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Question from Don, a student:

I have a t-test statistic of 1.28 and a degree of freedom of 6.
I know the p-value is .248 I cannot figure out how to calculate that value.
I have used software to get it but I want to know how to calculate it using the t-distribution table.

Hi Don,

An extremely nice feature of the normal distributions is that any random normal variable X with mean μ and standard deviation σ can be transformed to the standard normal random variable Z by the transformation Z = (X - μ)/σ. A consequence of this is that Pr(0 < X < k) = Pr(0 < (X - μ)/σ < (k - μ)/σ) = Pr(0 < Z < (k - μ)/σ) and hence Pr(0 < X < k) can be evaluated using the standard normal table.

There is no similar "standard t-distribution". If X has the t-distribution with 6 degrees of freedom and you want to evaluate Pr(0 < X < k) you need a table for the t-distribution with 6 degrees of freedom. If X has the t-distribution with 8 degrees of freedom and you want to evaluate Pr(0 < X < k) you need a table for the t-distribution with 8 degrees of freedom. There are reference books that have tables of t-distributions for various degrees of freedom but most statistics books have one table with a few values for degrees of freedom from 1 to 30.

If you want to find a p-value where the test statistic has a t-distribution there are two options commonly used. If the degrees of freedom are large (usually over 30) then the t-distribution and standard normal distribution are approximately equal so you can use the standard normal table. A better solution is the one you have found, that is to use software to calculate the p-value.

Harley

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