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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

An experiment was carried out to compare electrical resitivity for six different low-permeability concrete bridge deck mixtures. There were \(26\)measurements on concrete cylinders for each mixture; these were obtained days \(28\)after casting.The entries in the accompanying ANOVA table are based on information in the article 鈥淚n-place Resitivity of Bridge Deck Concrete Mixtures鈥(ACI Matrerials j.,2009: 114-122).Fill in the remaining entries and test appropriate hypothese.

it is common practice in many countries to destroy (shred)refrigerators at the end of their usefull lives.In this process material from insulating foam may be released into the atmosphere.The article 鈥淩elease of fluorocarbons from Insulation foam in Home Appliances During shredding鈥(J.of the Air and waste Mgmt.Assoc.,2007:1452-1460)gave the following data on foam density(g/L)For each of two refrigerators produced by four different manufactures:

\(\begin{aligned}{l}1.30.4,\,29.2\,\,\,\,\,2.27.7,27.1\\3.27.1,24.8\,\,\,\,\,\,4.25.5,28.8\end{aligned}\)

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A:

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8

B:

14

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9

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C:

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D:

17

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12

22

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It is believed that the number of flaws has approximately a Poisson distribution for each brand. Analyse the data at level .01 to see whether the expected number of flaws per reel is the same for each brand.

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

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Oats \(8.3\) \(6.1\) \(7.8\) \(7.0\) \(5.5\) \(7.2\)

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