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Magazine Chapter 10: Q46SE (page 436)
In Example \(10.11\), subtract \({\overline x _{i.}}\), from each observation in the \({i^{th}}\)sample \(\left( {i = 1,...,6} \right)\) to obtain a set of \(18\)residuals. Then construct a normal probability plot and comment on the plausibility of the normality assumption.
Short Answer
The probability plot is
And it is plausible to assume normality.
Step by step solution
Step 1: Analyzing the given data and its differences
The data given in the Table in Example \(10.11\) with the corresponding averages is shown in the below table.
\(\begin{aligned}{*{20}{c}}{S.NO}&{1:}&{2:}&{3:}&{4:}&{5:}&{6:}\\1&{55}&{26}&{78}&{92}&{49}&{80}\\2&{53}&{37}&{91}&{100}&{51}&{85}\\3&{54}&{32}&{85}&{96}&{50}&{83}\\{{{\overline x }_i}}&{54.00}&{31.67}&{84.67}&{96.00}&{50.00}&{82.67}\end{aligned}\)
Using the table above, the appropriate differences are,
\(\begin{aligned}{*{20}{c}}{1:}&{2:}&{3:}&{4:}&{5:}&{6:}\\{1.00}&{ - 5.67}&{ - 6.67}&{ - 4.00}&{ - 1.00}&{ - 2.67}\\{ - 1.00}&{5.33}&{6.33}&{4.00}&{1.00}&{2.33}\\{0.00}&{0.33}&{0.33}&{0.00}&{0.00}&{0.33}\end{aligned}\)
Plotting the probability graph
Firstly, let us arrange them in order from smallest to highest.
\(\begin{aligned}{l} - 6.67, - 5.67, - 4.00, - 2.67, - 1.00, - 1.00,0.00,0.00,0.00,0.33,0.33,0.33,\\1.00,1.00,2.33,4.00,5.33,6.33\end{aligned}\)
The corresponding \(z\)percentiles are,
\(\begin{aligned}{l} - 1.91, - 1.38, - 1.09, - 0.86, - 0.67, - 0.67, - 0.36, - 0.36, - 0.36,0.07,0.07,\\0.36,0.51,0.51,0.86,1.09,1.38,1.91\end{aligned}\)
The normal probability plot is shown in the figure below,
The above plot shows us that there is no doubt that the residuals are not normally distributed.
So, it is plausible to assume normality.
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