Question

A random sample of 1000 people was taken. Four hundred fifty of the people in the sample favored Candidate AT. The 95% confidence interval for the true proportion of people who favor Candidate A is a. .424 to .476. b. .419 to .481. c. .40 to .50. d. .45 to .55.

Answer 1

`0.45 - 1.96 \sqrt{(0.45(1-0.45))/(1000)}=0.419`

`0.45 + 1.96 \sqrt{(0.45(1-0.45))/(1000)}=0.481`

And the 95% confidence interval would be given (0.419;0.481). And the best option would be:

b. .419 to .481

Explanation:

We know the following info:

`n = 1000` sample size selected

`X= 450` represent the number of people who favored Candidate AT

The sample proportion would be:

`\hat p=(450)/(1000)=0.45`

The confidence interval would be given by this formula

`\hat p \pm z_(\alpha/2) \sqrt{(\hat p(1-\hat p))/(n)}`

For the 95% confidence interval the value of
`\alpha=1-0.95=0.05` and
`\alpha/2=0.025`, with that value we can find the quantile required for the interval in the normal standard distribution.

`z_(\alpha/2)=1.96`

And replacing into the confidence interval formula we got:

`0.45 - 1.96 \sqrt{(0.45(1-0.45))/(1000)}=0.419`

`0.45 + 1.96 \sqrt{(0.45(1-0.45))/(1000)}=0.481`

And the 95% confidence interval would be given (0.419;0.481). And the best option would be:

b. .419 to .481