Question

Consider the following hypothesis test.

H₀: σ₁² = σ₂²

H_a: σ₁² ≠ σ₂²

(a)

What is your conclusion if

n₁ = 21,

s₁² = 8.2,

n₂ = 26,

and

s₂² = 4.0?

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 =

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≥:_____

Answer 1

Solution:

Given:

n₁ = 21,

s₁² = 8.2,

n₂ = 26,

and

s₂² = 4.0

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(2.05,20,25)

=0.0452

Since this is two tailed test,

p-value = 2 X 0.0452

p-value = 0.0904

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

Decision:

Since F test statistic value = 2.05 is less > 0.41 and less than 2.30, we fail to reject H0.

That is: 0.41 < F test statistic value = 2.05 < 2.30, we fail to reject H0.