate.

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

a. Construct a 90​% confidence interval. Express the percentages in decimal form.

------------- <p<-------------

​(Round to three decimal places as​ needed.)

b. Compare the results from part​ (a) to this 90​% confidence interval for the percentage of orders that are not accurate at Restaurant​ B: 0.131 <p< 0.182.

Answer 1

Given that In a study of the accuracy of fast food​ drive-through orders, Restaurant A had 341 accurate orders and 57 that were not accurate.

So number of total orders, n = 341 + 57

= 398

Proportion of orders that were not accurate, = 57 / 398

= 0.143216

90% confidence interval for estimate of percentage of orders that are not accurate = Z/2 * * (1 - ) / n

Here given confidence level = 90% = 0.9

= 1 - confidence level

= 1 - 0.9

= 0.1

/2 = 0.05

So Z/2 will be z-score that has an area of 0.05 to its right which is 1.64485

90% confidence interval =  0.143216 1.64485 * 0.143216 * (1 - 0.143216) / 398

= 0.143216 1.64485 * 0.000308

  = 0.143216 1.64485 * 0.017559

=  0.143216 0.028881

= (0.114335, 0.172097)

So 90% confidence interval for estimate of the percentage of orders that are not accurate = (0.114335, 0.172097)

After roudning to 3 decimals we have

0.114 < p < 0.172

Question (b)

As you can observe the lower limit in the confidence interval of the Restaurant A is larger than that of lower limit in the confidence interval of the Restaurant B

The higher limit in the confidence interval of the Restaurant B is smaller than that of higher limit in the confidence interval of the Restaurant B

Margin of error = (Width of the confidence interval) / 2 = (upper limit - lower limit) / 2

Margin of error for restaurant A = (0.172 - 0.114) / 2

= 0.058 / 2

= 0.029

Margin of error for restaurant B = (0.182 - 0.131) / 2

= 0.051 / 2

= 0.0255

The Margin of error for Restaurant A is more than that of Restuarant B which implies the confidence in restaurant A is less than that of the confidence in restaurant B

In other terms the accuracy of the population parameters will be more in restaurant B than in restaurant A