Question

According to a dental association, 63% of all dentists use nitrous oxide (laughing gas) in their practice. In a random sample of 500 dentists, let Prepresent the proportion who use laughing gas in practice. Completeparts a through d belowa. Find E()EO) -(Round to two decimal places as needed.)b. Find os04- (Round to four decimal places as nooded.)c. What does the Central Limit Theorem say about the shape of the sampling distribution of ?O A. The Central Limit Theorem says that as the sample size grows largo, the sampling distribution of will approach the population distributionO B. The Central Limit Theorem says that the shape of the sampling distribution of is completely determined by the population parameter p.OC. The Central Limit Theorem says that as the sample size grows largo, the sampling distribution of will becomo approximately normal.OD. The Contral Limit Theorem does not say anything about the sampling distribution of because it is not a population mean,d. Find PC >046)P>0.48) - (Round to three decimal places as nooded.)

Answer 1

`\begin{gathered} a)\text{ Find }E(\hat{p}) \\ E(\hat{p})=0.63,\text{ from the 63\% of all dentist who use nitrous oxide} \\ b\mathrm{})\text{ Find }\sigma_{\hat{p}} \\ \sigma_{\hat{p}}=\sqrt[]{\frac{\hat{p}(1-\hat{p})}{n}} \\ \sigma_{\hat{p}}=\sqrt[]{(0.63(1-0.63))/(500)} \\ \sigma_{\hat{p}}=\sqrt[]{((0.63)(0.37))/(500)} \\ \sigma_{\hat{p}}=0.2331 \\ c)\text{ The Central Limit Theorem says that as the sample size grows large,} \\ \text{the sampling distribution of }\hat{p}\text{ will become approximately normal} \end{gathered}`
`\begin{gathered} d) \\ \text{Convert to a z-value} \\ z=\frac{0.46-0.63}{\sigma_{\hat{p}}} \\ z=(-0.17)/(0.2331) \\ z=-0.7293 \\ \text{Find }p(z>-0.7293) \\ P(z>-0.7293)=P\mleft(Z<-0.7293\mright)+0.5 \\ P(z>-0.7293)=0.233+0.5 \\ P(z>-0.7293)=0.733 \\ P(\hat{p}>0.46)=0.733 \end{gathered}`