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Repeat Exercise supposing that \(\,\,{\overline x _{2.}} = 502.8\)in addition to\(\,\,{\overline x _{3.}} = 427.5.\)

Short Answer

Expert verified

The T Method for Identifying Significantly Different μi ’s

Find value \({Q_{\alpha ,}}I,{I_{\left( {j - 1} \right)}}\)at the Table A.10. in the appendix of the book for given \(\alpha \).

\(\begin{aligned}{l}\,\,{\overline x _{3.}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\overline x _{1.}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\overline x _{4.}}\,\,\,\,\,\,\,\,\,\,\\\underline {\,\,\,437.5\,\,\,\,\,\,\,462.0\,\,\,\,\,\,\,\,\,\,\,\,469.3\,\,\,\,\,\,\,\,} \end{aligned}\)\(\begin{aligned}{l}\,\,{\overline x _{2.}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\overline x _{5.}}\\\underline {\,\,\,515.8\,\,\,\,\,\,\,532.1\,\,\,\,\,\,\,\,} \end{aligned}\)

Step by step solution

01

definition of mean

Two are more numbers of mathematical average called mean.

The T Method for Identifying Significantly Different μi ’s

Find value \({Q_{\alpha ,}}I,{I_{\left( {j - 1} \right)}}\)at the Table A.10. in the appendix of the book for given \(\alpha \).

Compute and list the sample means in increasing order. Calculate

\(w = {Q_{\alpha ,}}I,{I_{\left( {j - 1} \right)}}.\sqrt {\frac{{MSE}}{J}} \)

and underline pairs of the sample means that differ by less than W . The pair of sample which are not underscored by the same line corresponding of population or treatment means that they are significantly different.

From the mentioned table, and\(\alpha = 0.05\)

\(\begin{aligned}{l}{Q_{\alpha ,}}I,{I_{\left( {j - 1} \right)}} = {Q_{0.5,5,15,}}\\ = 4.37\end{aligned}\)

The value of

\(MSE = 278.8\)

Compute the w value as

\(w = {Q_{\alpha ,}}I,{I_{\left( {j - 1} \right)}}.\sqrt {\frac{{MSE}}{J}} = 4.37..\sqrt {\frac{{272.8}}{4}} = 36.09\)

First order the sample means

\({\overline x _{3.}} < {\overline x _{1.}}\,\,\, < {\overline x _{4.}}\,\, < {\overline x _{2.}} < {\overline x _{5.}}\)

The following table

Brand, i

Sample mean

\({\overline x _{i.}} - {\overline x _{3.}}\)

\({\overline x _{i.}} - {\overline x _{1.}}\)

\({\overline x _{i.}} - {\overline x _{4.}}\)

\({\overline x _{i.}} - {\overline x _{2.}}\)

\(3\)

\(427.5\)

\(1\)

\(462\)

\(34.5\)

\(4\)

\(469.3\)

\(41.8\)

\(7.3\)

\(2\)

\(515.8\)

\(75.3\)

\(40.8\)

\(33.5\)

\(5\)

\(532.1\)

\(104.6\)

\(70.1\)

\(62.8\)

\(16.3\)

The bold values are smaller than w

Start with mean third sample

\({\overline x _{1.}} - {\overline x _{3.}} = 462 - 427.5 = 34.5 < w = 36.09,\)

Indicated that pair \(\left( {3,1} \right)\)underline as pair:

\(\begin{aligned}{l}\,\,\,\,{\overline x _{3.}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\overline x _{1.}}\\\underline {\,\,\,427.5\,\,\,\,\,\,\,462\,\,\,\,\,\,\,\,} \,\end{aligned}\)

Do the same for all pairs

\({\overline x _{4.}} - {\overline x _{3.}} = \,469.3 - 427.5 = 41.8 > w = 36.09\)

So the pair \(\left( {4,3} \right)\) underline as a pair:

\({\overline x _{4.}} - {\overline x _{1.}} = \,469.3 - 462 = 7.3 < w = 36.09\)

SO The pair \(\left( {1,4} \right)\) underline together

\(\begin{aligned}{l}\,\,{\overline x _{1.}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\overline x _{4.}}\\\underline {\,\,\,462\,\,\,\,\,\,\,469.3\,\,\,\,\,\,\,\,} \end{aligned}\)

Next pair is \(\left( {1,2} \right)\)

\({\overline x _{2.}} - {\overline x _{1.}} = 502.8 - 469.3 = 33.5 < w = 36.09\)

The pair\(\left( {1,2} \right)\) should not be underline together.

\({\overline x _{2.}} - {\overline x _{4.}} = 502.8 - 469.3 = 33.5 < w = 36.09\)

02

first difference

This pair \(\left( {4,2} \right)\) should underline together.

