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

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

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

represent the real population proportion for telephone  

represent the estimated proportion for telephone

is the sample size required for telephone

represent the real population proportion for mail

represent the estimated proportion for mail

is the sample size required for mail

represent the critical value for the margin of error  

The population proportion have the following distribution  

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

For the 95% confidence interval the value of and , with that value we can find the quantile required for the interval in the normal standard distribution.  

And replacing into the confidence interval formula we got:  

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

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