Mathematics, 02.04.2020 01:46 littlebirdd987

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

Answers: 2

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Answer from: StacySams7361

$(0.35-0.25) - 1.96 \sqrt{\frac{0.35(1-0.35)}{200} +\frac{0.25(1-0.25)}{200}}=0.0107$

$(0.35-0.25) + 1.96 \sqrt{\frac{0.35(1-0.35)}{200} +\frac{0.25(1-0.25)}{200}}=0.1892$

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

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".

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

represent the critical value for the margin of error

Solution to the problem

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

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

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

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:

$(0.35-0.25) - 1.96 \sqrt{\frac{0.35(1-0.35)}{200} +\frac{0.25(1-0.25)}{200}}=0.0107$

$(0.35-0.25) + 1.96 \sqrt{\frac{0.35(1-0.35)}{200} +\frac{0.25(1-0.25)}{200}}=0.1892$

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

Answer from: Quest

Bi’m pretty sure hope it works

Answer from: Quest

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