Graduation. It's believed that as many as 25% of adults over 50 never graduated from high school. We wish to see this percentage is the same among the 25 to 30 age group.

a. How many of this younger age group must we survey in order to estimate the proportion of non-grads to within 6% with 90% confidence?
b. Suppose we want to cut the margin of error to 4%. What' s the necessary sample size?
c. What sample size would produce a margin of error of 3%?

a)  

And rounded up we have that n=141

b)  

And rounded up we have that n=316

c)  

And rounded up we have that n=561

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

Part a

The estimated proportion for this case is

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 90% 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)  

And replacing into equation (b) the values from part a we got:

And rounded up we have that n=141

Part b

And rounded up we have that n=316

Part c

And rounded up we have that n=561

i don't think there is enough