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11. Likehood Ratio and Other Tests 525

With α = .01 and 7 degrees of freedom, one has t 7,.01 = 2.9980. But, since u calc
does not go below –t 7,.01 , we do not reject the null hypothesis at 1% level. In
other words, at 1% level, we conclude that the job training has not been
effective. !
11.4.2 LR Test for the Correlation Coefficient

With fixed α ∈ (0, 1), we wish to construct a level α LR test for a null
hypothesis H : ρ = 0 against a two-sided alternative hypothesis H : ρ ≠ 0. We
0 1
denote θ = (µ , µ , ρ) and write Θ = {(µ , µ , 0) : µ ∈ ℜ, µ ∈
1 2 0 1 2 1 2
+
+
+
ℜ, σ ∈ ℜ , σ ∈ ℜ }, and Θ = {(µ , µ , ρ) : µ ∈ ℜ, µ ∈ ℜ, σ ∈ ℜ ,
1 2 1 2 1 2 1
+
σ ∈ ℜ , ρ ∈ (-1, 1)}. The likelihood function is given by
2

for all θ ∈ Θ. We leave it as Exercise 11.4.5 to show that the MLE’s for µ , µ ,
1 2
and ρ are respectively given by
and
These stand for the customary sample means, sample variances (not unbi-
ased), and the sample correlation coefficient. Hence, from (11.4.6) one has

Under the null hypothesis, that is when ? = 0, the likelihood function happens
to be

We leave it as the Exercise 11.4.4 to show that the MLE’s for µ , µ , and
1 2
are respectively given by
These again stand for the customary sample means and
sample variances (not unbiased). Hence, from (11.4.8) one has