Given:
Out of 200 people sampled, 174 had kids Based on this, construct a 90% confidence interval for the true population proportion of people with kids.
Required:
We need to find the confidence interval.
Explanation:
The proportion of the mean is
[tex]\frac{174}{200}=0.87[/tex]The standard error of the sample is
[tex]\sqrt{0.87\times\frac{(1-0.87)}{200}}=0.02378[/tex][tex]\text{alpha\lparen}\alpha\text{\rparen is }1-\frac{90}{100}=0.10[/tex][tex]\text{ critical probability }(P^*)=1-\frac{\alpha}{2}=0.95[/tex]Assuming a normal distribution, look for the z-score associated with a 0.95 cumulative probability.
z-score = 1.645
[tex]Margin\text{ of error =1.645}\times0.0238=0.0395[/tex]The confidence interval is 0.87+0.0395 and 0.87-0.0395.
[tex]0.8305Final answer:
[tex]0.8305
