Question

A simple random sample of size

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,

38

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,

Point estimate = sample proportion = = x / n = 38 / 400 = 0.095

1 - = 1 - 0.095 = 0.905

Z/2 = 2.576

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

= 2.576 * (((0.095 * 0.905) / 400)

= 0.038

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.095 - 0.038 < p < 0.095 + 0.038

0.057 < p < 0.133

The 99% confidence interval for the population proportion p is : 0.057 , 0.133