Question

An airline recently polled all of its loyalty card customers and discovered that 22.5% are planning on taking a vacation this year. A flight scheduler for the airline takes a simple random sample of 144 customers. With a 30% chance, the scheduler's sample proportion will be no greater than what value of p ^?

Answer 1

Given that

n=144, p = 0.225

The sampling distribution of sample proportion will be approximately normal with mean

$\mu_{\hat{p}}=p=0.225$

and standard deviation

$\sigma_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.225\cdot 0.775}{144}}=0.0348$

Here we need to find  p ^ such that

$P(\text{less than equal to } \hat{p})=0.30$

We need z-score that have 0.30 area to its left. Using z table, z-score -0.53 has 0.30 area to its left.

So required sample proportion is

$z=\frac{\hat{p}-\mu_{\hat{p}}}{\sigma_{\hat{p}}}=\frac{\hat{p}-0.225}{0.0348}$

$-0.53=\frac{\hat{p}-0.225}{0.0348}$

$\hat{p}=0.206556$

Answer: 0.2066