How do you find the 95% confidence interval?

Find the 95% confidence interval for a sample of size 39 with a mean of 20.3 and a standard deviation of 10.2.

Jan 28, 2017

A 95% confidence interval with the given data gives: $\left(17.1 , 23.5\right)$

Explanation:

The confidence interval is given by the formula:

$\overline{x} \pm z \left(\frac{\sigma}{\sqrt{n}}\right)$

Or you can write it as:

$\left(\overline{x} - z \left(\frac{\sigma}{\sqrt{n}}\right) , \overline{x} + z \left(\frac{\sigma}{\sqrt{n}}\right)\right)$

Where $\overline{x}$ is the sample mean, $z$ is the standardized value for the normal distribution, in relation to the percentage (don't really know how to explain this one that well), $\sigma$ is the standard deviation, and $n$ is the sample size.

We know all the values except for $z$. To find the $z$ value, we can imagine a normal distribution graph with 95% of it shaded, where the middle of this is the mean.

As you can see from this picture the $z$ value is 1.96. This can be found by using a normal distribution percentage points look up table.
I hope you can see that to the left and right of the shaded area, 2.5% is taken up by the white space each side.

So therefore you do 95%+2.5% = 97.5% then you can look that value up in the tables which is in fact: 1.96.

Now you can just substitute all the numbers into the expression:

$\left(20.3 - 1.96 \left(\frac{10.2}{\sqrt{39}}\right) , 20.3 + 1.96 \left(\frac{10.2}{\sqrt{39}}\right)\right)$

Enter this into your calculator and you get:

$\left(17.1 , 23.5\right)$

Hope this helps!