Question

A researcher wants to find a 90% confidence interval for the population proportion of those who support additional handgun control. She collects an SRS of 80 people, 50 of whom say they support additional controls. Which of these is the correct confidence interval?a. (.52, .73)b. (.54, .71)c. (.49, .76)d. (.51, .75)e. (.58, .68)

Answer 1

Answer: b. (.54, .71)

Explanation:

Confidence interval for population proportion is given by :-

`\hat{p}\pm z\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}`

,where
`\hat{p}` = sample proportion

z= Critical z-value

n= sample size.

Let p be the proportion of people who support additional handgun control.

As per given , we have

n= 80

`\hat{p}=(50)/(80)=0.625`

Critical z-value for 90% confidence interval is 1.645

Now , a 90% confidence interval for the population proportion of those who support additional handgun control will become:

`0.625\pm (1.645)\sqrt{(0.625(1-0.625))/(80)}`

`=0.625\pm (1.645)√(0.0029296875)`

`=0.625\pm 0.089\\\\=(0.625-0.089, 0.625+0.089)\\\\=(0.536,\ 0.714)\approx(0.54,\ 0.71)`

So the correct answer is : b. (.54, .71)