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consider the following summary data on the modulus of elasticity\(\left( {x\,\,{{10}^{6\,}}psi} \right)\)for lumber of three different grades (in close agreement with values in the article 鈥淏ending strength stiffness of second-Growth Douglas-fir Dimension Lumber鈥 (Forest products J., 1991:35-43), except that the sample size there were larger):

Use this data and a significance level of to test the null hypothesis of no difference in mean modulus of elasticity for the three grades.

Short Answer

Expert verified

consider the following summary data on the modulus of elasticity.

The hypotheses of interest are

\({H_0}\,:\,{\mu _i} = {\mu _j},i \ne j\,\)

versus alternative hypothesis

\({H_a}:\)at least two of the are different,

do not reject null hypothesis

Step by step solution

01

definition of hypotheses

Hypotheses are typically written in the form of if/then statements, such as if someone consumes a lot of sugar, they will develop cavities in their teeth.

The hypotheses of interest are

\({H_0}\,:\,{\mu _i} = {\mu _j},i \ne j\,\)

versus alternative hypothesis

\({H_a}:\)at least two of the are different,

where \(\,{\mu _1},\,{\mu _2},\,{\mu _3}\) are corresponding true averages modulus of elasticity grade \(1,2,3\)respectively.

Denote with

\(\begin{aligned}{l}{x_i}\sum\limits_{j = 1}^J {{x_{ij}}} \\{x_{..}}\sum\limits_{i = 1}^I {{x_{ij}}} .\end{aligned}\)

The total sum of squares Upgrade

\(\left( {SST} \right),\)

\(\left( {SSTr} \right)\),and

Error sum of squares

\({\rm{ }}\left( {SSE} \right)\) are given by

\(SST = \sum\limits_{i = 1}^I {\sum\limits_{j = 1}^J {{{\left( {{x_{ij}} - \overline x ..} \right)}^2}} = } \sum\limits_{i = 1}^I {\sum\limits_{j = 1}^J {{x^2}_{ij} - \frac{1}{{I\,\,.\,J}}} {x^2}_{..;}} \)

\(SSTr = \sum\limits_{i = 1}^I {\sum\limits_{j = 1}^J {{{\left( {\overline {{x_{i.}}} - \overline x ..} \right)}^2}} = \frac{1}{J}.} \sum\limits_{j = 1}^J {{x^2}_{i.} - \frac{1}{{I\,\,.\,J}}{x^2}_{..;}} \)

\(SSE = \sum\limits_{i = 1}^I {{{\sum\limits_{j = 1}^J {\left( {{x_{ij}} - \overline {{x_{i.}}} } \right)} }^2}} .\)

The mean squares are

\(\begin{aligned}{l}MSTr = \frac{1}{{I - 1}}.SSTr;\\MSE = \frac{1}{{I.\left( {J - 1} \right)}}.SSE.\end{aligned}\)

F is ratio of the two mean

\(F = \frac{{MSTr}}{{MSE}}.\)

The degree of freedom

\(\begin{aligned}{l}I - 1 = 3 - {\rm{ }}1{\rm{ }} = {\rm{ }}2;\\I.\left( {J - 1} \right) = 3 \cdot \left( {10 - 1} \right) = 27;\\I\,\,.\,J - 1 = 3 \cdot 10 - 1 = 29.\end{aligned}\)

\(\overline X ..\)grand mean is

\(\overline X .. = \frac{1}{{I\,\,.\,\,J}}.\,x.. = \frac{1}{I}\sum\limits_{i = 1}^I {\overline {{x_i}} } {\rm{ }} = \frac{1}{3} \cdot \left( {1.63 + 1.56 + 1.42} \right)\)

\( = 1.3567,\)

02

The treatment sum of squares

The treatment sum of squares is

\(SSTr = \sum\limits_{i = 1}^I {\sum\limits_{j = 1}^J {{{\left( {\overline {{x_i}} . - \overline {x..} } \right)}^2}} = J.} {\rm{ }}\sum\limits_{i = 1}^I {{{\left( {\overline {{x_i}} . - \overline x ..} \right)}^2}} \)

\( = 10.{\left( {\left( {1.63 - 1.5367} \right)} \right)^2} + {\left( {1.56 - 1.5367} \right)^2} + {\left. {\left( {1.42 - 1.5367} \right)} \right)^2}\)

\( = 0.2286.\)

Error sum of squares is

\(SSE = \sum\limits_{i = 1}^I {\sum\limits_{j = 1}^J {{{\left( {{x_{ij}} - \overline {{x_i}.} } \right)}^2}} = \left( {J - 1} \right).} \left( {{s_1}^2 + {s_2}^2 + {s_3}^3} \right)\)

\(\begin{aligned}{l} = \left( {10 - 1} \right).\left( {{{0.27}^2} + {{0.24}^2} + {{0.26}^2}} \right)\\ = 1.7829.\end{aligned}\)

The mean squares can be computed now as

\(\begin{aligned}{l}MSTr = \frac{1}{{3 - 1}}.SSTr = \frac{1}{2}.0.2286 = 0.1143;\\MSE = \frac{1}{{3.\left( {10 - 1} \right)}}.1.7829 = 0.066.\end{aligned}\)

The value of F Statistic

\(f = \frac{{MSTr}}{{MSE}} = \frac{{0.1143}}{{0.066}} = 1.73.\)

can either make conclusion about the hypotheses look at the F critical value or a P value. Remember that the hypotheses of interest are

\({H_0}\,:\,{\mu _i} = {\mu _j},i \ne j\,\)

versus alternative hypothesis

\({H_a}:\)at least two of the are different

The P value is the area to the right of f value under the curve where F has Fisher's distribution with degrees of freedom \(2\)and\(27\) ;

\(P = P\left( {F > f} \right) = P\left( {F > 1.3} \right) = 0.1964.{\rm{ }}\)

thus which was computed using software (you could estimate it using the table in the appendix). Because \({\rm{P = 0}}{\rm{.1964 > }}\alpha \)

do not reject null hypothesis

at given significance level. There is no statistically significance difference in true averages among three grades.

Hence,

do not reject null hypothesis.

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