Question

Researchers measured skulls from different time periods in an attempt to determine whether interbreeding of cultures occurred. Results are given below. Assume that both samples are independent simple random samples from populations having normal distributions. Use a 0.05 significance level to test the claim that the variation of maximal skull breadths in 4000 B.C. is the same as the variation in A.D. 150.
n x s
4000 B.C. 30 131.62 mm 5.19 mm
A.D. 150 30 136.07 mm 5.35 mm
What are the null and alternative​ hypotheses?Identify the test statistic, F=?The P-value is ?What is the concluion for this hypothesis test?A. Fail to reject Upper H0. There is sufficient 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.B. 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.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.D. Reject Upper H 0. There is sufficient 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.

Answer 1

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$_{n1-1;n2-1}$

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$F_{n1-1;n2-1;\alpha /2} = \frac{1}{F_{n2-1;n1-1;1-\alpha /2} } = \frac{1}{F_{29;29;0.95} } = \frac{1}{2.10} } =0.47$

Right: $F_{n1-1; n2-1; 1-\alpha /2} = F_{29; 29; 0.975} = 2.10$

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!