Question

Using Table I, test the significance at α = 0.01 of the correlation coefficient r = −0.944, obtained in Exercise 2. n= 6-1=5

Answer 1

To Test :-

H0 :- ρ = 0
H1 :- ρ ≠ 0

Test Statistic :-
t = (r * √(n - 2) / (√(1 - r2))
t = ( -0.944 * √(6 - 2) ) / (√(1 - 0.8911) )
t = -5.7212

Test Criteria :-
Reject null hypothesis if t < -t(α,n-2)
t(α/2,n-2) = t(0.01/2 , 6 - 2 ) = 4.6041
t < -t(α/2, n-2) = -5.7212 < -4.6041
Result :- Reject null hypothesis

Decision based on P value
P - value = P ( t > 5.7212 ) = 0.0046
Reject null hypothesis if P value < α = 0.01 level of significance
P - value = 0.0046 < 0.01 ,hence we reject null hypothesis
Conclusion :- We reject H0

There is statistically linear correlation between variables.