/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q16E Reconsider the axial stiffness d... [FREE SOLUTION] | 91影视

91影视

Reconsider the axial stiffness data given in Exercise ANOVA output from Minitab follows:

Analysis of variance for stiffness

Source DF SS MS F p

Length \(4\) \(43993\) \(10998\) \(10.48\) \(0.000\)

Error \(30\) \(31475\) \(1049\)

Total \(34\) \(75468\)

Level \(N\) Mean St Dev

\(4\) \(7\) \(333.21\) \(36.59\)

\(6\) \(7\) \(368.06\) \(28.57\)

\(8\) \(7\) \(375.13\) \(20.83\)

\(10\) \(7\) \(407.36\) \(44.51\)

\(12\) \(7\) \(437.17\) \(26.00\)

Pooled st Dev \( = 32.39\)

Tukey鈥檚 pairwise comaparisons

Family error rate \( = 0.0500\)

Individual error rate\( = 0.00693\)

Critical value\( = 4.10\)

Intervals for (column level mean)-(row level mean)

\(4\) \(6\) \(8\) \(10\)

\(6\) \(\begin{aligned}{l} - 85.0\\15.4\end{aligned}\)

\(8\) \(\begin{aligned}{l} - 92.1\\\,\,\,\,8.3\end{aligned}\) \(\begin{aligned}{l} - 57.3\\43.1\end{aligned}\)

\(10\) \(\begin{aligned}{l} - 124.3\\\, - 23.9\end{aligned}\) \(\begin{aligned}{l} - 89.5\\10.9\end{aligned}\) \(\begin{aligned}{l} - 82.4\\18.0\end{aligned}\)

\(12\) \(\begin{aligned}{l}\, - 15.2\\\, - 53.8\end{aligned}\) \(\begin{aligned}{l} - 119.3\\ - 18.9\end{aligned}\) \(\begin{aligned}{l} - 112.2\\ - 11.8\end{aligned}\) \(\begin{aligned}{l} - 80.0\\20.4\end{aligned}\)

  1. Is it plausible that the variances of the five axial stiffness index distributions are identical?Explain

b. Use the output(without references to our F table)to test the relevant hypotheses.

c.Use the Tukey intervals given in the output to determine which means differ,and construct the corresponding underscoring pattern.

Short Answer

Expert verified

Reconsider the axial stiffness data given.

The fact that the following is true

\({S_4} = 44.51 > 2.20.83 = 2.{S_3}\)

The hypotheses of interest are

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

\(\begin{aligned}{l}\,\,\,\,\,{\overline x _{1.}}\,\,\,\,\,\,\,\,\,\,\,\,\,{\overline x _{2.}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\overline x _{3.}}\\\underline {333.21\,\,\,\,\,368.06\,\,\,\,\,\,375.13\,\,\,\,} \end{aligned}\) \(\underline \begin{aligned}{l}\,\,\,\,{\overline x _{4.}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\overline x _{5.}}\,\\407.36\,\,\,\,437.17\,\,\,\end{aligned} \,\,\,\)

Step by step solution

01

Step1:Definition of hypothese.

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.

a) plausible that the variances of the five axial stiffness index.

The fact that the following is true

\({S_4} = 44.51 > 2.20.83 = 2.{S_3}\)

b) To test the relevant hypotheses.

you can conclude that the population variances are equal (this stands).

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 value of F statistic is given, and it is \(f = 10.48;\)

Thus the P value is the area under the F curve to the right of f which can be found at the appendix.

\(P(F > f) = 0\)

e has degrees of freedom \(4\,\,and\,30\) (given in the exercise). Hence,

reject null hypothesis

there is difference in axial stiffness for the different plate lengths.

c) construct the corresponding underscoring pattern.

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\(\alpha = 0.05\)\(I = 5,J = 7\)

\({Q_{\alpha ,}}I,{I_{\left( {J - 1} \right)}} = {Q_{0.05,5,30}} = 4.1\)

The value of of estimate using the table.

