Out of 400 people sampled, 276 had kids. Based on this, construct a 95% confidence interval for the true population proportion of people with kids.

Answer 1

Relevant formulas:

Confidence Interval = p ± Zcrit×(sqrt (pq/n))

Population proportion (p) = X/N

q = 1 - p

Margin of Error (E) = Zcrit×(sqrt (pq/n))

Step 1: State Your Givens

X = 276

N = 400

p = 0.69 (**We got this by taking the amount of people who had kids / total people sampled**)

q = 0.31 (**We got this answer by taking 1 - .69.**)

Z_α/2=Z_0.025=1.96

Step 2: Calculate Margin of Error with Given Information

Margin of Error (E) = Zcrit×(sqrt (pq/n))

=1.96 × sq rt ((.69)(.31)/400)

=0.045

Step 3: Calculate Confidence Interval

A 95% confidence interval for the population proportion (p) is,

p-E < p < p+E

.69 - 0.045 < p < .69 + 0.045

0.645 < p < 0.735

Step 4: Interpret Your Results

We are 95% confident that the proportion of the population of people with kids will be between 64.5% and 73.5%.