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prohojiy [21]
3 years ago
8

A relief fund has been set up to collect donations for the families affected by a recent hurricane. A random sample of 400 peopl

e 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
Mathematics
2 answers:
Yuki888 [10]3 years ago
7 0

Answer:

(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

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{\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

Minchanka [31]3 years ago
5 0

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

Step-by-step explanation:

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