The formula for margin of error is e = (zσ)/√n. We have everything we need in the beginning except the z-statistic for the 95% confidence level. This can be obtained by various methods with standard normal tables and/or technologies. For a total area of 0.95 that is symmetrical around the mean, each half would have an area of 0.475. If we have the capability of finding the cumulative left-tail area, then we need the z score to be 0.5 + 0.475 = 0.975. So for a 95% confidence level, the z-statistic is 1.96. So we can use the formula now with these numbers:
(1.96·32)/√84 = 6.843.
For the second question, we can calculate the margin of error, which I'll do down below. But using the concepts and the mathematical relationships, we can intuit the answer as well. We would expect a smaller sample size to give us less reliable information. But if the confidence level d.oesn't change, then the margin of error would have to change. So when our information is less reliable, then we can intuitively see that the margin of error would have to be larger.
But we can also see from the mathematical relationships that, if n is smaller, then we have a smaller denominator, which means a larger quotient, and therefore a larger margin of error.
This new, larger margin of error would be (1.96·32)/√75 = 7.242.
