Question

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

Answer 1

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

Explanation:

Answer 2

`(0.35-0.25) - 1.96 \sqrt{(0.35(1-0.35))/(200) +(0.25(1-0.25))/(200)}=0.0107`

`(0.35-0.25) + 1.96 \sqrt{(0.35(1-0.35))/(200) +(0.25(1-0.25))/(200)}=0.1892`

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

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".

`p_A` represent the real population proportion for telephone

`\hat p_A =0.35` represent the estimated proportion for telephone

`n_A=200` is the sample size required for Brand A

`p_B` represent the real population proportion for emali

`\hat p_B =0.25` represent the estimated proportion for email

`n_B=200` is the sample size required for Brand B

`z` represent the critical value for the margin of error

Solution to the problem

The population proportion have the following distribution

`p \sim N(p,\sqrt{(p(1-p))/(n)})`

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

`(\hat p_A -\hat p_B) \pm z_(\alpha/2) \sqrt{(\hat p_A(1-\hat p_A))/(n_A) +(\hat p_B (1-\hat p_B))/(n_B)}`

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

`z_(\alpha/2)=1.96`

And replacing into the confidence interval formula we got:

`(0.35-0.25) - 1.96 \sqrt{(0.35(1-0.35))/(200) +(0.25(1-0.25))/(200)}=0.0107`

`(0.35-0.25) + 1.96 \sqrt{(0.35(1-0.35))/(200) +(0.25(1-0.25))/(200)}=0.1892`

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