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

s_{1} = $587 million.

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

s_{2} = $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.

H_{0}: σ_{1}^{2} ≠
σ_{2}^{2}

H_{a}: σ_{1}^{2} =
σ_{2}^{2}

H_{0}: σ_{1}^{2} ≤
σ_{2}^{2}

H_{a}: σ_{1}^{2} >
σ_{2}^{2}

H_{0}: σ_{1}^{2} >
σ_{2}^{2}

H_{a}: σ_{1}^{2} ≤
σ_{2}^{2}

H_{0}: σ_{1}^{2} =
σ_{2}^{2}

H_{a}: σ_{1}^{2} ≠
σ_{2}^{2}

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 *H*_{0}. We cannot conclude there is a
statistically significant difference between the variances for the
two companies. Reject *H*_{0}. We can conclude there
is a statistically significant difference between the variances for
the two companies. Do not reject
*H*_{0}. We cannot conclude there is a statistically
significant difference between the variances for the two companies.
Do not reject *H*_{0}. 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 H_{0}. We cannot conclude
there is a statistically significant difference between the
variances for the two companies**

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