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Chapter 10: Q1E (page 419)

In an experiment to compare the tensile strengths of different type of\({\bf{I = 5}}\) copper wire, \({\bf{J = 4}}\) samples of each type were used. The between-samples and within- sample estimates of \({{\bf{\sigma }}^{\bf{2}}}\) were computed as \({\bf{MS}}{{\bf{T}}_{\bf{r}}}{\bf{ = 2673}}{\bf{.3}}\) and respectively. Use the F test at level. to test \({{\bf{H}}_{\bf{0}}}\,{\bf{:}}\,\,{{\bf{\mu }}_{\bf{1}}}{\bf{ = }}{{\bf{\mu }}_{\bf{2}}}{\bf{ = }}{{\bf{\mu }}_{\bf{3}}}{\bf{ = }}{{\bf{\mu }}_{\bf{4}}}{\bf{ = }}{{\bf{\mu }}_{\bf{5}}}\)versus \({{\bf{H}}_{\bf{a}}}{\bf{:}}\)at least two \({{\bf{\mu }}_{\bf{i}}}{\bf{'s}}\) are unequal.

Short Answer

Expert verified

Do not reject null hypothesis and

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

\({H_0}\,:\,\) at least two of the are different.

F is ratio of the two mean squares

\(F = \frac{{MS{T_r}}}{{MSE}}.\)

The two values are given in the exercise

\(\begin{aligned}{l}MS{T_r} = 2673;\\MSE = 1094.2.\end{aligned}\)

02

Value of statistic

Thus, the values of the statistic F is

\(\begin{aligned}{c}f = \frac{{MS{T_r}}}{{MSE}}\\ = \frac{{267.3}}{{1094.2}}\\ = 2.44.\end{aligned}\)

The degrees of freedom are \(l - 1 = 5 - 1 = 4\) and \(I\left( {J - 1} \right) = 5 \times 3 = 15.\)

Using degrees of freedom for \(\alpha = 0.05\) , the P value is

\(P = P\left( {F > 2.44} \right) = 0.09\)

This value was computed using software. You can estimate it using the table in the appendix of the book using software is recommended

03

Decision

Since \(P = 0.09 > 0.05,\) do not reject null hypothesis

At significance level \({\rm{0}}{\rm{.5}}{\rm{.}}\) From the given value it can be concluded that there is no statistical difference in the mean of tensile strengths of the given types.

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Most popular questions from this chapter

Let \({c_1},{c_2},...{c_l}\) be numbers satisfying \(\sum {{c_i} = 0} \). Then \(\sum {{c_i}{\mu _i} = {c_1}{\mu _1} + .. + {c_l}{\mu _l}} \) is called a contrast in the \({\mu _i}'s\). Notice that with \({c_1} = 1,{c_2} = - 1,{c_3} = ... = {c_1} = 0,\sum {{c_{i{\mu _i}}} = {\mu _1} - {\mu _2}} \)which implies that every pairwise difference between \({\mu _i}'s\)is a contrast. A method attributed to Scheffe鈥檚 gives simultaneous CI鈥檚 with simultaneous confidence level \(100\left( {1 - \alpha } \right)\% \) for all possible contrast. The interval for \(\sum {{c_i}{\mu _i}} \)is

\({\sum {{c_i}{{\overline x }_{i.}} \pm \left( {\sum {{\raise0.7ex\hbox{\({{c_i}^2}\)} \!\mathord{\left/

{\vphantom {{{c_i}^2} {{J_i}}}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{\({{J_i}}\)}}} } \right)} ^{{\raise0.7ex\hbox{\(1\)} \!\mathord{\left/

{\vphantom {1 2}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{\(2\)}}}} \times {\left( {\left( {I - 1} \right) \times MSE \times {F_{\alpha ,I - 1,n - I}}} \right)^{{\raise0.7ex\hbox{\(1\)} \!\mathord{\left/

{\vphantom {1 2}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{\(2\)}}}}\)

Using the critical flicker frequency data of Exercise \(42\), calculate the Scheffe interval for the contrast \({\mu _1} - {\mu _2},{\mu _1} - {\mu _3},{\mu _2} - {\mu _3}and5{\mu _1} + 5{\mu _2} - {\mu _3}\). Which contrast appear to differ significantly from \(0\), and why?

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 \(.0.5\)

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

A study of the properties of mental plate 鈥揷onnected trusses used for roof support (鈥淢odeling joints made with Light-Gauge metal connector plates,鈥� Forest products J., 1979:39-44) yielded the following observations on axial-stiffness index(kips/in.) for plate lengths \(4,6,8.10,12\)in:

\(\begin{aligned}{l}4:309.2\\6:402.1\\8:392.4\\10:346.7\\12:407.4\end{aligned}\) \(\begin{aligned}{l}409.5\\347.2\\366.2\\452.9\\441.8\end{aligned}\) \(\begin{aligned}{l}311.0\\361.0\\351.0\\461.4\\419.9\end{aligned}\) \(\begin{aligned}{l}326.5\\404.5\\357.1\\433.1\\410.7\end{aligned}\) \(\begin{aligned}{l}316.8\\331.0\\409.9\\410.6\\473.4\end{aligned}\) \(\begin{aligned}{l}349.8\\348.9\\367.3\\384.2\\441.2\end{aligned}\) \(\begin{aligned}{l}309.7\\381.7\\382.0\\362.6\\465.8\end{aligned}\)

Does variation in plate length have any effect on true average axial stiffness? state and test the relevant hypotheses using analysis of variance with Display your results in an ANOVA table.(Hint : \({\sum x ^2}_{ij.} = 5,241,420.79.)\)

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 \({\mu _4} = {\mu _2} + \sigma \)?

An experiment to compare the spreading rates of five brands of yellow interior latex paint available in a particular area used \(4\)gallons \(\left( {J = 4} \right)\)of each paint. The sample average spreading rates \(\left( {f{t^2}/gal} \right)\) for the five brands were \(\,{\overline x _{1.}} = 462.3,\,{\overline x _{2.}} = 512.8,\,{\overline x _{3.}} = 437.5,\,{\overline x _{4.}} = 469.3\,and\,\,{\overline x _{5.}} = 532.1\,\) the computed value of F was found to be significant at level \(\alpha = .05.\) with MSE= \(272.8\)use Tukey鈥檚 procedure to investigate significant differences in the true average spreading rates between brands.
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