MGMT2263 midterm solutions

 

Question 1

a)     Here is the table:

Source

SS

Df

MS

F

Factor

584.4

3

194.8

8.14

Error

382.8

16

23.93

 

Total

967.2

19

 

 

b)     Ho: µ1 = µ2 = µ3 = µ4

Ha: not all equal

Reject Ho if F > 3.24

F = 8.14

Reject Ho and conclude that there is a significant difference in the mean selling price of at least two suppliers.

c)     Ho: s1 = s2 = s3 = s4

Ha: not all equal

Reject Ho if H > 20.6

H = 56.5/10.7 = 5.28

Do not reject Ho

Conclude there is no significant difference in the variances. The assumption of equal variances is satisfied.

d)     Lower limit = (64.8 – 53) – (2.12)sqrt(23.93/5 + 23.93/5) = 11.8 – 6.559 = 5.24

Upper limit = 11.8 + 6.559 = 18.36

We are 95% confident that the white brand is 5.24 to 18.36 cents more expensive on average than the red brand.

e)     Since µ1 - µ4 = 0 falls outside the confidence interval, we conclude there is a significant difference between the means of these two brands; this is consistent with ANOVA in which we conclude that at least two means are significantly different.

f)      From Tukey’s test, we observe that the mean for the white brand is significantly different from all the other brands. This is consistent with ANOVA in which we conclude that at least two means are significantly different.

 

Question 2

a)     Ho: µd ≤ 0

Ha: µd > 0

This is based on the assumption that d = electronic – paper

Reject Ho if t > 2.015

T = 2.5/(2.881/sqrt(6)) = 2.13

Reject Ho

Conclude the workers can process more electronic returns on average than paper.

b)     Lower limit = 2.5 – (4.032)(2.881/sqrt(6)) = 2.5 – 4.74 = -2.24

Upper limit = 2.5 + 4.74 = 7.24

c)     With 99% confidence, the average number of returns processed electronically ranges from 2.24 less to 7.24 more per day than by paper.

d)     Since µd = 0 falls inside the 99% confidence interval, we do not reject Ho at a 1% level of significance and conclude there is no significant difference in the two processing methods.

e)     Ho: d ≤ 0

Ha: d > 0

Reject Ho if T- ≤ 1

T- = 1

Reject Ho

Same conclusion as part a.

 

Question 3

a)     Ho: s1 = s2

Ha: s1 ≠ s2

Reject Ho if F > 11.39

F = 11.657/3.0777 = 3.79

Do not reject Ho.

Conclude there is no significant difference in the variability in growth rates of the mutual funds.

b)     Ho: µ1 ≤ µ2

Ha: µ1 > µ2

This is assuming that fund A is group 1.

Sp2 = 6.8907

T = (7.32 – 4.52)/sqrt(6.8907/5 + 6.8907/6) = 1.76

The p-value is between 5% and 10%.

The results are inconclusive.

 

Question 4

a)     Ho: p1 ≤ p2

Ha: p1 > p2

This is assuming the more expensive system is group 1.

Reject Ho if Z > 1.85

Phat = 326/450 = 0.7244

Z = (0.76 – 0.68)/sqrt(0.7244(0.2756)/200 + 0.7244(0.2756)/250) = 1.89

Reject Ho

Conclude the expensive system is significantly more effective in reducing pollutants to acceptable levels than the inexpensive system.

b)     P-value = P(Z > 1.89) = 0.5 – 0.4706 = 0.0294. Since the p-value is less than the level of significance, we reject Ho.

c)     Since the p-value is between 1% and 10%, the results would be inconclusive.

d)     Ho: p1 – p2 ≤ 0.02

Ha: p1 – p2 > 0.02

Reject Ho if Z > 1.645

Z = [(0.76 – 0.68) – 0.02]/sqrt(0.76(0.24)/250 + 0.68(0.32)/200) = 1.41

Do not reject Ho.

Conclude the expensive system is not at least 2% more effective.

 

Question 5

Ho: µ1 - µ2 ≤ 10

Ha: µ1 - µ2 > 10

This is assuming that Edmonton is group 1.

Z = [(455 – 428) – 10]/sqrt(64.22/58 + 41.752/42) = 1.60

P-value = P(Z > 1.6) = 0.5 – 0.4552 = 0.0548

Since the p-value is between 1% and 10%, the results are inconclusive.