Question

The United Nations wants to determine what the percentage of females in the Canadian labour force is. They select a random sample of 951 employment records, and find that 349 of the people are females. What is the margin of error in the 90% confidence interval for the population proportion based on this sample?

Note:

1- Round any intermediate numbers to 5 decimal places.
2- Round your final answer to 3 decimal places. Enter your final answer with 3 decimal places.

Answer 1

Solution :

Given that,

Point estimate = sample proportion = = x / n = 349 / 951 = 0.36698

1 - = 0.36698 = 0.63302

Z_/2 = 1..645

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

= 1.645 * (((0.36698 * 0.63302) / 951)

= 0.02571

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.36698 - 0.02571 < p < 0.36698 + 0.02571

0.341 < p < 0.393

The 90% confidence interval for the population proportion p is : 0.341 , 0.393