How do you construct a 99 percent confidence interval estimate of the mean?

Answer 1

First of all, we need to find the margin of error which we will call E.

This will be the greatest possible distance between our point estimate and the value of the parameter, the mean, we are estimating.

We need the critical z score, (or possibly critical t score) that is to say the z score corresponding to 99% confidence interval. This means 99% of the area under the standard normal curves falls within 2.576 standard deviations of the mean.

Here is how you find 2.576. First, divide the confidence level by two, and then look up the area in the the Z-table and look up the z-score on the outside. .99/2 =.495 The z table or the calculator give you 2.576.

(The area in one tail is 0.0050, and some people find this easier to see)

Next, use your sample mean, x bar, as your point estimate. The lower limit of the confidence interval will be x bar- E and the upper limit will be x bar + E.

( By the way, x bar is used here to indicate the letter x with a bar over it which is the standard notation for the sample mean)

Now we need to find E.

We need the critical z score we found to do this. The notation zc is often used to denote this and is used below.

If sigma, the population standard deviation is known, we use it. Otherwise if the sample size is greater than or equal to 30, we find the sample standard deviation s and use it as an estimate for sigma. (When we do this, we will not be able to use a z score if n is less than 30, we will need to use s/√n instead of s)

E will equal the product of the critical z score and ( sigma divided by the square root of n)

E=zc (Sigma/√n) or if we use s, E=zc (s/√n)

If n is less than 30, we use a t-distribution and instead of s we use s/√n .

So E will be( tc s/√n ) and once again we will add an subtract this value from x bar.

There will be n-1 degrees of freedom. The critical t will be talpha/2

(It is important to understand that the true mean, MU exists. There are only two possibilities. Either it is in the interval or not. It is not accurate to state that there is a 99% probability that the actual mean is in the interval. You should say there is a 99% probability that the confidence interval contains MU.)

Here is a summary of which test to use:

First is sigma known? If it is not and the population is normally distributed, use the t distribution. If the population is not normally distributed and n>30, we also use the t distribution. If the population is not normally distributed and n is less than or 30 we need nonparametric methods.

If sigma is known, and the population is normally distributed, use the z scores. If sigma is known and the population is not normally distributed, you can use the z distribution if n>30. If not, once again we need nonparametric methods.