Suppose we wanted the margin of error for the 99% confidence level to be about 2.75%. What is the smallest sample size we could take to achieve this? Note: For consistency's sake, round your z* value to 3 decimal places before calculating the necessary sample size.

And rounded up we have that n=2194

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

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 99% 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)  

Since we don't have prior information for the true proportion we can use as an estimator. And replacing into equation (b) the values from part a we got:

And rounded up we have that n=2194

answer:

c. the third graph

stops on an even number.