Question

A relief fund is set up to collect donations for the families affected by recent storms. A random sample of 400 people shows that 28% of those 200 who were contacted by telephone actually made contributions compared to only 18% of the 200 who received first class mail requests. Which is the correct 95% confidence interval for the difference in the proportions of people who make donations if contacted by telephone or first class mail

Answer 1

Answer:

$(0.28-0.18) - 1.96 \sqrt{\frac{0.28(1-0.28)}{200} +\frac{0.18(1-0.18)}{200}}=0.0181$  

$(0.28-0.18) - 1.96 \sqrt{\frac{0.28(1-0.28)}{200} +\frac{0.18(1-0.18)}{200}}=0.182$  

And the 95% confidence interval would be given (0.0181;0.181).  

We are confident at 95% that the difference between the two proportions is between $0.0181 \leq p_B -p_A \leq 0.182$

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".  

Solution to the problem

$p_A$ represent the real population proportion for telephone  

$\hat p_A =0.28$ represent the estimated proportion for telephone

$n_A=200$ is the sample size required for telephone

$p_B$ represent the real population proportion for mail

$\hat p_B =0.18$ represent the estimated proportion for mail

$n_B=200$ is the sample size required for mail

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

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:  

$(0.28-0.18) - 1.96 \sqrt{\frac{0.28(1-0.28)}{200} +\frac{0.18(1-0.18)}{200}}=0.0181$  

$(0.28-0.18) - 1.96 \sqrt{\frac{0.28(1-0.28)}{200} +\frac{0.18(1-0.18)}{200}}=0.182$  

And the 95% confidence interval would be given (0.0181;0.181).  

We are confident at 95% that the difference between the two proportions is between $0.0181 \leq p_B -p_A \leq 0.182$