Question

A random sample of 860 births in a state included 429 boys. Construct a 95% confidence interval estimate of the proportion of boys in all births. It is believed that among all births, the proportion of boys is 0.512. Do these sample results provide strong evidence against that belief?

Answer 1

Step-by-step explanation

Given

`\begin{gathered} x=429 \\ n=860 \\ c=95\text{\%=0.95} \end{gathered}`

We can fid

Part A: Construct a 95% confidence interval

The best point estimate of the population proportion p is the sample proportion. The sample proportion is the number of successes divided by the sample size

`\hat{p}=(x)/(n)=(429)/(860)\approx0.4989`

`For\text{ confidence level 1-}\alpha=0.95,\text{ we have z}_{(\alpha)/(2)}=z_(0.025)=1.96`

Therefore, we can calculate the margin of error as

`\begin{gathered} E=z_{(\alpha)/(2)}*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}=1.96*\sqrt{(0.4989(1-0.4989))/(860)}=1.96*\sqrt{(0.4989*0.5011)/(860)} \\ =1.96*\sqrt{2.90696*10^(-4)} \\ =1.96*0.0170498\approx0.0334 \end{gathered}`

The boundaries of the confidence interval are then

Answer

`\begin{gathered} \hat{p}-E=0.4989-0.0334=0.4655 \\ \hat{p}+E=0.4989+0.0334=0.5323 \end{gathered}`

Part B:

We are 95% confident that the true proportion of boys in all births is between 0.4655 and 0.5323

By observation, we can see that 0.512 is between 0.4655 and 0.5323, thus these sample results do not provide strong evidence against the belief that the proportion is 0.512.

Answer: No