A study is conducted comparing ALT liver enzyme values in individuals who consume alcohol regularly and those who do not.  The data are shown below. Alcohol use N Mean SD Yes 26 56.5 12.5 No 35 49.9 9.6   Generate a 99% confidence interval for the difference in mean ALT levels between those who consume alcohol regularly and those who do not.  I am trying to find out if there is a way to do this in Excel?

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2-Proportion Test

In a 2-proportion test, two separate proportions are measured at the same time, and their similarity is evaluated. It is also known as a 2-proportion z-test, and a comparison of two different proportions is done in this method.