Question

A recent study by the Campanotto/Campos Consumer Group randomly surveyed 1025 women; 29% of them stated that they shopped online several times throughout the year.

a. Construct the 99% confidence interval for the proportion of women that shopped online several times throughout the year. (Triola, 9th. Ed)

b. What is the point estimate? If we want to estimate the proportion of women that shop online to determine the maximum impact on its sales, what is the maximum percentage of women that they should choose?

Answer 1

Solution :

Given that

n = 1025

= 0.290

1 - = 1 - 0.290 = 0.710

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576

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

= 2.576* (((0.290 * 0.710) / 1025)

= 0.036

A 99 % confidence interval for population proportion p is ,

- E < P < + E

0.290 - 0.036 < p < 0.290 + 0.036

0.254 < p < 0.326