Suppose a repairman wants to determine the current percentage of customers who keep up with regular house maintenance. how many customers should the repairman survey in order to be 98% confident that the estimated (sample) proportion is within 5 percentage points of the true population proportion of customers who keep up with regular house maintenance?

n=543

Step-by-step explanation:

1) Notation and definitions

random sample (variable of interest)

estimated proportion of interest

true population proportion of interest

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

2) Solution to the problem

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 (5% points means 0.05 on fraction) and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

  (b)  

Since we don't have a prior estimate for we can use 0.5 as the prior estimate, and replacing into equation (b) the values from part a we got:

And rounded up we have that n=543

sure i'm up for it

step-by-step explanation: