Question

T Score: 1.049
DF: 14

Significance Level: .05

One-tailed

The P-Value is .155975.

The result is not significant at p < .05.

Here's my Question. Can someone explain in detail how this calculator comes up with this P- value. Can it be solved by hand? None of the calculators online show any work.

Thank you

Answer 1

Here, T-score is 1.049. That is, it is the value of the test statistic and the distribution of test statistic is student t distribution with 14 degrees of freedom. Since the hypothesis is one-tailed, we compute the p-value by using following formula.

P-value = P(T > value of test statistic) =  P(T < value of test statistic)

-----------------------------------(since student t distribution is symmetric around 0)

where T is t variate with desired degrees of freedom.

Here T follows student t distribution with 14 degrees of freedom.

Therefore,

P-value = P(T>1.049) =  P(T<1.049) = 0.155975

The above probability is obtained in Excel using "TDIST(1.049,14,1)".