Question

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Data were collected on the top 1,000 financial advisers. Company A had 239 people on the list and another company, Company B, had 121 people on the list. A sample of 16 of the advisers from Company A and 10 of the advisers from Company B showed that the advisers managed many very large accounts with a large variance in the total amount of funds managed. The standard deviation of the amount managed by advisers from Company A was

s₁ = $587 million.

The standard deviation of the amount managed by advisers from Company B was

s₂ = $485 million.

Conduct a hypothesis test at

α = 0.10

to determine if there is a significant difference in the population variances for the amounts managed by the two companies. What is your conclusion about the variability in the amount of funds managed by advisers from the two firms?

State the null and alternative hypotheses.

H₀: σ₁² ≠ σ₂²

H_a: σ₁² = σ₂²

H₀: σ₁² ≤ σ₂²

H_a: σ₁² > σ₂²

     

H₀: σ₁² > σ₂²

H_a: σ₁² ≤ σ₂²

H₀: σ₁² = σ₂²

H_a: σ₁² ≠ σ₂²

Find the value of the test statistic. (Round your answer to two decimal places.)

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

p-value =

State your conclusion.

Reject H₀. We cannot conclude there is a statistically significant difference between the variances for the two companies. Reject H₀. We can conclude there is a statistically significant difference between the variances for the two companies.     Do not reject H₀. We cannot conclude there is a statistically significant difference between the variances for the two companies. Do not reject H₀. We can conclude there is a statistically significant difference bet

Answer 1

we have to test whether there is a significant difference in the population variances for the amounts managed by the two companies. So, it is a two tailed hypothesis test

Test statistic F = s1^2/s2^2

setting s1 = 587 and s2 = 485

so, we get

F = 587^2/485^2

= 1.46 (2 decimals)

degree of freedom numerator (df1) = n1 - 1 = 16-1 = 15

degree of freedom denominator (df2) = n2 - 1 = 10-1 = 9

p value = F.DIST.RT(F,df1,df2)

= F.DIST.RT(1.46,15,9)

= 0.2878

it is clear that the p value is greater than the significance level of 0.10, so we failed to reject the null hypothesis

Do not reject H₀. We cannot conclude there is a statistically significant difference between the variances for the two companies