Restaurants Blocks of Evaluators X 18 Y 18 12 13 11 19 17 16 16 10 18 10 14 20 9. Use a level of significance of a = 0.05, is there evidence of a difference in evaluation between restaurants? Solution: Step 1: State the hypotheses. Step 2: The level of significance is Step 3: Determine the degrees of freedom and the critical value of F. 234r

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Author:Amos Gilat
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Chapter1: Starting With Matlab
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Suppose that a Restaurant chain specialized with pasta having three branches in a
particular geographic location would like to evaluate the service at this restaurant. The
researcher for the restaurant hires 18 investigators (evaluator) with varied experiences in
food- service evaluations. After the initial consultations, the 15 investigators are stratified
into five blocks of three-based on food service evaluation experience- so that three most
experience evaluators are placed in block 1, the next three most experienced evaluators
are placed in block 2, and so on. Within each of the five homogeneous blocks, the three
evaluators are then randomly assigned to evaluate the service at a particular restaurant
using a rating scale from 0 (low) to 20 (high). The results are summarized in the table

Restaurants
Blocks of Evaluators
X
Y
1
18
18
12
2
19
17
13
3
16
16
11
4
10
18
10
14
20
Use a level of significance of a = 0.05, is there evidence of a difference in evaluation
between restaurants?
Solution:
Step 1: State the hypotheses.
Himmen-
Haid-
Step 2: The level of significance is
Step 3: Determine the degrees of freedom and the critical value ofF.
Transcribed Image Text:Restaurants Blocks of Evaluators X Y 1 18 18 12 2 19 17 13 3 16 16 11 4 10 18 10 14 20 Use a level of significance of a = 0.05, is there evidence of a difference in evaluation between restaurants? Solution: Step 1: State the hypotheses. Himmen- Haid- Step 2: The level of significance is Step 3: Determine the degrees of freedom and the critical value ofF.
dfsas =
dfen =
Filock.
Treatment
Step 4: Complete the table and compute for the value of F.
Degrees of
Freedom
Source of
Variation
Treatment
(Bonding
Agent)
Sum of
Squares
Mean of
Squares
F
Blocks
| (Materials)
Error
Total
Step 5: Decision Rule.
Treatments:
Blocks:
Step 6: Conclusion.
Treatments:
Blocks:
Transcribed Image Text:dfsas = dfen = Filock. Treatment Step 4: Complete the table and compute for the value of F. Degrees of Freedom Source of Variation Treatment (Bonding Agent) Sum of Squares Mean of Squares F Blocks | (Materials) Error Total Step 5: Decision Rule. Treatments: Blocks: Step 6: Conclusion. Treatments: Blocks:
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