please help! Due tonight!

Question

Out of 300 people sampled, 162 had kids. Based on this, construct a 90% confidence interval for the true population proportion of people with kids.

Give your answers as decimals, to three places

 < p < 

Answer 1

The question requires you to compute a confidence interval (CI) for the true population proportion. Let p be the sample proportion, then the true population proportion belongs to the confidence interval p +/- (Z_critical)*sqrt([p*(1-p)]/(n)).

The sample proportion is given by p = 162/300 =.54, n = 300.

The standard deviation can be calculated by sqrt([p*(1-p)]/(n)) = sqrt [(0.54*(1-0.54))/300]=0.0288

From the Z-Table, the Z_critical for 90% CI is 1.645.

The CI is given by 0.54 +/- 1.645*0.0288, which gives 0.4926 to 0.5874.