Answer:
Since the calculated t= 7.8120 falls in the critical region ≥ 2.306 we reject H0 and conclude that there is a linear correlation between the two variables.
Step-by-step explanation:
We see that r²= 0.981 and r= 0.995 from the figure .
The P- value is less than 0.05 which indicates that the two variables are linearly correlated.
Set up the hypothesis as
H0: β= 0 i.e the two variables X and Y are not related
Ha: β≠0 i.e the two variables are related
The significance Level is set at ∝=0.05
The test statistic , if H0 is true is
t= b/sb
where sb² = s²yx / ∑( x-x`)² = ∑( y-y`)²/n-2 ∑( x-x`)²
assuming that the distribution of Yi for each Xi is normal with the same mean and the same standard deviation , the statistic t conforms to the Student's t distribution with n-2= 8 degrees of freedom
(X - x`)2 (X - x`)(Y - y`) (Y - y`)2
265.69 259.17 15.9²=252.81
7174.09 6953.87 -82.1²=6740.41
75.69 87.87 -10.1=102.01
2052.09 1671.57 36.9²=1361.61
4529.29 5848.37 86.9²=7551.61
8335.69 7294.87 79.9²=6384.01
4998.49 5380.27 -76.1²=5791.21
1069.29 1016.97 -31.1=967.21
1043.29 1159.57 35.9²=1288.81
3102.49 3124.77 -56.1²=3147.21
SS: 32646.1 SP: 32797.3 33,586.9
Sum of X = 3877
Sum of Y = 3991
Mean X = 387.7
Mean Y = 399.1
Sum of squares (SSX) = 32646.1
Sum of products (SP) = 32797.3
b = SP/SSX = 32797.3/32646.1 = 1.00463
a = Y` - bX `= 399.1 - (1*387.7) = 9.60437
Syx= √∑(y-y`)²/n-2= 33,586.9/8= 4198.3625
sb= syx/ √∑(x-x`)²=4198.3625/32646.1 =0.12860
Putting the values in the regression equation
ŷ = bX + a
ŷ = 1.00463X + 9.60437
The critical region is t ≥ t (0.025) (8)= 2.306
t= b/ sb= 1.00463/ 0.12860=7.8120
Since the calculated t= 7.8120 falls in the critical region ≥ 2.306 we reject H0 and conclude that there is a linear correlation between the two variables.
