Question

A survey of households in a small town showed that in 850 of 1,200 sampled households, at least one member attended a town meeting during the year. Using the 99% level of confidence, what is the confidence interval for the proportion of households represented at a town meeting?

Answer 1

Answer:

Hence the confidence interval is ( 0.6745, 0.7422).

Step-by-step explanation:

Now the given are  

Sample size = n = 1200  

x = 850  

Sample proportion is  

$\hat{p}=\frac{x}{n}=\frac{850}{1200}=0.7083$  

We have to construct 99% confidence interval for the population proportion.  

Formula Used:  

$(\hat{p}-E , \hat{p}+E)$  

Here E is a margin of error.  

$E =Zc\times\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$  

Zc = 2.58  

$E =2.58\times\sqrt{\frac{0.7083*(1-0.7083)}{1200}}\\\\E=2.58\times\sqrt{0.000172}=0.0339$  

So confidence interval is ( 0.7083 - 0.0339 , 0.7083 + 0.0339)

                                            = ( 0.6745 , 0.7422).