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

Step-by-step 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".  

represent the real population proportion of gun owners  

represent the estimated proportion of gun owners

is the sample size required for Brand A

represent the real population proportion of not gun owners  

represent the estimated proportion of not gun owners

is the sample size required for Brand B

represent the critical value for the margin of error  

The population proportion have the following distribution  

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

For the 95% confidence interval the value of and , with that value we can find the quantile required for the interval in the normal standard distribution.  

And replacing into the confidence interval formula we got:  

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

i did it myself and i got 4.2 (b) hope its right

step-by-step explanation: