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An article in the material

13-12. An article in the Materials Research Bulletin (1991, Vol. 26(11)] investigated four different methods of preparing the
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Answer:

a)

For the given data, Using one way ANOVA

Treatments Total w 5 5 ΣΧ 61.5 66.9 276.4 Mean 14.8 14.8 12.3 13.38 13.82 x2 1095.22 1095.3 757.31 902.09 3849.92 Std.Dev. 0.Source SS MS Between- treatments 22.124 7.3747 F= 14.84583 Within- treatments 7.948 0.4967 Total 30.072 19 The f-ratio value

b) P- value = 0.00007

c)

Normal Probability Plot (response is Response) +---------------------------------------------------- + --------- ---------- -Versus Fits (response is Response) Residual -1.5- -2.0-4 12.0 12.5 13.0 13.5 Fitted Value 14.0 14.5 15.0 The normal probabili

d)

formula for the 100(1-a) percent confidence interval on the mean of the ith treatment is Jottakam),/MS, From the data we have

Therefore, the 95% confidence interval on the mean Te when method 1 is used to prepare the material is 7.10.025.16MS/n = 14.8

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