Suppose we have a sample of salaries for a job title and we know the population’s variance. In cases where we don’t know what the population variance is, we’ll use the Student’s t statistic. In this case we have the population’s standard deviation, even though that might seem to be a rare privilege. Suppose we have the following information.
- Sample size is 30 (n)
- Sample mean of $80,000
- Population standard deviation of $12,000 (σ) it’s rare to have this information
- Standard Error of 2.6
- Confidence Interval of 95% – this is something we set
In statistics, the standard deviation of a sample statistic is called the standard error. Does this make sense?
With a CI of 95%, we have an alpha of 0.05 (five percent), which is 1 minus 95%. using the z statistic table we get 1.96. Critical value is a common term used for z. The critical value for the 95% confidence interval is 1.96. Now we can substitute into the formula.
When we plug in our numbers we get the following.
The interval is $75,705.86 to $84,294.14. So we are 95% confident that the average salary is between $75,705 and $84,294.
