Question

A simple random sample of size n=300 individuals who are currently employed is asked if they work at home at least once per week. Of the 300 employed individuals​ surveyed, 27 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

olution :

Given that,

n = 300

x = 27

Point estimate = sample proportion = = x / n = 27/300=0.09

1 -   = 1- 0.09 =0.91

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* (((0.09*0.91) /300 )

E = 0.043

A 99% confidence interval for proportion p is ,

- E < p < + E

0.09-0.043 < p < 0.09+0.043

0.047< p < 0.133