Question

Consider the following hypothesis test.

H₀: σ₁² = σ₂²

H_a: σ₁² ≠ σ₂²

(a)

What is your conclusion if

n₁ = 21,

s₁² = 2.2,

n₂ = 26,

and

s₂² = 1.0?

Use

α = 0.05

and the p-value approach.

Find the value of the test statistic.

Find the p-value. (Round your answer to four decimal places.)

p-value =

State your conclusion.

Reject H₀. We cannot conclude that σ₁² ≠ σ₂².Do not reject H₀. We cannot conclude that σ₁² ≠ σ₂².    Reject H₀. We can conclude that σ₁² ≠ σ₂².Do not reject H₀. We can conclude that σ₁² ≠ σ₂².

(b)

Repeat the test using the critical value approach.

Find the value of the test statistic.

State the critical values for the rejection rule. (Round your answers to two decimal places. If you are only using one tail, enter NONE for the unused tail.)

test statistic≤test statistic≥

State your conclusion.

Reject H₀. We cannot conclude that σ₁² ≠ σ₂².Do not reject H₀. We cannot conclude that σ₁² ≠ σ₂².    Reject H₀. We can conclude that σ₁² ≠ σ₂².Do not reject H₀. We can conclude that σ₁² ≠ σ₂².

Answer 1

Solution:

Given:

n₁ = 21,

s₁² = 2.2,

n₂ = 26,

and

s₂² = 1.0

the following hypothesis are given:

H₀: σ₁² = σ₂²

H_a: σ₁² ≠ σ₂²

Level of Significance = α = 0.05

Part a) Find the value of the test statistic.

Find the p-value

Find df_numerator =n₁-1 = 21-1=20

df_denominator =n₂-1 = 26-1=25

Use following Excel command:

=F.DIST.RT( F , df_numerator , df_denominator )

=F.DIST.RT(2.20,20,25)

=0.0317

Since this is two tailed test,

p-value = 2 X 0.0317

p-value = 0.0634

State your conclusion.

Since p-value = 0.0634 > 0.05 significance level , we do not reject H0.

Thus correct answer is:

Do not reject H₀. We cannot conclude that σ₁² ≠ σ₂².  

Repeat the test using the critical value approach.

Find the value of the test statistic.

State the critical values for the rejection rule.

Level of Significance = α = 0.05

Use following Excel command:

=F.INV(α / 2 , df_numerator , df_denominator )

=F.INV(0.05/2,20,25)

=0.41

and

=F.INV.RT(α / 2 , df_numerator , df_denominator )

=F.INV.RT(0.05/2,20,25)

=2.30

Thus rejection rule is:
test statistic≤: 0.41

test statistic≥: 2.30

State your conclusion.

Since F test statistic value = 2.20 is less > 0.41 and less than 2.30, we fail to reject H0.

That is: 0.41 < F test statistic value = 2.20 < 2.30, we fail to reject H0.

Thus correct answer is:

Do not reject H₀. We cannot conclude that σ₁² ≠ σ₂².