Question

Part A

For this and the next 2 parts: A recently hired MBA on the staff of Ace Systems

thinks she can improve the training of production workers, who until now have been

trained only on the job. She has developed three new alternative programs involving

classroom sessions. These are compared, along with the existing procedure, as four

methods or treatments:

[1] On-the-job training only.

[2] Classroom training only.

[3] Heavy on-the-job training with light classroom work.

[4] Light on-the-job training with heavy classroom work.

Each method is conducted with a different sample group of 8 randomly chosen

trainees, who are given a placement examination upon completion. The investigator

uses the scores on that test as the response variable. Statistical analysis provides

the following sample means:

X-bar1 = 50.2. X-bar2 = 44.9. X-bar3 = 78.5. X-bar4 = 64.0

The SSE for the study is 8,431.31. Calculate the mean square treatment (MSTR).

1. 8,803.12
2. 8,431.31
3. 301.12
4. None of the above

Part B

If the computed F statistic for the study 6.03, how should the investigator conclude?

1. At alpha = 0.01, the critical value of F is 4.57. Therefore, the null hypothesis that

all training methods result in identical outcomes should be rejected

2. At alpha = 0.01, the critical value of F is 10.85. Therefore, the null hypothesis that

all training methods result in identical outcomes should be rejected

3. At alpha = 0.01, the critical value of F is 4.57. Therefore, the null hypothesis that

all training methods result in identical outcomes should NOT be rejected

4. At alpha = 0.01, the critical value of F is 10.85. Therefore, the null hypothesis that

all training methods result in identical outcomes should NOT be rejected

Part C

At alpha = 0.05, compute the Tukey Criterion appropriate for a pairwise comparison

test. Choose the closest answer.

1. 6.1351
2. 23.8044
3. 125.9605
4. 33.6645

Answer 1

a)

count, ni =	8	8	8	8
mean , x̅ i =	50.200	44.900	78.500	64.000

grand mean , x̅̅ =    Σni*x̅i/Σni =                 59.4

square of deviation of sample mean from grand mean	84.64	210.25	364.81	21.16
					TOTAL
SS(between)= SSB = Σn( x̅ - x̅̅)² =	677.12	1682	2918.48	169.28	5446.88

df between = k-1 = 4-1=

3

mean square between groups(treatment) , MSB = SSB/k-1 =    1815.6267

so, answer is none of above(option d)

b)

F-stat>f-critical,so, reject Ho

option a) is answer

c)

MSE=SSE/df = 8431.31/28=301.118

tukey critical range=q_0.05,4,28*√(MSE/n)=3.8612*√(301.118/8) = 23.6890

so, option b) 23.8044 is answer

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please revert if have any doubts