# Solved – p-value of t-test versus F-test(joint hypothesis)

SAS code :

``data pr;     input y x1 x2 @@;     cards;     5 2 2 7 3 3 3 2 0     5 2 4 4 3 3 7 2 4     ;   run;   proc reg data = pr;     model y = x1 x2; run; quit;   ``

When I run the regression procedure,
I got the result

``[F-test : p-value for model = 0.3757]   [t-test : p-value for x1 = 0.9238]   [t-test : p-value for x2 = 0.2045]   ``

I'm wondering why the p-value of F-test is bigger than that of t-test for x2?

F test has the joint hypothesis H0 : beta1 = beta2 = 0
so, it has the more complicated hypothesis than t test (H0 : beta2 = 0)

My logic is this:
when hypothesis has the more restrictions, then it is far difficult to realize,
so p-value gets lower.

Where am i misunderstood?

Contents

Good question.
Now obviously the results are not comparable in the sense that the F-Test you test (in this case) two linear combinations. It is therefore outside of the scope of the t-Test.
I will therefore use \$H0_F\$ and \$H0_t\$ to discriminate the two hypothesis.

With that being said, an intuitive approach to understand what is going on is to look at what you are testing.
Your p value for the F-Tests gives you the probability for the event that the value from the F-Statistic from your data would have been observed the way it is, if the the \$H0_F\$ hypothesis was true (ie. if the F-Statistic really had been F-distributed), ie. \$P(Data|H0_F)\$. For that to be an F-Distributed statistic, your model obviously has to be correctly specified (ie. if your error is non normal, no dice) because the F-Test compares the residual variation of the two models. So we are saying, in essence, the model without our \$beta\$ we like better than the model with.

The p-value for the t-test assesses the probability of the the t-statistic giving you a value as such, if your \$beta_i\$ would really have been zero and your model was truly correct. Therefore it is essentially also \$P(Data|H0_t)\$
But careful! This is not a t-Test where you assess the mean of the x2 variable. It is a t-test for the estimated influence of said variable. As such, it is based on the combined distribution of \$x_i\$ and \$y_i\$, the difference being that for \$H0_t\$ to be true, your complete model has to be true. So your \$H0_t\$ really says \$beta_2=0\$ AND also the random influence comes from a non systematic, gaussian error term.