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

Hence the confidence interval is ( 0.6745, 0.7422).

Explanation:

Now the given are

Sample size = n = 1200

x = 850

Sample proportion is

`\hat{p}=(x)/(n)=(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*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}`

Zc = 2.58

`E =2.58*\sqrt{(0.7083*(1-0.7083))/(1200)}\\\\E=2.58*√(0.000172)=0.0339`

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

= ( 0.6745 , 0.7422).