MGMT2263 Midterm practice solutions

Item 1

a) Let Operator 1 be group 1.

Ho: m_d = 0

Ha: m_d ¹ 0

The degrees of freedom is 5. Reject Ho if t < -2.571 or > 2.571

t = (0.1667 – 0)/(2.9269/sqrt(6)) = 0.139.

Do not reject Ho. Conclude there is no significant difference between the operators in their job performance.

b) Lower limit = 0.1667 – 2.571(2.9269)/sqrt(6) = 0.1667 – 3.0721 = -2.9054

Upper limit = 0.1667 + 3.0721 = 3.2388

We would reach the same conclusion since the hypothesized difference of zero falls inside the confidence interval.

c) If we could no longer assume normality, we would need to conduct the Wilcoxon test.

Ho: d = 0

Ha: d ¹ 0

Reject Ho if T £ 1 (using Table A.11 with n=6 and 2-sided alpha = 0.05)

We rank the absolute values of the differences from lowest to highest:

	1	2	3	4	5	6
|d|	1	2	2	3	3	4
sign	+	+	+	+	-	-
rank	1	2.5	2.5	4.5	4.5	6

Then T+ = 1 + 2.5 + 2.5 + 4.5 = 10.5 and T- = 4.5 + 6 = 10.5. Note than T+ plus T- = 21 = 6(7)/2. T = the smaller of T+ and T- = 10.5.

Do not reject Ho.

Same conclusion as part a.

Item 2

a) First the ANOVA table:

Source of Variation	SS	df	MS	F
Between Groups	63.2855	3	21.0952	3.4616
Within Groups	97.5040	16	6.094
Total	160.7895	19

b) Ho: s1 = s2 = s3 = s4

Ha: not all the standard deviations are equal

Reject Ho if H > 20.6

H = 8.903/3.683 = 2.42

Do not reject Ho.

Conclude the standard deviations are equal.

c) Ho: m1 = m2 = m3 = m4

Ha: not all the means are equal

Reject Ho if F > 3.24

F = 3.4616

Reject Ho. Conclude that at least 2 means are not equal.

d) Tukey’s test is useful since we rejected the null hypothesis in part c.

Q = 4.05; D = 4.05sqrt(6.094/5) = 4.4712

Here are the means from smallest to largest:

Groups	Average
Supplier 1	19.52
Supplier 4	21.16
Supplier 3	22.84
Supplier 2	24.26

Supplier 2 – Supplier 1 = 24.26 – 19.52 = 4.74. So, the means of these 2 suppliers are significantly different.

Supplier 2 – Supplier 4 = 24.26 – 21.16 = 3.1. So these means are not significantly different.

Supplier 3 – Supplier 1 = 22.84 – 19.52 = 3.32. So these means are not significantly different.

So, the only suppliers whose means are significantly different are Suppliers 2 and 1.

e) The t value is 2.12 since we are working with 16 degrees of freedom.

Lower limit = (24.26 – 22.84) – 2.12sqrt(6.094/5 + 6.094/5) = 1.42 – 3.31 = -1.89

Upper limit = 1.42 + 3.31 = 4.73

-1.89 < m2 - m3 < 4.73

Item 3

a) Since B has the larger standard deviation, we will make this group 1.

Ho: s1 = s2

Ha: s1 ¹ s2

In this case n1 = 5 and n2 = 7. So the degrees of freedom are 4 and 6. Reject Ho if F > 6.23 or if F < 1/ 9.2 = 0.1087.

F = 6.5422²/3.9881² = 2.691

Do not reject Ho.

Conclude the standard deviations are equal.

b) We should conduct the t test assuming equal variances.

c) Ho: m1 = m2

Ha: m1 ¹ m2

The degrees of freedom is 5 + 7 – 2 = 10. Reject Ho if the p-value < 1%. Do not reject Ho if p-value > 10.

Sp² = [4(6.5422²) + 6(3.9881²)]/10 = 26.6629

t = (19.2857 – 12.6)/sqrt(26.6629/7 + 26.6629/5) = 2.2113

When we look at the t table to see where the test statistic falls, we see it falls between the 5% critical value of 1.812 and the 2.5% critical value of 2.228. This means 2.5% < p-value/2 < 5% or 5% < p-value < 10%. Since the p-value is between 5% and 10%, the results are inconclusive.

d) If we used a 1% level of significance, we would reject Ho if t < -3.169 or t > 3.169. In this case, the test statistic falls in the acceptance region. Thus, we do not reject Ho and conclude there is no significant difference in the average number of ads between the 2 magazines.

e) The t value is 2.228.

Lower limit = (19.2857 – 12.6) – 2.228 sqrt(26.6629/7 + 26.6629/5) = 6.6857 – 6.7364 = -0.0507

Upper limit = 6.6857 + 6.7364 = 13.4221

Item 4

a) The appropriate test would be the Mann-Whitney test

Ho: median 1 = median 2

Ha: median 1 ¹ median 2

Reject Ho if p-value < 5%

Value	8	9	10	12	15	16	17	18	20	23	24	26
Group	B	B	B	B	A	A	A	A	A	A	B	A
Rank	1	2	3	4	5	6	7	8	9	10	11	12

We have n1=5 and n2=7. The total rank sum is 12(13)/2 = 78

T1 = 1 + 2 + 3 + 4 + 11 = 21. This means T2 = 78 – 21 = 57.

U1 = 5(7) + (5)(6)/2 – 21 = 29

U2 = 5(7) + (7)(8)/2 – 57 = 6

Then U = 6. From Table A.10 we find p-value/2 = 0.0366, meaning p-value = 0.0732.

Do not reject Ho. Same conclusion as in the previous question.

Item 5

a) We will let the 25 or under group be group 1.

Ho: p1 – p2 £ 10%

Ha: p1 – p2 > 10%

Reject Ho if Z > 1.645

p1 = 0.75; p2 = 0.6

Z = [(0.75 – 0.6) – 0.1]/sqrt(0.75(0.25)/1200 + 0.6(0.4)/1500) = 2.81

Reject Ho. Conclude the difference in the percentages is more than 10%

b) P-value = P(Z > 2.81) = 0.5 – 0.4975 = 0.0025. Since this p-value is less than the 5% level of significance, we reject the null hypothesis.

c) Lower limit = (0.75 – 0.6) – 1.96 sqrt(0.75(0.25)/1200 + 0.6(0.4)/1500) = 0.15 – 0.0349 = 0.1151

Upper limit = 0.15 + 0.0349 = 0.1849

11.51% < p1 – p2 < 18.49%