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

-2.9054 < m_d < 3.2388

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

c) P-value = 2P(t > 0.139). From the t table, the 10% critical value is 1.476. Since P(t > 1.476) = 10%, P(t > 0.139) > 10% and the p-value > 20%. The interpretation is that since the p-value is greater than the 5% level of significance, we do not reject Ho.

d) Under the general rule of thumb, since the p-value is greater than 10%, we do not reject Ho.

Item 2

a) First the ANOVA table:

Source of Variation	df	SS	MS	F
Between Groups	3	63.2855	21.0952	3.4616
Within Groups	16	97.504	6.094
Total	19	160.7895

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 there is no significant difference in the standard deviations. It is appropriate to conduct ANOVA since the assumption of equal variances is satisfied.

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 there is a significant difference in average tensile strengths between at least 2 suppliers.

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

Technically, there are 4C2 = 6 comparisons to be made.

Supplier 2 – Supplier 1 = 24.26 – 19.52 = 4.74. Conclude µ1 ≠ µ2.

Supplier 2 – Supplier 4 = 24.26 – 21.16 = 3.1. Conclude µ2 = µ4.

Supplier 3 – Supplier 1 = 22.84 – 19.52 = 3.32. Conclude µ1 = µ3.

The remaining conclusions logically follow from these. For example, if suppliers 2 and 4 are not significantly different, suppliers 2 and 3 would not be either. So, the only suppliers whose means are significantly different are Suppliers 2 and 1.

a) The margin of error is the value of D from above.

Lower limit = (24.26 – 19.52) – 4.4712 = 4.74 – 4.4712 = 0.2688

Upper limit = 4.74 +4.4712 = 9.2112

0.2688 < m2 - m1 < 9.2112

Note that m2 - m1 = 0 falls outside the 95% simultaneous confidence interval. This indicates a significant difference between m2 and m1. This supports the results of ANOVA in which we concluded a significant difference between at least 2 means.

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. As an alternate solution, we noted from the previous question that the p-value is greater than 5% from which we infer that the p-value is greater than 1% and we would not reject Ho.

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

-0.0507 < m1 - m2 < 13.4221

Note that the hypothesis difference of zero falls inside the 95% confidence interval. Therefore, do not reject Ho at a 5% level of significance.

Item 4

a) The appropriate test would be the Wilcoxon rank sum test since the scale is interval/ratio and there are no assumptions of normality.

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. Therefore Magazine B is group 1. Reject Ho if T1 £ 20 or T1 ³ 45.

T1 = 1 + 2 + 3 + 4 + 11 = 21

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%

Item 6

We choose system B as group 1 in order to set up a right tail test.

Ho: d £ 0

Ha: d > 0

First we compute the differences:

Person	1	2	3	4	5	6	7	8	9	10
System A	5	7	7	4	6	7	7	7	7	6
System B	8	7	6	7	6	8	9	8	8	8
d = B-A	3	0	-1	3	0	1	2	1	1	2

Since there are two differences of zero, n = 8. Reject Ho if T- £ 6

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

Observation	1	2	3	4	5	6	7	8
|d|	1	1	1	1	2	2	3	3
Sign	-	+	+	+	+	+	+	+
Rank	2.5	2.5	2.5	2.5	5.5	5.5	7.5	7.5

T- = 2.5. Reject Ho. Conclude that system B is rated significantly better than system A.

Item 7

a) Choose region 1 as group 1.

Ho: m1 £ m2

Ha: m1 > m2

Z = (67.5 – 64.4)/sqrt(7.8²/100 + 8.2²/100) = 2.74

P-value = P(Z > 2.74) = 0.5 – 0.4969 = 0.0031 = 0.31%

Since the p-value < 1%, reject Ho. Conclude that the average score for region 1 is significantly higher than that of region 2.

b) Lower limit = (67.5 – 64.4) – 2.24 sqrt(7.8²/100 + 8.2²/100) = 3.1 – 2.5 = 0.6

Upper limit = 3.1 + 2.5 = 5.6

0.6 < m1 - m2 < 5.6

c) Ho: m1 - m2 £ 1 (with region 1 as group 1)

Ha: m1 - m2 > 1

Reject Ho if Z > 1.76

Z = [(67.5 – 64.4) – 1]/sqrt(7.8²/100 + 8.2²/100) = 1.86

Reject Ho. Conclude that the average grade for Region 2 is more than 1 mark less than the average grade in Region 1.

Item 8

a) Choose downtown as group 1 since it has the larger mean.

Ho: m1 = m2

Ha: m1 ¹ m2

df = (153.9808²/12 + 29.8339²/12)²/((153.9808²/12)/11 + (29.8339²/12)/11) = 11.8 which we round to 11. Reject Ho if t < -2.201 or > 2.201

t = (503.0833 – 476.3333)/sqrt(153.9808²/12 + 29.8339²/12) = 0.5908

Do not reject Ho. Conclude there is no significant difference in average daily rates between downtown and suburban hotels.

b) P-value = 2P(t > 0.5908) with 11 degrees of freedom. The 10% critical value is 1.363 meaning that P(t > 1.363) = 10%. By inference, P(t > 0.5908) > 10% and the p-value > 20%. Under the general rule of thumb, since the p-value is greater than 10%, we do not reject Ho.

c) Lower limit = (503.0833 – 476.3333) – 2.201sqrt(153.9808²/12 + 29.8339²/12)

= 26.75 – 99.66 = -72.91

Upper limit = 26.75 + 99.66 = 126.41

-72.91 < m1 - m2 < 126.41

Since the hypothesis difference of zero falls inside the 95% confidence interval, we do not reject Ho at a 5% level of significance.