Question

1. Given with the following sample information

n1 = 125, n2 = 120; s1 = 31, s2 = 38; x1-bar = 116, x2-bar = 105

Test the following hypotheses, assuming a significance level of 0.05 is to be used with equal variances.

Ho : µ1 - µ2 = 0

HA : µ1 - µ2 ≠ 0

What is your conclusion?

a) Since t test statistic = 1.82 < t-critical value = 2.49, we do not reject Ho.

b) Since t test statistic = 2.49 > t-critical value = 1.97, we reject Ho.

c) Since t test statistic = 1.82 < t-critical value = 1.97, we do not reject Ho.

d) Since t test statistic = 2.49 > t-critical value = 1.54, we reject Ho.

2. You are given two random samples with the following information:

Item	Sample 1	Sample 2
1	19.6	21.3
2	22.1	17.4
3	19.5	19
4	20	21.2
5	21.5	20.1
6	20.2	23.5
7	17.9	18.9
8	23	22.4
9	12.5	14.3
10	19	17.8

Based on these samples, test at alpha = 0.10 whether the true difference in population variances is equal to zero. What is the lower F critical value?

a) 0.31

b) 1.15

c) 2.44

d) 0.42

3. Assume that you are testing the difference in the means of two independent populations at the 0.05 level of significance.

The null hypothesis is: Ho : µA - µB >=0 and you have found the test statistic is z = -1.92.

What should you conclude?

a) The mean of population B is greater than the mean of population A because p-value < alpha.

b) The mean of population A is greater than the mean of population B because p-value > alpha.

c) The mean of population A is greater than the mean of population B because p-value < alpha.

d) The mean of population B is greater than the mean of population A because p-value > alpha.

Answer 1

Q1:

For Sample 1 : x̅1 = 116, s1 = 31, n1 = 125

For Sample 2 : x̅2 = 105, s2 = 38, n2 = 120

Null and Alternative hypothesis:

Ho : µ1 - µ2 = 0

H1 : µ1 - µ2 ≠ 0

Pooled variance :

S²p = ((n1-1)*s1² + (n2-1)*s2² )/(n1+n2-2) = ((125-1)*31² + (120-1)*38²) / (125+120-2) = 1197.5309

Test statistic:

t = (x̅1 - x̅2) / √(s²p(1/n1 + 1/n2 ) = (116 - 105) / √(1197.5309*(1/125 + 1/120)) = 2.49

df = n1+n2-2 = 243

Critical value :

Two tailed critical value, t crit = T.INV.2T(0.05, 243) = 1.970

Reject Ho if t < -1.97 or if t > 1.97

Decision: Reject the null hypothesis

Answer : b) Since t test statistic = 2.49 > t-critical value = 1.97, we reject Ho.

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Q2.

n₁ = 10, n₂ = 10

Degree of freedom:

df₁ = n₁-1 = 9

df₂ = n₂-1 = 9

Critical value(s):

Lower tailed critical value, Fα/₂ = F.INV(0.1/2, 9, 9) = 0.31

Answer : a) 0.31

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Q3. z = -1.92

p-value = NORM.S.DIST(-1.92, 1) = 0.0274

Answer : a) The mean of population B is greater than the mean of population A because p-value < alpha.