/*! 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} Q1E In an experiment to compare the ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

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.

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

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 : \({\sum x ^2}_{ij.} = 5,241,420.79.)\)

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.

The following data refers to yield of tomatoes (kg/plot) for four different levels of salinity. Salinity level here refers to electrical conductivity (EC), where the chosen levels were EC = 1.6, 3.8, 6.0, and 10.2 nmhos/cm.

Use the F test at level\(\alpha \)=.05 to test for any differences in true average yield due to the different salinity levels.

The critical flicker frequency \(\left( {cff} \right)\) is the highest frequency at which a person can detect the flicker in a flickering light source. At frequencies above the cff, the light source appear to be continuous even though it is actually flickering. An investigation carried out to see whether true average cff depends on iris color yielded the following data (based on the article 鈥淭he Effects of Iris Color on Critical Flicker Frequency鈥�.

Iris color

1.Brown

2.Green

3.Blue

\({\bf{26}}.{\bf{8}}\)

\({\bf{26}}.{\bf{4}}\)

\({\bf{25}}.{\bf{7}}\)

\({\bf{27}}.{\bf{9}}\)

\({\bf{24}}.{\bf{2}}\)

\({\bf{27}}.{\bf{2}}\)

\({\bf{23}}.{\bf{7}}\)

\({\bf{28}}.{\bf{0}}\)

\({\bf{29}}.{\bf{9}}\)

\({\bf{25}}.{\bf{0}}\)

\({\bf{26}}.{\bf{9}}\)

\({\bf{28}}.{\bf{5}}\)

\({\bf{26}}.{\bf{3}}\)

\({\bf{29}}.{\bf{1}}\)

\({\bf{29}}.{\bf{4}}\)

\({\bf{24}}.{\bf{8}}\)

\({\bf{28}}.{\bf{3}}\)

\({\bf{25}}.{\bf{7}}\)

\({\bf{24}}.{\bf{5}}\)

\({J_i}\)

\({\bf{8}}\)

\({\bf{5}}\)

\({\bf{6}}\)

\({x_i}\)

\({\bf{204}}.{\bf{7}}\)

\({\bf{134}}.{\bf{6}}\)

\({\bf{169}}.{\bf{0}}\)

\({\overline x _i}\)

\({\bf{25}}.{\bf{59}}\)

\({\bf{26}}.{\bf{92}}\)

\({\bf{28}}.{\bf{17}}\)

\(n = 19,{x_{..}} = 508.3\)

  1. State and test the relevant hypotheses at significance level\(.05\)(Hint:\(\sum {\sum {{x_{ij}}^2} = 13659.67,CF = 13598.36} \))
  2. Investigate difference between iris colors with respect to mean cff.

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:

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

Does it appear that true average foam density is not the same for all these manufacture?carry out an appropriate test of hypotheses by obtaining as much p-value information as possible,and summarize your analysis in an ANOVA table.
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Decision Maths

Read Explanation

Logic and Functions

Read Explanation

Applied Mathematics

Read Explanation

Calculus

Read Explanation

Geometry

Read Explanation
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.