91影视

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.

Given,

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

\({\mu _i}\)is the true average growth hormone type \(i,i = 1,2,...5\)

The data given the table

This table summarizes everything needed to carry out F teat. Here is explanation How to obtain those values.

\(I = 5\)Column-treatments

And

\(J = 4\)Row,

The following table needs to be filled with corresponding values:

The degrees of freedom are

\(\begin{aligned}{l}I - 1 = 5 - 1 = 4\\I.\left( {J - 1} \right)5.\left( {4.1} \right) = 15\\I\,\,.\,J - 1 = 5.4 - 1 = 19\end{aligned}\)

Denote with

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

The total sum of squares

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

And treatment sum of squares

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

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

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

Compute all value:

\(\begin{aligned}{l}{X_{1.}} = 13 + 17 + 7 + 14 = 51;\\{X_{2.}} = 21 + 13 + 20 + 17 = 71;\\{X_{3.}} = 18 + 15 + 20 + 17 = 20;\\{X_{4.}} = 7 + 11 + 18 + 10 = 46;\\{X_{5.}} = 6 + 11 + 15 + 8 = 40;\end{aligned}\)

Values of

\(\overline {{x_{i.}}} = \frac{1}{J}.{x_{i.}}\)

Are given by

\(\begin{aligned}{l}\overline {{x_{1.}}} = \frac{1}{4}.51 = 12.75\\\overline {{x_{2.}}} = \frac{1}{4}.71 = 17.75;\\\overline {{x_{3.}}} = \frac{1}{4}70 = 17.5;\\\overline {{x_{4.}}} = \frac{1}{4}.46 = 11.5;\\\overline {{x_{5.}}} = \frac{1}{4}.40 = 10;\end{aligned}\)

The grand mean is \(\overline {x..} \)

\(\begin{aligned}{l}\overline {x..} = \frac{1}{{I\,\,.\,\,J}}.x.. = \sum\limits_{i = 1}^I {\sum\limits_{j = 1}^J {{x_{ij}}} = \frac{1}{{5.4}}.\left( {13 + 21.. + 10 + 8} \right)} \\ = 13.9\end{aligned}\)

And

\(\begin{aligned}{l}x.. = \sum\limits_{j = 1}^J {{x_{ij}} = } \left( {13 + 21... + 10 + 8} \right)\\ = 278\end{aligned}\)

Total sum square

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

\(\begin{aligned}{l} = \left( {{{13}^2} + {{21}^2} + {{...10}^2} + {{18}^2}} \right) - \frac{1}{{5\,.\,4}}{.278^2}\\ = 4280 - 3864.2\\ = 415.8.\end{aligned}\)

The treatment sum of square is

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

\(\begin{aligned}{l} = \frac{1}{4}.\left( {{{51}^2} + {{71}^2} + {{70}^2} + {{46}^2} + {{40}^2}} \right) - \frac{1}{{5\,.\,4}}{.278^2}\\ = 4064 - 3864.2\\ = 200.3\end{aligned}\)

Fundamental Identify

SST = SSTr + SSE.

Error sum of squares is

\(SST - SSTr = 415.8 - 200.3 = 215.5\)

The mean computed

\(\begin{aligned}{l}MSTr = \frac{1}{{I - 1}}.SSTr = \frac{1}{4}.200.3 = 50.075;\\MSE = \frac{1}{{I\,.\left( {J - 1} \right)}}.SSE = \frac{1}{{5.\left( {4 - 1} \right)}}.215.5 = 14.3667.\end{aligned}\)

The value of F statistic is

\(f = \frac{{MSTr}}{{MSE}} = \frac{{50.075}}{{14.3667}} = 3.485.\)

ANOVA table now

As for usual tests, you 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 F curve where F has Fisher's distribution with degrees of freedom \(4\) and \(30\) ; thus

\(P = P\left( {F > f} \right) = P\left( {F > 3.485} \right) = 0.033,\)

which was computed using software

\(p = 0.033 < 0.05 = \alpha \)

Reject null hypothesis

at given significance level. There is no statistically significance difference in true averages among the four types of iron formation.

Using the table, you could use e.g. \({F_{0,1,2,3,36}}\,\,is\,0.1\)value for which the area under the curve to the right of. \({F_{0,05,4,15}}\,is0.05\) The value is

\({F_{0,05,4,15}} = 3.06 < 3.485 = f\)

which indicates to reject null hypothesis

significant level \(0.05\)

Hence,

The T Method for Identifying Significantly Different 渭i 鈥檚

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

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

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\(\alpha = 0.05\)\(I = 6,J = 26\)

\({Q_{\alpha ,}}I,{I_{\left( {J - 1} \right)}} = {Q_{0.05,5,15}} = 4.37\)

The value of of estimate using the table.

\(MSE = 14.3667\)

Compute the w value as

\(w = {Q_{\alpha ,}}I,{I_{\left( {j - 1} \right)}}.\sqrt {\frac{{MSE}}{J}} = 4.37.\sqrt {\frac{{14.3667}}{7}} = 8.28\)

First order the sample means

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

The bold value are smaller than w

Every single difference is smaller than \(w = 8.28\)

Conclusion is that there is no significant difference between any treatments.F test and Turkey鈥檚 method are at odds.This can be represented using lines as

\(\begin{aligned}{l}{\overline x _{5.\,\,\,\,}}\,\,\,\,{\overline x _{4.\,\,\,\,}}\,\,\,\,{\overline x _{1.}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\overline x _{3.\,\,\,\,}}\,\,\,\,\,\,\,\,\,\,\,\,\,{\overline x _{2.}}\\\underline {10\,\,\,\,\,\,11.5\,\,\,12.75\,\,\,\,\,17.5\,\,\,\,\,\,17.75} \end{aligned}\)

Hence,

a. Reject null hypothesis

b. There are no significant differences