Question

An article in Fortune magazine reported on the rapid rise of fees and expenses charged by mutual funds. Assuming that stock fund expenses and municipal bond fund expenses are each approximately normally distributed, suppose a random sample of 12 stock funds gives a mean annual expense of 1.63 percent with a standard deviation of 0.45 percent, and an independent random sample of 12 municipal bond funds gives a mean annual expense of 0.93 percent with a standard deviation of 0.20 percent. Let µ₁ be the mean annual expense for stock funds, and let µ₂ be the mean annual expense for municipal bond funds. Do parts a, b, and c by using the equal variances procedure.

(a) Set up the null and alternative hypotheses needed to attempt to establish that the mean annual expense for stock funds is larger than the mean annual expense for municipal bond funds. Test these hypotheses at the 0.05 level of significance. (Round your s_p² answer to 4 decimal places and t-value to 2 decimal places.)

H0: µ_1-µ_{2 <\= _____ versus Ha: ____ _____
s2p = _________ t= _________
(reject/do not reject) H0 with a = 0.05}

(b) Set up the null and alternative hypotheses needed to attempt to establish that the mean annual expense for stock funds exceeds the mean annual expense for municipal bond funds by more than 0.5 percent. Test these hypotheses at the 0.05 level of significance. (Round your t-value to 2 decimal places and other answers to 1 decimal place.)

H0: µ_1-µ_{2 _____ _______ versus Ha:} µ_1-µ_{2 ____ _______
t= _________
(reject/do not reject) H0 with a = 0.05}

(c) Calculate a 95 percent confidence interval for the difference between the mean annual expenses for stock funds and municipal bond funds. Can we be 95 percent confident that the mean annual expense for stock funds exceeds that for municipal bond funds by more than .5 percent? (Round your answers to 3 decimal places.)

The interval = [_____, _____] (Yes/No), the interval is (above/below) 0.05.

Answer 1

For Sample 1 :

x̅1 = 1.63, s1 = 0.45, n1 = 12

For Sample 2 :

x̅2 = 0.93, s2 = 0.2, n2 = 12

a)

Null and Alternative hypothesis:

Ho : µ1 ≤ µ2

H1 : µ1 > µ2

Pooled variance :

S²p = ((n1-1)*s1² + (n2-1)*s2² )/(n1+n2-2) = ((12-1)*0.45² + (12-1)*0.2²) / (12+12-2) = 0.1213

Test statistic:

t = (x̅1 - x̅2) / √(s²p(1/n1 + 1/n2 ) = (1.63 - 0.93) / √(0.1213*(1/12 + 1/12)) = 4.92

df = n1+n2-2 = 22

Critical value :

Right tailed critical value, t crit = ABS(T.INV(0.05, 22)) = 1.717

Reject Ho if t > 1.717

p-value :

Right tailed p-value = T.DIST.RT(4.9242, 22) = 0.0000

Decision:

p-value < α, Reject the null hypothesis

b)

Null and Alternative hypothesis:

Ho : µ1 - µ2 ≤ 0.5

H1 : µ1 - µ2 > 0.5

Pooled variance :

S²p = ((n1-1)*s1² + (n2-1)*s2² )/(n1+n2-2) = ((12-1)*0.45² + (12-1)*0.2²) / (12+12-2) = 0.1213

Test statistic:

t = (x̅1 - x̅2 - (µ1 - µ2)) / √(s²p(1/n1 + 1/n2 ) = (1.63 - 0.93 - 0.5) / √(0.1213*(1/12 + 1/12)) = 1.41

df = n1+n2-2 = 22

Critical value :

Right tailed critical value, t crit = ABS(T.INV(0.05, 22)) = 1.717

Reject Ho if t > 1.717

p-value :

Right tailed p-value = T.DIST.RT(1.4069, 22) = 0.0867

Decision:

p-value > α, Do not reject the null hypothesis.

c)

95% Confidence interval for the difference:

At α = 0.05 and df = n1+n2-2 = 22, two tailed critical value, t-crit = T.INV.2T(0.05, 22) = 2.074

Lower Bound = (x̅1 - x̅2) - t-crit*√(S²p*(1/n1 +1/n2))

= (1.63 - 0.93) - 2.074*√(0.1213*(1/12 + 1/12)) = 0.405

Upper Bound = (x̅1 - x̅2) + t-crit*√(S²p*(1/n1 +1/n2))

= (1.63 - 0.93) + 2.074*√(0.1213*(1/12 + 1/12)) = 0.995

0.405 < µ1 - µ2 < 0.995