Answer:
C. Fail to reject Upper H 0. There is insufficient evidence to warrant rejection of the claim that the variation of maximal skull breadths in 4000 B.C. is the same as the variation in A.D. 150
Step-by-step explanation:
Hello!
You have two different independent samples and are asked to test if the population variances of both variables are the same.
Sample 1 (4000 B.C)
X₁: Breadth of a skull from 4000 B.C. (mm)
X₁~N(μ₁;σ₁²)
n₁= 30 skulls
X[bar]₁= 131.62 mm
S₁= 5.19 mm
Sample 2 (A.D. 150)
X₂: Breadth of a skull from 150 A.D. (mm)
X₂~N(μ₂;σ₂²)
n₂= 30 skulls
X[bar]₂= 136.07 mm
S₂= 5.35 mm
Since you want to test the variances, the proper test to do is an F-test for the population variance ratio. The hypothesis can be established as equality between variances or as a quotient between them.
The hypothesis is:
H₀: σ₁²/σ₂² = 1
H₁: σ₁²/σ₂² ≠ 1
Remember, when you express the hypothesis as a quotient of variances, if it's true that they are the same, the result will be 1, this is the number you'll use to replace in the F-statistic.
α: 0.05
F= (S₁²/S₂²) * (σ₁²/σ₂²) ~F[tex]_{n1-1;n2-1}[/tex]
F= (5.19/5.35)*1 = 0.97
The p-value = 0.5324
Since the p-value is greater than the level of significance, the decision is to not reject the null hypothesis.
Using critical values:
Left: F[tex]F_{n1-1;n2-1;\alpha /2} = \frac{1}{F_{n2-1;n1-1;1-\alpha /2} } = \frac{1}{F_{29;29;0.95} } = \frac{1}{2.10} } =0.47[/tex]
Right: [tex]F_{n1-1; n2-1; 1-\alpha /2} = F_{29; 29; 0.975} = 2.10[/tex]
The calculated F-value (0.97) is in the not rejection zone (0.47<F<2.10) ⇒ Don't reject the null hypothesis.
I hope this helps!
