Question

Out of 200 people sampled, 110 had kids. Based on this, construct a 95% confidence interval for the true population proportion of people with kids.

Answer 1

Answer:

From the sample of n= 200 people,

The proportion of people with kids is:

p'= 110/200= 0.55

The confidence interval for population proportion is given by:

$p' - Z_{\alpha /2} \sqrt[]{\frac{p'(1-p')}{n} } } \leq P\leq p' + Z_{\alpha /2}\sqrt[]{\frac{p'(1-p')}{n} } }$

$= p' - Z_{\alpha /2} \sqrt[]{\frac{0.55(0.45)}{200} } } \leq P \le p' + Z_{\alpha /2} \sqrt[]{\frac{0.55(0.45)}{200} } }$

$= p' - Z_{\alpha /2} 0.0352} } } \leq P\le p' + Z_{\alpha /2} 0.0352} }$

Where P is the population proportion and Z is the critical value at a given level of significance α.