Question

A simple random sample of size n equals 200 individuals who are currently employed is asked if they work at home at least once per week. Of the 200 employed individuals​ surveyed, 41 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

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

//0.20113,0.20887[/tex]

Step-by-step explanation:

step 1:-

Given sample  size n=200

of the 200 employed individuals surveyed 41 responded that they did work at home at least once per week

 Population proportion of employed individuals who work at home at least once per week  P = $\frac{x}{n} =\frac{41}{200} =0.205$

Q=1-P= 1-0.205 = 0.705

step 2:-

Now  $\sqrt{\frac{P Q}{n} } =\sqrt{\frac{(0.205)(0.705)}{200} }$

=0.0015

step 3:-

Confidence intervals

using formula

$(P - Z_∝} \sqrt{\frac{P Q}{n},} (P + Z_∝} \sqrt{\frac{P Q}{n},$

$(0.205-2.58(0.0015),0.205+2.58(0.0015)\\0.20113,0.20887$

=0.20113,0.20887[/tex]

conclusion:-

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

//0.20113,0.20887[/tex]