A relief fund has been set up to collect donations for the families affected by a recent hurricane. A random sample of 400 people shows that 35% of the 200 who were contacted by telephone made contributions, compared with only 25% of the 200 who received email requests. Which of the formulas calculates the 95% confidence interval for the difference in the proportions of people who make donations when contacted by telephone versus those contacted by email?

a. (0.35- 0.25)±1.96 √ (0.35)(0.25)/200

b. (0.35- 0.25)±1.96 √ (0.35)(0.25)/400

c. (0.35- 0.25)±1.96 √ (0.35)(0.25)/200+ √ (0.35)(0.25)/200

d. (0.35- 0.25)±1.96 √ (0.35)(0.65)/200+ √ (0.35)(0.75)/200

e. (0.35- 0.25)±1.96 √ (0.35)(0.65)/400+ √ (0.35)(0.75)/400

And the 95% confidence interval would be given (0.0107;0.1892).  

And the best answer would be:

d. (0.35- 0.25)±1.96 √ (0.35)(0.65)/200+ √ (0.35)(0.75)/200

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 for telephone

represent the estimated proportion for telephone

is the sample size required for Brand A

represent the real population proportion for emali

represent the estimated proportion for email

is the sample size required for Brand B

represent the critical value for the margin of error  

Solution to the problem

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 (0.0107;0.1892).  

And the best answer would be:

d. (0.35- 0.25)±1.96 √ (0.35)(0.65)/200+ √ (0.35)(0.75)/200

529 and 531

step-by-step explanation:

lol.

it's ok we all do this sometimes.

-aparri