The question is incomplete! Complete question along with answer and step by step explanation is provided below.
Question:
Most individuals are aware of the fact that the average annual repair cost for an automobile depends on the age of the automobile. A researcher is interested in finding out whether the variance of the annual repair costs also increases with the age of the automobile. A sample of 26 automobiles 4 years old showed a sample standard deviation for annual repair costs of $120 and a sample of 23 automobiles 2 years old showed a sample standard deviation for annual repair costs of $100. Let 4 year old automobiles be represented by population 1.
State the null and alternative versions of the research hypothesis that the variance in annual repair costs is larger for the older automobiles.
At a 0.01 level of significance, what is your conclusion? What is the p-value?
Answer:
Null hypotheses = H₀ = σ₁² ≤ σ₂²
Alternative hypotheses = Ha = σ₁² > σ₂²
Test statistic = 1.44
p-value = 0.1954
0.1954 > 0.01
Since the p-value is greater than the given significance level therefore, we cannot reject the null hypothesis.
We can conclude that there is no sufficient evidence to support the claim that the variance in annual repair costs is larger for older automobiles.
Explanation:
Let σ₁² denotes the variance of 4 years old automobiles
Let σ₂² denotes the variance of 2 years old automobiles
State the null and alternative hypotheses:
The null hypothesis assumes that the variance in annual repair costs is smaller for older automobiles.
Null hypotheses = H₀ = σ₁² ≤ σ₂²
The alternate hypothesis assumes that the variance in annual repair costs is larger for older automobiles.
Alternative hypotheses = Ha = σ₁² > σ₂²
Test statistic:
The test statistic is given by
Test statistic = σ₁²/σ₂²
Test statistic = 120²/100²
Test statistic = 1.44
p-value:
The degree of freedom corresponding to 4 years old automobiles is given by
df₁ = n - 1
df₁ = 26 - 1
df₁ = 25
The degree of freedom corresponding to 2 years old automobiles is given by
df₂ = n - 1
df₂ = 23 - 1
df₂ = 22
Using Excel to find out the p-value,
p-value = FDIST(F-value, df₁, df₂)
p-value = FDIST(1.44, 25, 22)
p-value = 0.1954
Conclusion:
When the p-value is less than the significance level then we reject the Null hypotheses
p-value < α (reject H₀)
But for the given case,
p-value > α
0.1954 > 0.01
Since the p-value is greater than the given significance level therefore, we cannot reject the null hypothesis.
We can conclude that there is no sufficient evidence to support the claim that the variance in annual repair costs is larger for older automobiles.