Question

Kay Mary, senior vice president for marketing at Terrapin Cosmetics, asked you to estimate, with 96% confidence, the proportion of all of the company’s customers who placed “large” orders, meaning over $200, in both January and February of 2019. To estimate the proportion, you selected a random sample of 845 customers. Of the customers in the sample, 165 placed large orders in both January and February of 2019. Please write the interval boundaries to THREE decimal places and interpret the interval.

Answer 1

Solution:

Given:

Sample size = n = 845

x = Number of customers placed large orders in both January and February of 2019= 165

We have to construct a 96% confidence interval for  the proportion of all of the company’s customers who placed “large” orders, meaning over $200, in both January and February of 2019.

Formula:

where

We need to find z_c value for c=96% confidence level.

Find Area = ( 1 + c ) / 2 = ( 1 + 0.96) /2 = 1.96 / 2 = 0.9800

Look in z table for Area = 0.9800 or its closest area and find z value.

Area = 0.9798 is closest to 0.9800 and it corresponds to 2.0 and 0.05 , thus z critical value = 2.05

That is : Z_c = 2.05

Thus

Thus

Thus we are 96% confident that the true population proportion of all of the company’s customers who placed “large” orders, meaning over $200, in both January and February of 2019 is between 0.167 and 0.223 .