Question

A simple random sample of size n equals 400 individuals who are currently employed is asked if they work at home at least once per week. Of the 400 employed individuals​ surveyed, 34 responded that they did work at home at least once per week. Construct a​ 99% confidence interval for the population proportion of employed individuals who work at home at least once per week.

The upper bound is (Round to three decimal places as​ needed.)

Answer 1

Solution :

Given that,

n = 400

x = 34

Point estimate = sample proportion = = x / n = 0.085

1 - =0.915

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)

= 2.576 * (((0.085*0.915) / 400)

= 0.036

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.085 - 0.036 < p < 0.085 + 0.036

0.049 < p < 0.121

The 99% confidence interval for the population proportion p is : ( 0.049 , 0.121 )