MGMT2263 Midterm
solutions
Question 1 – Version A; Question 3 – Version B
Part a
Ho: σ1 = σ2
Ha: σ1 ≠ σ2
Reject Ho if F > 3.18
F = 1.252/0.8482 = 2.17
Do not reject Ho.
Conclude no significant difference in variances.
Assume σ1 = σ2.
Part b
Ho: µ1 ≤ µ2
Ha: µ1 > µ2
Sp2 = [12(0.848)2 + 15(1.25)2]/27
= 1.19
t = (5.01 – 3.97)/sqrt(1.19/13 + 1.19/16) =
2.55
P-value = P(t > 2.55)
based on 27 degrees of freedom.
0.005 < p-value < 0.01
P-value < 1%. Reject Ho
Conclude the new computer system reduced the customer
waiting time.
Question 2 – Version A; Question 5 – Version B
Ho: p1 – p2 ≤ 0.2
Ha: p1 – p2 > 0.2
Z = [(0.67 – 0.30) – 0.2]/sqrt(0.67(0.33)/200 +
0.32(0.68)/150) = 2.97
P-value = 0.5 – 0.4985 = 0.0015
P-value is less than 1%. Reject Ho
Conclude the retention rate for Masters Degree
employees is more than 20% less than that of the Bachelor Degree employees.
Question 3 – Version A; Question 6 – Version B
Part a
We have dependent samples with no assumptions of
normality – use the Wilcoxon signed-ranks test.
Part b
This is set up with d = reduced – regular
Ho: d ≤ 0
Ha: d > 0
Reject Ho if T- ≤ 4
T- = 8
Do not reject Ho.
Conclude the price reduction did not increase sales of
the unit.
Question 4 – Version A; Question 2 – Version B
Part a
Ho: µd ≤ 0
Ha: µd > 0
Reject Ho if t > 1.415
t = 0.825/(1.5473/sqrt(8)) = 1.58
Reject Ho
Conclude the average consumer debt per household had
significantly increased from December 2008 to December 2009.
Ha: µ1 ≠ µ2
Reject Ho if Z < -1.96 or Z > 1.96
Z = (24.88 – 23.5)/sqrt(2.12/40 +
2.922/40) = 2.43
Reject Ho.
Conclude there is a significant difference in average
completion times between the two teams.
Part b
P-value = 2P(Z > 2.43) =
2(0.0075) = 0.015
Part c
P-value < α(5%) ->
reject Ho.
Part d
The results are inconclusive since the p-value is
between 1% and 10%.
Part e
(24.88 – 23.5) ± 1.96 sqrt(2.12/40 +
2.922/40)
= 1.38 ± 1.1146
0.27 < µ1 - µ2 < 2.49
With 95% confidence, the average completion time for
team B is 0.27 to 2.49 minutes more than that of team A.
Part f
d0 = 0 falls outside the confidence interval. Conclude
µ1 ≠ µ2.
Question 5 – Version A; Question 4 – Version B
Part a
This is done with d = December 2009 – December 2008
Ho: µd ≤ 0
Ha: µd > 0
Reject Ho if t > 1.415
t = 0.825/(1.5473/sqrt(8)) = 1.58
Reject Ho
Conclude the average consumer debt per household had
significantly increased from December 2008 to December 2009.
Part b
P-value = P(t > 1.58)
based on 7 degrees of freedom.
0.05 <
p-value < 0.1.
Therefore, p-value > α(3.5%) -> do not
reject Ho
One other solution posed by a number of students is
that the 3.5% critical value must lie between the 5% critical value of 1.895
and the 2.5% critical value of 2.365. Since the test statistic is to the left
of the 5% critical value, we do not reject Ho.
Part c
Ho: µd ≤ 0.2 (200 was accepted)
Ha: µd > 0.2
Reject Ho if t > 1.895
t = (0.825 – 0.2)/(1.5473/sqrt(8)) = 1.21
Do not reject Ho
Conclude the average consumer debt per household had
not significantly increased from December 2008 to December 2009 by more than $200.
Question 6 – Version A; Question 1 – Version B
Part a
|
Source
|
SS |
DF |
MS |
F |
|
Factor |
5180.93 |
2 |
2590.47 |
10.67 |
|
Error |
2914.4 |
12 |
242.87 |
|
|
Total |
8095.33 |
14 |
|
|
Part b
Ho: µ1 = µ2 = µ3
Ha: not all means equal
Reject Ho if F > 3.89
F = 10.67
Reject Ho.
Conclude there is a significant difference in the
distance travelled by the different designs of golf balls.
Part c
P-value = P(F > 10.67)
based on 2 and 12 degrees of freedom.
The 1% critical value is 6.93. Therefore p-value <
1% which in turn means p-value < α(5%). Reject
Ho.
Part d
This should be done with groups 1 and 3 (Cruncher and
Woods)
(315 – 269.6) ± 3.77sqrt(242.87/5)
= 45.4 ± 26.28
19.13 < µ3 - µ1 < 71.68
Part e
µ3 - µ1 = 0 falls outside the confidence interval.
Conclude µ3 ≠ µ1. This supports the conclusion in part b in which we
conclude a significant difference between the means.