Question

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

Step 1 of 2:

Suppose a sample of 2737 new car buyers is drawn. Of those sampled, 930 preferred foreign over domestic cars. Using the data, estimate the proportion of new car buyers who prefer foreign cars. Enter your answer as a fraction or a decimal number rounded to three decimal places.

Step 2 of 2:

Suppose a sample of 2737 new car buyers is drawn. Of those sampled, 930 preferred foreign over domestic cars. Using the data, construct the 98% confidence interval for the population proportion of new car buyers who prefer foreign cars over domestic cars. Round your answers to three decimal places.

Please find lower and upper endpoints for Step 2!

Thank you!!!

Answer 1

Solution :

Given that,

n = 2737

x = 930

= x / n = 930 / 2737 = 0.340

1 - = 1 - 0.340 = 0.660.

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

Z_/2 = Z_0.01 = 2.326

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

= 2.326* (((0.340 * 0.660) /2737) = 0.021

A 98 % confidence interval for population proportion p is ,

- E < P < + E

0.340 - 0.021 < p < 0.340 + 0.021

0.319 < p < 0.361

Lower  endpoint = 0.319

Upper endpoint = 0.361