8.3 confidence intervals for proportion (z-table)
out of 300 people sampled, 27 had kids. based on this, construct a 95% confidence interval for the true population
proportion of people with kids.
give your answers as decimals, to three places

(0.058, 0.122)

Step-by-step explanation:

The confidence interval of a proportion is:

CI = p ± SE × CV,

where p is the proportion, SE is the standard error, and CV is the critical value (either a t-score or a z-score).

We already know the proportion:

p = 27/300

p = 0.09

But we need to find the standard error and the critical value.

The standard error is:

SE = √(p (1 − p) / n)

SE = √(0.09 × (1 − 0.09) / 300)

SE = 0.0165

To find the critical value, we must first find the alpha level and the degrees of freedom.

The alpha level for a 95% confidence interval is:

α = (1 − 0.95) / 2 = 0.025

The degrees of freedom is one less than the sample size:

df = 300 − 1 = 299

Since df > 30, we can approximate this with a normal distribution.

If we look up the alpha level in a z score table, we find the z-score is 1.96.  That's our critical value.  CV = 1.96.

Now we can find the confidence interval:

CI = 0.09 ± 0.0165 * 1.96

CI = 0.09 ± 0.0324

CI = (0.058, 0.122)

So we are 95% confident that the proportion of people with kids is between 0.058 and 0.122.

i think asking google will you get the answer to that question