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.