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# Interchanging two rows in a contingency table will not have an effect on the chi- - page 12 / 13

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c)

Cannot conclude proportion of acceptances differ by individuals in segments A-E

d)

Conclude proportion of acceptances differ by individuals in segments A-E

Conclude that as segment increases, mean number of acceptances increases

21.

For the following tables of observed and expected cell counts for a test of whether the proportions of individuals selecting brand A (versus Brand B) is the same for 3 different label types for Brand A, give the chi-square statistic, and P-value.

Observed (f)

Label \ Selection

Brand A

Brand B

1

200

100

2

400

200

3

300

150

Expected (e)

Label \ Selection

Brand A

Brand B

1

200

100

2

400

200

3

300

150

a)  chi-square statistic = 0 ,      p-value = 0

b)  chi-square statistic = 0.67 , p-value > 0.05

c)  chi-square statistic = 2.00 , p-value > 0.05

d)  chi-square statistic = 0 ,      p-value = 1

e)  chi-square statistic = 0  ,     p-value = 0.50

1.

An experiment is conducted to measure percent shrinkage in a=2 types of fabric at each of b=3 drying temperatures. The following table gives the sample means (standard deviations) based on 5 replicates at each of the combinations of fabric and temperature.

## Fabric \ Drying Temp

210o

220o

230o

1

2.0  (0.2)

4.0  (0.4)

9.0  (1.0)

2

3.0  (0.3)

4.0  (0.5)

8.0  (0.7)

You wish to fit the model:

a)

Give the ANOVA table.

b)

Test whether there is a temperature/fabric interaction

i.

Null hypothesis:

ii.

Alternative hypothesis:

iii.

Test Statistic:

iv.

Decision Rule:

c)

Test whether there is a dryer termperature main effect at =0.05 significance level.

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