Question

An automotive manufacturer wants to know the proportion of new car buyers who prefer foreign cars over domestic.

Step 2 of 2:

Suppose a sample of 1453 new car buyers is drawn. Of those sampled, 363 preferred foreign over domestic cars. Using the data, construct the 99% confidence interval for the population proportion of new car buyers who prefer foreign cars over domestic cars. Round your answers to three decimal places.

Lower Endpoint:

Upper Endpoint:

Answer 1

Solution :

Given that,

n = 1453

x = 363

Point estimate = sample proportion = = x / n = 363 / 1453 = 0.250

1 - = 1 - 0.250 = 0.750

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

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

= 1.96 * (((0.250 * 0.750) / 1453)

= 0.029

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.250 - 0.029 < p < 0.250 + 0.029

0.221 < p < 0.279

Lower endpoint = 0.221

Upper endpoint = 0.279