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

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 for telephone

[tex]\hat p_A =0.35[/tex] represent the estimated proportion for telephone

[tex]n_A=200[/tex] is the sample size required for Brand A

[tex]p_B[/tex] represent the real population proportion for emali

[tex]\hat p_B =0.25[/tex] represent the estimated proportion for email

[tex]n_B=200[/tex] is the sample size required for Brand B

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

Solution to the problem

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](0.35-0.25) - 1.96 \sqrt{\frac{0.35(1-0.35)}{200} +\frac{0.25(1-0.25)}{200}}=0.0107[/tex]

[tex](0.35-0.25) + 1.96 \sqrt{\frac{0.35(1-0.35)}{200} +\frac{0.25(1-0.25)}{200}}=0.1892[/tex]

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