Question

A researcher estimates the 90% CI for a sample with a mean of M = 8.9 and a standard error (sigma_M) of 2.1. What is the confidence interval at this level of confidence?

Answer 1

The confidence interval at this level of confidence is between 5.4455 and 12.3545.

Explanation:

We have that to find our
`\alpha` level, that is the subtraction of 1 by the confidence interval divided by 2. So:

`\alpha = (1-0.9)/(2) = 0.05`

Now, we have to find z in the Ztable as such z has a pvalue of
`1-\alpha`.

So it is z with a pvalue of
`1-0.05 = 0.95`, so
`z = 1.645`

Now, find M as such

`M = z*s`

In which s is the standard deviation of the sample, which is also called standard error. So

`M = 1.645*2.1 = 3.4545`

The lower end of the interval is the sample mean subtracted by M. So it is 8.9 - 3.4545 = 5.4455

The upper end of the interval is the sample mean added to M. So it is 8.9 + 3.4545 = 12.3545

The confidence interval at this level of confidence is between 5.4455 and 12.3545.