Question

Question 5. (Two-factor ANOVA) A manufacturer had problems with of its catalyst production plant. The problem was that the factory was not have the same production rates under identical conditions. After discussion with the research unit of which you are part, it was agreed to analyze the effects twelve (12) different combinations of four (4) reagents and three (3) catalysts. The recorded production rates were collected in the following table (2 obs / case).

Catalyseur Réactif 11,7 13, 15 15,9 12, 14 9,7 13, 13 6,4 13, 15 12, 12

1) Construct Two-Factor ANOVA Without Replication

2) Construct Two-Factor ANOVA With Replication

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Answer #1

C17 2REACTIF 2 3 3 A 6 7 13 9 6 13 15 12 12 13 9 12 14 9 D 7 10 9​​​​​​Enter data in excel as given above.

1)Go to Data option of excel placed on the top ribbon.Then at the right last cornner of excel sheet there is Data Analysis option.choose option.

2)After choosing one dilog boX is opened name is Data Analysis=> Analysis:Tools

3)For two factor ANOVA without replication

Choose option from the list name is

Anova:Two factor without replication=>Another dilog box is opened

a)select input range.Tick Labels option.

b)select output range=> click ok

Output is on screen as given below

1 Anova: Two-Factor Without Replication SUMMARY CountSum Average Variance 20 6.666667 14.33333 22 7.333333 2.333333 26 8.6666Here,

F1=2.021< Fc=2.7641 and F2=2.6808<Fc=3.7388 at 0.05 significance level then we accept both null(H0) Hypothesis.

4)For Two factor Anova with replication

Follow the same procedure as given above upto Data Analysis dilog box is opened

Choose option from the list name is

Anova:Two factor with replication=>Another dilog box is opened

a)select input range

b)select rows per sample in our case it is true.

c)select output range=> click ok output is on screen.

as given below

2 Anova: Two-Factor With Replication SUMMARY 3 Total Count Sum Average Variance 2 10 2 14 7 拷 2 6 181 42 2 6.8 212 10 Count SCount Sum Average 2 26 13 2 2 16 24 12 0 8l 11 Variance 2 6.4 Total Count Sum Average Variance 8 72 9 96172 12 19.71429 8.285Here,

F1=10>Fc=3.49

F2=6>Fc=3.88

F3=3.5>2.99

We reject all null hypothesis (H0) at 0.05 significance level.

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