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

Answer:

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

Step-by-step explanation:

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

$\alpha = \frac{1-0.9}{2} = 0.05$

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.