Question

A population proportion is 0.5. A sample of size 300 will be taken and the sample proportion will be used to estimate the population proportion. Use z-table. Round your answers to four decimal places. Do not round intermediate calculations. a. What is the probability that the sample proportion will be within +0.04 of the population proportion? b. What is the probability that the sample proportion will be within 20.06 of the population proportion?

Answer 1

Given : n=300 , p=0.5

Since , the sampling distribution of the sample proportion is , $\bar{p}\to N(\mu_{\bar{p}}=p,\sigma_{\bar{p}}=\sqrt{\frac{p(1-p)}{n}})$

a. We have ,

$P(-0.04<\bar{p}-\mu_{\bar{p}}<0.04)=P(-\frac{0.04}{\sqrt{\frac{0.5(1-0.5)}{300}}}<\frac{\bar{p}-\mu_{\bar{p}}}{\sigma_{\bar{p}}}<\frac{0.04}{\sqrt{\frac{0.5(1-0.5)}{300}}})$

$=P(-\frac{0.04}{0.04}<Z<\frac{0.04}{0.04})=P(-1<Z<1)=P(Z<1)-P(Z<-1)$

$=\Phi(1)-\Phi(-1)=\Phi(1)-[1-\Phi(1)]=2*\Phi(1)-1$

$=2*0.8413-1=0.6828$ ; From standard normal distribution table

b. We have ,

$P(-0.06<\bar{p}-\mu_{\bar{p}}<0.06)=P(-\frac{0.06}{\sqrt{\frac{0.5(1-0.5)}{300}}}<\frac{\bar{p}-\mu_{\bar{p}}}{\sigma_{\bar{p}}}<\frac{0.06}{\sqrt{\frac{0.5(1-0.5)}{300}}})$

$=P(-\frac{0.06}{0.04}<Z<\frac{0.06}{0.04})=P(-1.5<Z<1.5)$

$=P(Z<1.5)-P(Z<-1.5)=\Phi(1.5)-\Phi(-1.5)=\Phi(1.5)-[1-\Phi(1.5)]$

$=2*\Phi(1.5)-1=2*0.9332-1=0.8664$ ; From standard normal distribution table