Question

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. Give your answers as decimals, to three places

Answer 1

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.

Step-by-step explanation:

The proportion of the mean is

`(174)/(200)=0.87`

The standard error of the sample is

`\sqrt{0.87*((1-0.87))/(200)}=0.02378`
`\text{alpha\lparen}\alpha\text{\rparen is }1-(90)/(100)=0.10`
`\text{ critical probability }(P^*)=1-(\alpha)/(2)=0.95`

Assuming a normal distribution, look for the z-score associated with a 0.95 cumulative probability.

z-score = 1.645

`Margin\text{ of error =1.645}*0.0238=0.0395`

The confidence interval is 0.87+0.0395 and 0.87-0.0395.

[tex]0.8305

Final answer:

[tex]0.8305