Question

Time, March 29, 1993, reported the proportions of adult Americans who favor "stricter gun-control laws." A telephone poll of 800 adult Americans, of who 374 were gun owners and 426 did not own guns, showed that 206 gun owners and 338 non-gun owners favor stricter gun-control laws. Let p1 and p2 be the respective proportions of gun owners and non-gun owners who favor stricter gun-control laws. Find a 95% confidence interval for p1 - p2.

A.

negative 0.306 less than p subscript 1 minus p subscript 2 less than negative 0.179

B.

0.225 less than p subscript 1 minus p subscript 2 less than 0.775

C.

negative 0.300 less than p subscript 1 minus p subscript 2 less than negative 0.170

D.

negative 0.332 less than p subscript 1 minus p subscript 2 less than negative 0.154

Answer 1

A. negative 0.306 less than p subscript 1 minus p subscript 2 less than negative 0.179

Explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

`p_A` represent the real population proportion of gun owners

`\hat p_A =(206)/(374)=0.551` represent the estimated proportion of gun owners

`n_A=374` is the sample size required for Brand A

`p_B` represent the real population proportion of not gun owners

`\hat p_B =(338)/(426)=0.793` represent the estimated proportion of not gun owners

`n_B=426` is the sample size required for Brand B

`z` represent the critical value for the margin of error

The population proportion have the following distribution

`p \sim N(p,\sqrt{(p(1-p))/(n)})`

The confidence interval for the difference of two proportions would be given by this formula

`(\hat p_A -\hat p_B) \pm z_(\alpha/2) \sqrt{(\hat p_A(1-\hat p_A))/(n_A) +(\hat p_B (1-\hat p_B))/(n_B)}`

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:

`(0.551-0.793) - 1.96 \sqrt{(0.551(1-0.551))/(374) +(0.793(1-0.793))/(426)}=-0.306`

`(0.551-0.793) + 1.96 \sqrt{(0.551(1-0.551))/(374) +(0.793(1-0.793))/(426)}=-0.179`

And the 95% confidence interval would be given (-0.306;-0.179).

We are confident at 95% that the difference between the two proportions is between
`-0.306 \leq p_A -p_B \leq -0.179`