Answer:
Since the calculated t value does not fall in the critical region we accept H0 and conclude that one model is not significantly better than the other at 1% level of significance.
Step-by-step explanation:
When the observations from two samples are paired either naturally or by design , we find the difference between two observations of each pair. Treating the differences as a random sample from a normal population with mean μd =μ1-μ2 and unknown standard deviation σd we perform one sample t- test on them. This is called paired difference t- test.
M1 M2 Difference d²
d= (M1-M2)
30.5, 22.4, 8.1 65.61
32.2, 14.5, 17.7 313.29
20.7, 22.4, -1.7 8.3521
20.6, 19.6, 1.0 1
31.0, 20.7, 10.3 106.09
41.0, 20.4, 20.6 424.36
27.7, 22.1, 5.6 31.36
26.0, 19.4, 6.6 43.56
21.5, 16.2, 5.3 28.09
26.0. 35.0. -9.0 81.0
∑ 329.2 213.1 64.5 1102.7121
Now
d` = ∑di/n = 64.5/10= 6.45
sd² = ∑(di-d`)²/n-1= 1/n-1 [ ∑di²- (∑di)²/n]
=1/9[ 1102.7121 - (64.5)²/10 ]
=[1102.7121 - 416.025/9]
= 79.298
sd= 8.734
We state our null and alternate hypotheses as
H0 : μd= 0 and Ha: μd≠0
The significance level is set at ∝ = 0.01
The test statistic under H0 is
t= d`/ sd/√n
which has a t distribution with n-1 degrees of freedom.
The critical region is t ≥ t (0.005,9)= 3.250
Calculating t
t= 6.45 / 8.734/ √10
t = 6.45 / 8.734/3.162
t= 6.45 / 2.7621
t= 2.335
Since the calculated t value does not fall in the critical region we accept H0 and conclude that one model is not significantly better than the other at 1% level of significance.