Answer:
(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.
