Question

2. Newsweek performed a poll in which 567 American parents were asked the question and found that only 12% agreed with the question , “Would you prefer to have your child taught by a male for grades K-12?”. Find the 95% C.I. for the true proportion of Americans agreeing with their child being taught by a male for grades K-12.

Answer 1

Let $\hat{p}$ be the sample proportion. When $n$  is large and $p$ is not close to zero or one, we can use the normal distribution to approximate the binomial.

The sampling distribution of $\hat{p}$ is normal.

The mean of the sampling distribution is $p.$

The standard deviation of the sampling distribution (called the standard error) is

$$SE=\sqrt{{p(1-p) \over n}}$$

Given that $n=576, p=0.12.$

$$SE=\sqrt{{0.12(1-0.12) \over 576}}\approx0.013540$$

Confidence Level 95%: $z^*-value=1.96$

Confidence interval for $p$ is

$$\hat{p}\pm z^*\times SE$$

$$0.12\pm 1.96\times 0.013540\approx0.12\pm0.0265$$

$$95\%\ CI=(0.0935, 0.1465)$$