Question

a) Bargain Town is a large discount chain. The management wishes to compare the performance of its credit managers in Ohio and Illinois, by comparing the mean dollar amount owed by customers with charge accounts in both two states. A small mean is desirable.

Suppose that Bargain Town randomly selected the following small samples.

Sample of Ohio accounts Sample of Illinois accounts

n_o=10 n_o= 6

yo = $124 y₁ = $68

s²_o= 1681 s²_o= = 484

Assuming that equal variances, independent samples, and normality assumptions hold, compute a 95% confidence interval for μo-μ1.

b) Assuming that only the independent samples and normality assumptions hold, test H_o: μ_o-μ₁=0 versus H₁: μ_o-μ₁≠0 by setting α=0.05. Based on the test we cannot reject the Null Hypothesis.

True or False?

c) To carry our the F-test for equality of population variances, you need to first calculate the values of F, r₁, r₂,F_α^(r₁,r₂)

Compute the following: F, r₁, r₂,F_α^(r₁,r₂)

Carry out the F -test for equality of variances, H_o: σ²_o=σ²₁versus H₁: σ²_o≠σ²₁ Do we reject the null hypothesis?

Select one:

a. [F, r₁, r₂, F_α^(r₁,r₂) , reject H_o versus H₁ ] = [2.565, 6, 9, 6, reject]

b. [F, r₁, r₂, F_α^(r₁,r₂) , reject H_o versus H₁ ] = [2.565, 9, 4, 6, reject]  

c. [F, r₁, r₂, F_α^(r₁,r₂) , reject H_o versus H₁ ] = [3.473, 9, 4, 6, reject]

d. [F, r₁, r₂, F_α^(r₁,r₂) , reject H_o versus H₁ ] = [2.565, 8, 5, 5, we cannot reject]

e. [F, r₁, r₂, F_α^(r₁,r₂) , reject H_o versus H₁ ] = [3.473, 9, 4, 6, we cannot reject]

Answer 1

(a)95% confidence interval =(16.79,95.21)

SE((mean1-mean2)=((s_p*(1/n1 +1/n2)^1/2)=18.283

s_p²=((n1-1)s1²+(n2-1)s2²)/n=1253.5 with n=n1+n2-2=10+6-2=14 and

(1-alpha)*100% confidence interval for μo-μ1=difference of sample mean ±t(alpha/2,n)*SE(difference of mean)

95% confidence interval =56±2,1448*18.283=56±39.21=(16.79,95.21)

	sample	mean	s	s2	n	(n-1)s2
	Ohio	124.0000	41.0000	1681.0000	10	15129.0000
	Illinois	68.0000	22.0000	484.0000	6	2420.0000
	difference=	56.0000	sum=	2165.0000	16	17549.0000
	sp2=	1253.5000
	sp=	35.4048
	SE=	18.2830

(b) answer is False,

since here 0 values does not lies in the 95% confidence interval (16.79,95.21), so we reject null hypothesis H0.

Based on the test we cannot reject the Null Hypothesis.: False

(c) e. [F, r₁, r₂, F_α^(r₁,r₂) , reject H_o versus H₁ ] = [3.473, 9, 4, 6, we cannot reject]

test statistic F=s1²/s2²=1681/484=3.473 with df (10-1,6-1)=(9, 5)

typical critical F(0.05,9,6)=4.099 is more than calculated F=3.473, so we fail to reject H0( or accept H0)