# What is the 99% confidence interval?

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))

• You have to use square root of 53.6
– user83346
Commented Oct 19, 2017 at 4:23
• 99% CI's are in many cases meaningless, you could as well use 100% interval: $(-\infty, \infty)$, in both cases it would be very wide and would not provide you much information.
– Tim
Commented Oct 19, 2017 at 7:54

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.

• So just to clarify: If you are given an estimate you ALWAYS have to use the t-statistic? Also, when do you seek 0.01/2 or 0.005 vs 0.01? Because some questions I simply use the value of 2.576. Is that only without degrees of freedom? Commented Oct 19, 2017 at 2:38
• Yes, if the sample is 'small'. The t-distribution converges on the normal distribution as the number of samples grows. By the time you have 50 samples from an infinite population, the difference is usually irrelevant. Commented Oct 19, 2017 at 5:28