Question

A simple random sample of size n equals n=400 individuals who are currently employed is asked if they work at home at least once per week. Of the 400 employed individuals​ surveyed, 40 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.

Answer 1

Solution :

Given that,

sample size n = 400

x = 40

Point estimate = sample proportion = = x / n = 40/400=0.1

1 - = 1-0.1=0.9

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 ( Using z table )

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

= 2.576 (((0.1*0.9) /400 )

= 0.03864

​ 99% confidence interval for the population proportion of employed individuals who work at home at least once per week

- E < p < + E

0.1 - 0.03864 < p < 0.1 + 0.03864

0.06136< p < 0.13864

confidence interval (0.06136 , 0.13864 )