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

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".

[tex]p_A[/tex] represent the real population proportion of gun owners

[tex]\hat p_A =\frac{206}{374}=0.551[/tex] represent the estimated proportion of gun owners

[tex]n_A=374[/tex] is the sample size required for Brand A

[tex]p_B[/tex] represent the real population proportion of not gun owners

[tex]\hat p_B =\frac{338}{426}=0.793[/tex] represent the estimated proportion of not gun owners

[tex]n_B=426[/tex] is the sample size required for Brand B

[tex]z[/tex] represent the critical value for the margin of error

The population proportion have the following distribution

[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]

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

[tex](\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}[/tex]

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

[tex]z_{\alpha/2}=1.96[/tex]

And replacing into the confidence interval formula we got:

[tex](0.551-0.793) - 1.96 \sqrt{\frac{0.551(1-0.551)}{374} +\frac{0.793(1-0.793)}{426}}=-0.306[/tex]

[tex](0.551-0.793) + 1.96 \sqrt{\frac{0.551(1-0.551)}{374} +\frac{0.793(1-0.793)}{426}}=-0.179[/tex]

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 [tex]-0.306 \leq p_A -p_B \leq -0.179[/tex]