Suppose a realtor wants to determine the current percentage of customers who have a family of five or more. how many customers should the realtor survey in order to be 98% confident that the estimated (sample) proportion is within 2 percentage points of the true population proportion of customers who have a family of five or more?

n=3394

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".  

The population proportion have the following distribution

Solution to the problem

The confidence interval for the mean is given by the following formula:  

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 98% of confidence, our significance level would be given by and . And the critical value would be given by:

The margin of error for the proportion interval is given by this formula:  

   (a)  

And on this case we have that and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

  (b)  

We can use as estimation of since we don't have any other info provided. And replacing into equation (b) the values from part a we got:

And rounded up we have that n=3394

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