\(\begin{aligned}{l}\,\,{\overline x _{2.}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\overline x _{4.}}\\\underline {\,\,\,502.8\,\,\,\,\,\,\,469.3\,\,\,\,\,\,\,\,} \end{aligned}\)

The next pair is\(\left( {5,4} \right)\)

\({\overline x _{5.}} - {\overline x _{4.}} = 532.8 - 469.3 = 62.8 < w = 36.09\)

So the pair \(\left( {4,5} \right)\)should not be underline together.

\({\overline x _{5.}} - {\overline x _{2.}} = 532.1 - 502.8 = 29.3 < w = 36.09\)

So the pair \(\left( {2,5} \right)\)should underline together.

\(\begin{aligned}{l}\,\,{\overline x _{2.}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\overline x _{5.}}\\\underline {\,\,\,515.8\,\,\,\,\,\,\,532.1\,\,\,\,\,\,\,\,} \end{aligned}\)

The connection can be represented

\(\begin{aligned}{l}\,\,{\overline x _{3.}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\overline x _{1.}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\overline x _{4.}}\,\,\,\,\,\,\,\,\,\,\\\underline {\,\,\,437.5\,\,\,\,\,\,\,462.0\,\,\,\,\,\,\,\,\,\,\,\,469.3\,\,\,\,\,\,\,\,} \end{aligned}\)\(\begin{aligned}{l}\,\,{\overline x _{2.}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\overline x _{5.}}\\\underline {\,\,\,515.8\,\,\,\,\,\,\,532.1\,\,\,\,\,\,\,\,} \end{aligned}\)

The conclusion is that there are no significant differences between pairs the a are which are connected with a line: there is significant difference pair that are not connected.

Hence,

\(\begin{aligned}{l}\,\,{\overline x _{3.}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\overline x _{1.}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\overline x _{4.}}\,\,\,\,\,\,\,\,\,\,\\\underline {\,\,\,437.5\,\,\,\,\,\,\,462.0\,\,\,\,\,\,\,\,\,\,\,\,469.3\,\,\,\,\,\,\,\,} \end{aligned}\)\(\begin{aligned}{l}\,\,{\overline x _{2.}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\overline x _{5.}}\\\underline {\,\,\,515.8\,\,\,\,\,\,\,532.1\,\,\,\,\,\,\,\,} \end{aligned}\)

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Most popular questions from this chapter

The following data refers to yield of tomatoes (kg/plot) for four different levels of salinity. Salinity level here refers to electrical conductivity (EC), where the chosen levels were EC = 1.6, 3.8, 6.0, and 10.2 nmhos/cm.

Use the F test at level\(\alpha \)=.05 to test for any differences in true average yield due to the different salinity levels.

Four laboratories \(\left( {1 - 4} \right)\)are randomly selected from a large population, and each is asked to make three determinations of the percentage of methyl alcohol in specimens of a compound taken from a single batch. Based on the accompanying data, the difference among laboratories a source of variation in the percentage of methyl alcohol? State and test the relevant hypothesis using significance level \(.0.5\)

\(\begin{aligned}{*{20}{c}}{1:}&{85.06}&{85.25}&{84.87}\\{2:}&{84.99}&{84.28}&{84.88}\\{3:}&{84.48}&{84.72}&{85.10}\\{4:}&{84.10}&{84.55}&{84.05}\end{aligned}\)

Repeat Exercise supposing that \(\,\,{\overline x _{2.}} = 502.8\)in addition to\(\,\,{\overline x _{3.}} = 427.5.\)

When sample sizes are equal\(\left( {{J_i} = J} \right)\), the parameters\({\alpha _1},{\alpha _2},...{\alpha _I}\,\)of the alternative parameterization are restricted by\(\Sigma {\alpha _i} = 0\). For unequal sample sizes, the most natural restriction is\(\Sigma {J_i}{\alpha _i} = 0\). Use this to show that

\(E\left( {MSTr} \right) = {\alpha ^2} + \frac{1}{{I - 1}}\Sigma {J_i}\alpha _i^2\)

What is\(E\left( {MSTr} \right)\)when\({H_0}\)is true? (This expectation is correct if\(\Sigma {J_i}{\alpha _i} = 0\,\)is replaced by the restriction\(\Sigma {\alpha _i} = 0\)(or any other single linear restriction on the ai ’s used to reduce the model to I independent parameters), but\(\Sigma {J_i}{\alpha _i} = 0\)simplifies the algebra and yields natural estimates for the model parameters (in particular,\({\mathop {\,\,\,\,\alpha }\limits^{\,\,\,\,\,\^} _i} = {\mathop X\limits^\_ _i}\, - \,\mathop X\limits^\_ ..\,\,\,\,{H_0}\)).)

When sample sizes are not equal, the noncentrality parameter is \(\sum {{J_i}\alpha _i^2/{\sigma ^2}} \)and \({\phi ^2} = (1/I)\sum {{J_i}\alpha _i^2/{\sigma ^2}} \) .Referring to Exercise 22, what is the power of the test when \({\mu _2} = {\mu _3},{\mu _1} = {\mu _2} - \sigma \), and \({\mu _4} = {\mu _2} + \sigma \)?

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