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# Variability - ANOVA - Mean Sum

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1. Variability in a dataset may be caused by various sources. Provide an example sample dataset where are more than one sources causing its variability. Explain briefly?with one or two lines explanations of those sources. Also, calculate the sample variance for the example data set that you have provided.

2. Organize the sample data you have presented in DQ 1 into groups according to the levels of one source of (or cause of variation) variation. Now compute the Mean Sum of Square Between and Mean Sum of Square Within. How do you characterize the Mean Sum of Square Between in your example? Is it too large, why or why not?

3. While doing and ANOVA you found the following:

The Mean Sum of Square Between (MSSB) = 2.76 with df 5 and the Mean Sum of square Within (MSSW)= 10.89 with df 11. What conclusion can you draw regarding the null hypothesis of the equality of the means?

Note: The test statistic in this case, F = MSSB/MSSW with numerator df = 5 and the denominator df = 11.

https://brainmass.com/statistics/analysis-of-variance/variability-anova-mean-sum-214536

## SOLUTION This solution is FREE courtesy of BrainMass!

1. Variability in a dataset may be caused by various sources. Provide an example sample dataset where are more than one sources causing its variability. Explain briefly—with one or two lines explanations of those sources. Also, calculate the sample variance for the example data set that you have provided.

Variability could be a result of 2 factors. For example, there are three ways of drying green concrete (say A, B, and C), and you want to find the one that gives you the best compressive strength. The concrete is mixed in batches that are large enough to produce exactly three cylinders. However, your production engineer believes that there is substantial variation in the quality of the concrete from batch to batch. Therefore, now the variability could be because of the drying treatment applied as well as because of the batch selected. The differences due to batch selection may be so high that the effect of drying treatment is not visible. In such a case, randomized block test should be used to check for the effects.

Batch (j)
Treatment (i) 1 2 3 4 5 Treatment Average Variance
A 52 47 44 51 42 47.2 18.7
B 60 55 49 52 43 51.8 40.7
C 56 48 45 44 38 46.2 43.2
Batch Mean 56.0 50.0 46.0 49.0 41.0 48.4 30.3

You have data from J = 5 batches on each of the I = 3 drying processes.

2. Organize the sample data you have presented in DQ 1 into groups according to the levels of one source of (or cause of variation) variation. Now compute the Mean Sum of Square Between and Mean Sum of Square Within. How do you characterize the Mean Sum of Square Between in your example? Is it too large, why or why not?

SSBlock = = 363.6 (Calculations in excel)

SSTreatment = = 89.2 (Calculations in excel)

SSTotal = = 499.6 (Calculations in excel)

SSError = SSTotal - SSBlock - SSTreatment
= 499.6 - 363.6 - 89.2 = 46.8

From excel calculation,
MSSWith = SST/(I-1) = 44.6
MSSBet = SSB/(J-1) = 90.9

3. While doing and ANOVA you found the following:

The Mean Sum of Square Between (MSSB) = 2.76 with df 5 and the Mean Sum of square Within (MSSW)= 10.89 with df 11. What conclusion can you draw regarding the null hypothesis of the equality of the means?

Note: The test statistic in this case, F = MSSB/MSSW with numerator df = 5 and the denominator df = 11.

F = = 2.76/10.89 = 0.2534

p-value = 0.9293 > 0.05 [Using excel function: FDIST(0.2534, 5, 11)]

Since the p-value is greater than 0.05, the null hypothesis is not rejected. Hence, at 5% significance level, there is not sufficient evidence to conclude that the means are different.

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