Question

Barron's has collected data on the top 1,000 financial advisers. Company A and Company B have many of their advisers on this list. A sample of 16 of the Company A advisers and 10 of the Company B advisers 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 the Company A advisers was s₁ = $583 million. The standard deviation of the amount managed by the Company B advisers was s² = $481 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?

a) State the null and alternative hypotheses.

b) 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.)

State your conclusion.

Do not reject H₀. We can 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.

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

Answer 1

Answer:

Given,

s₁ = $583 million

s² = $481 million

n1 = 16

n2 = 10

Null hypothesis can be given as follows

Ho :   =

Alternative hypothesis

Ha :

Here may be greater or less than for the variances to differ. Here it is a two tailed test.

Now we have to consider the sample variance for given samples

s1^2 = 583^2

= 339889

s2^2 = 481^2

= 231361

So now we have to give the test statistic

we know that,

F = s1^2 / s2^2 [here n1 > n2 so the first sample variance must be numerator & then the second sample be in denominator]

= 339889 / 231361

= 1.4691

F = 1.47

degree of freedom for numerator = n1 - 1 = 16 - 1 = 15

degree of freedom for denominator = n2 - 1 = 10 - 1 = 9

At the given degrees of freedom & 90% confidence interval, p = 0.2841 [since from the f table & f calculator]

Here we observed that p value is greater than significance level.

So that the investigator failed to reject the Ho. So there is no sufficient evidence to conclude that there is a significant difference in population variance of given samples.