\(MSE = 10.43\)

Compute the w value as

\(w = {Q_{\alpha ,}}I,{I_{\left( {j - 1} \right)}}.\sqrt {\frac{{MSE}}{J}} = 4.1.\sqrt {\frac{{1049}}{7}} = 50.79\)

The differences can be computed manually very easy; they are also given in the table in the exercise.

02

differences between means

Difference between means that are smaller than\(w = 50.79\)it is clear which are smaller the part. conclusion is that three group have been created, first group containing axial stiffness\(3,6\,\,and\,\,8\) a second group containing axial stiffness\(6,8\,\,and\,10\)third group containing axial stiffness \(10,12\) between the group there is significant difference but within the groups there are no significant differences. This can be represented using line .

\(\begin{aligned}{l}\,\,\,\,\,{\overline x _{1.}}\,\,\,\,\,\,\,\,\,\,\,\,\,{\overline x _{2.}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\overline x _{3.}}\\\underline {333.21\,\,\,\,\,368.06\,\,\,\,\,\,375.13\,\,\,\,} \end{aligned}\) \(\underline \begin{aligned}{l}\,\,\,\,{\overline x _{4.}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\overline x _{5.}}\,\\407.36\,\,\,\,437.17\,\,\,\end{aligned} \,\,\,\)

Hence,

\(\begin{aligned}{l}\,\,\,\,\,{\overline x _{1.}}\,\,\,\,\,\,\,\,\,\,\,\,\,{\overline x _{2.}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\overline x _{3.}}\\\underline {333.21\,\,\,\,\,368.06\,\,\,\,\,\,375.13\,\,\,\,} \end{aligned}\) \(\underline \begin{aligned}{l}\,\,\,\,{\overline x _{4.}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\overline x _{5.}}\,\\407.36\,\,\,\,437.17\,\,\,\end{aligned} \,\,\,\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Apply the modified Tukey鈥檚 method to the data in Exercise 22 to identify significant differences among the\({\mu _i}\)鈥檚.

In an experiment to compare the quality of four different brands of magnetic recording tape, five 2400-ft reels of each brand (A鈥揇) were selected and the number of flaws in each reel was determined.

A:

10

5

12

14

8

B:

14

12

17

9

8

C:

13

18

10

15

18

D:

17

16

12

22

14

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.

The article 鈥渙rigin of Precambrian iron formations鈥(Econ,Geology,1964;1025-1057)reports the following data on total Fe for four types of iron formation\((1 = carbonate,\,2 = silicate,\,3 = magnetite,\,4 = hematite).\)

\(\begin{aligned}{l}1:\,\,\,20.5\,\,\,\,\,\\\,\,\,\,\,\,25.2\\2:26.3\\\,\,\,\,\,\,34.0\\3:29.5\\\,\,\,\,\,\,\,26.2\\4:36.5\\\,\,\,\,\,\,33.1\end{aligned}\)\(\begin{aligned}{l}28.1\\25.3\\24.0\\17.1\\34.0\\29.9\\44.2\\34.1\end{aligned}\) \(\begin{aligned}{l}27.8\\27.1\\26.2\\26.8\\27.5\\29.5\\34.1\\32.9\end{aligned}\) \(\begin{aligned}{l}27.0\\20.5\\20.2\\23.7\\29.4\\30.0\\30.3\\36.3\end{aligned}\) \(\begin{aligned}{l}28.0\\31.3\\23.7\\24.9\\27.9\\35.6\\31.4\\25.5\end{aligned}\)

Carry out analysis of variance F test at significance level \(.01,\) and summarize the results in an ANOVA table.

Refer to Exercise \(19\) and suppose\(\overline {{X_{1.}}} = 10,\,\overline {{X_{2.}}} = 15,\overline {{X_{3.}}} = 20\) can you now find a values of SSE that produces such a contradiction between the F test and Tukey鈥檚 procedure?

Refer to Exercise 38. What is b for the test when true average DNA content is identical for three of the diets and falls below this common value by 1 standard deviation (s) for the other two diets?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.