Researchers are interested in determining whether more women than men prefer the beach to the mountains. In a random sample of 200 women, 45% prefer the beach, whereas in a random sample of 300 men, 52% prefer the beach. What is the 99% confidence interval estimate for the difference between the percentages of women and men who prefer the beach over the mountains?a. − 0.07 ± 0.1172b. − 0.07 ± 0.0557c. − 0.07 ± 0.0053d. 0.07 ± 0.01172e. 0.07 ± 0.0053

The margin of error would be:

So then the correct option for this case would be:

a. − 0.07 ± 0.1172

Step-by-step explanation:

Previous concepts

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 men

represent the estimated proportion of men

the random sample for male

represent the real population proportion of female

represent the estimated proportion of female

the random sample for female

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 99% confidence interval the value of and , with that value we can find the quantile required for the interval in the normal standard distribution.  

The standard error is given by:

The margin of error would be:

And replacing into the confidence interval formula we got:  

So then the correct option for this case would be:

a. − 0.07 ± 0.1172

answer: i guess 5 units?

step-by-step explanation: