Question

a simple random sample size n=250 individuals who are currently employed is asked if they work at home at least once per week. of the 250 employed individuals surveyed, 28 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 from home atleast once per week. the lower bound is? the upper bound is?

Answer 1

Given

n = 250

X = 28

Now calculate p

p = x /n = 28/250 = 0.1120

Now 99% of confidence interval using formula is given by

CI = p $\pm$ z $\sqrt$ (p*(1-p)/n)

z value for 99% confidence interval is 2.58

z = 2.58

CI = 0.112 $\pm$ 2.58*$\sqrt$((0.112*0.888)/250)

= 0.112 $\pm$ 0.0515

CI = 0.112 - 0.0515 and CI = 0.112 + 0.0515

= 0.0605 and CI = 0.1635

The Lower bound = 0.0605

Upper bound = 0.1635