 How to Use the CONFIDENCE Function in Excel - dummies

# How to Use the CONFIDENCE Function in Excel

The CONFIDENCE.NORM and CONFIDENCE.T functions in Excel calculate a value that you can use to create a confidence intervals for population means based on the sample mean. These definitions amount to a mouthful, but in practice what these functions do is straightforward.

Suppose that, based on a sample, you calculate that the mean salary for a chief financial officer for a particular industry equals \$100,000. You might wonder how close this sample mean is to the actual population mean. Specifically, you might want to know what range of salaries, working at a 95-percent confidence level, includes the population mean.

The CONFIDENCE.NORM function calculates the number that you use to create this interval using the syntax

`=CONFIDENCE.NORM(alpha,standard_dev,size)`

where alpha equals 1 minus the confidence level, standard_dev equals the standard deviation of the population, and size equals the number of values in your sample.

If the standard deviation for the population equals \$20,000 and the sample size equals 100, use the formula

`=CONFIDENCE.NORM(1-.95,20000,100)`

The function returns the value \$3920 (rounded to the nearest dollar). This interval suggests that if the average chief financial officer’s salary in your sample equals \$100,000, there’s a 95-percent chance that the population mean of the chief financial officers’ salaries falls within the range \$96,080 to \$103,920.

The CONFIDENCE.T function works in roughly the same way arguments do, but uses a Student T-distribution rather than a normal distribution. The CONFIDENCE.T function uses the following syntax:

`=CONFIDENCE.T(alpha,standard_dev,size)`

where alpha equals 1 minus the confidence level, standard_dev equals the standard deviation of the population, and size equals the number of values in your sample.

If the standard deviation for the population equals \$20,000 and the sample size equals 100, use the formula

`=CONFIDENCE.T(1-.95,20000,100)`

The function returns the value \$3968 (rounded to the nearest dollar). This interval suggests that if the average chief financial officer’s salary in your sample equals \$100,000, there’s a 95-percent chance that the population mean of the chief financial officers’ salaries falls within the range \$96,032 to \$103,968.