What is the 99% confidence interval?

Question

I'm having some trouble with this problem. The way I'd go about solving it is:

$56.2 +- 2.576 * \frac{53.6}{\sqrt5}$

However this does not give me the correct answer which is:

$56.2 +- 15.07$.

Can some one help me out?

EDIT:

I worked the problem out with the guidance from below:

56 +- 4.604 * (7.3212/sqrt(5))

Answer 1

The key to this problem is that they say you are working with $s^2$, which is an estimate of the population variance. You have applied a z-distribution to the problem, which would be correct if you had the actual population variance ($\sigma^2$) . However, when you are working with an estimate of the population variance you must apply a student's t-distribution to the problem.

The form of the t-distribution equation is similar to the z-distribution equation except that you replace the "z" with a "t", and the population standard deviation with the sample standard deviation (s):

$t = \frac{xbar - \mu}{\frac{s}{\sqrt{n}}}$

where $s$ is the square root of the sample variance, $n$ is the sample size, and $t$ is a value to be calculated from a t-table. You will need two things to find the t-value to plug into your equation:

1. Confidence interval - in this case you are asked for 99% confidence, meaning that your total acceptable Type I error rate ($\alpha$) is 1-.99 = .01. That error rate must be split over both ends of the t-distribution, thus the heading of the t - table column you seek should read .005.
2. Degrees of freedom ($n$). In this case, your Degrees of Freedom will be 5-1 = 4

Go to a t-table with a t($\alpha$/2=.005, DOF=4), and you will have enough information to find your answer.