Solve the problem involving Anova -Statistics

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Homework Help Overview

The discussion revolves around a statistical problem involving ANOVA, specifically testing hypotheses related to means of different groups. Participants are analyzing variance calculations and hypothesis formulations in the context of a statistical test.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to clarify the formulation of null and alternative hypotheses, with some suggesting different interpretations of what constitutes the alternative hypothesis. There are also discussions on the calculations of variances and the implications of the results.

Discussion Status

The discussion is ongoing, with participants providing corrections to variance calculations and questioning the definitions of hypotheses. Some guidance has been offered regarding the proper interpretation of the alternative hypothesis, but no consensus has been reached on the correct formulation.

Contextual Notes

There are indications of confusion regarding the definitions of hypotheses and the implications of statistical results. Participants are also addressing potential errors in calculations, which may affect the overall understanding of the problem.

chwala
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Homework Statement
See attached
Relevant Equations
stats
1648549036446.png


Kindly note that i do not have the solution or mark scheme to this question.

My take on this,
From stats...my points are in summary... ( am assuming the reader is literate on this), we shall have;
##k=3, N=15##
##d.f.N=2##
##d.f.D=12##
##α=0.05##

The critical value = ##3.89##
Let ##H_0: μ_1=μ_2=μ_3##
##H_A: μ_1≠μ_2=μ_3## or ##μ_1=μ_2≠μ_3##

##X_{am}##= ##\dfrac{333.1}{15}##=##22.207##

Between Group Variance;

##S^2_B##=##\dfrac{5(19.52-22.207)^2+5(24.26-22.207)^2+5(22.84-22.207)^2}{2}##

=##\dfrac{36.099845+21.074045+2.003445}{2}=## ##\dfrac{59.177335}{2}##=##29.5886##

Within Group Variance;

##S^2_W##=##\dfrac{4(7.237+3.683+1.9062)}{12}##=##\dfrac{12.8262×4}{12}##=##4.2754##

##F##=##\dfrac{S^2_B}{S^2_W}##=##\dfrac{29.5886}{4.2754}##=##6.9266>3.89##

Conclusion;
We reject the Null Hypothesis and accept the Alternative Hypothesis that there is a difference between the supplier's parachutes.
 
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Oops ! :wink:

##\dfrac{4(7.237+3.683+{\bf \color{red}{1.9062}})}{12}##

I have 4.553 there ...

##\ ##
 
BvU said:
Oops ! :wink:

##\dfrac{4(7.237+3.683+{\bf \color{red}{1.9062}})}{12}##

I have 4.553 there ...

##\ ##
true error there... It is supposed to be,

##\dfrac{18.212}{4}##

Let me amend last part here...no need to interfere with original post...to make it easier for viewers to follow;

Within Group Variance;

##S^2_W##=##\dfrac{4(7.237+3.683+4.553)}{12}##=##\dfrac{15.473×4}{12}##=##5.1576666##

##F##=##\dfrac{S^2_B}{S^2_W}##=##\dfrac{29.5886}{5.1576666}##=##5.73681>3.89##

Conclusion;
We reject the Null Hypothesis and accept the Alternative Hypothesis that there is a difference between the supplier's parachutes.
 
Last edited:
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Explain why your alternative hypothesis is as given. That is not the opposite of ##\mu_1=\mu _2=\mu _3##.
 
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I may need your input there...i tried to follow literature on such like questions, should it be

Let ##H_0: μ_1=μ_2=μ_3##
##H_A: μ_1≠μ_2≠μ_3##
 
The setup is as follows. You start with your initial hypothesis. If that turns out to be false with some level of confidence, then the opposite statement is accepted. In this case the null hypothesis is: ##\mu _1 = \mu _2## and ##\mu _2=\mu _3## and ##\mu _1=\mu _3## (it is redundant to write it like this, but it makes what follows clearer). The alternative is the negation, which is
<br /> \mu _1\neq \mu _2 \quad\mathrm{or}\quad \mu _2\neq\mu _3\quad\mathrm{or}\quad \mu_1\neq \mu _3.<br />
 
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nuuskur said:
The setup is as follows. You start with your initial hypothesis. If that turns out to be false with some level of confidence, then the opposite statement is accepted. In this case the null hypothesis is: ##\mu _1 = \mu _2## and ##\mu _2=\mu _3## and ##\mu _1=\mu _3## (it is redundant to write it like this, but it makes what follows clearer). The alternative is the negation, which is
<br /> \mu _1\neq \mu _2 \quad\mathrm{or}\quad \mu _2\neq\mu _3\quad\mathrm{or}\quad \mu_1\neq \mu _3.<br />
but how different is this;

\mu _1 = \mu _2## and ##\mu _2=\mu _3## and ##\mu _1=\mu _3## <br /> \mu _1\neq \mu _2 \quad\mathrm{or}\quad \mu _2\neq\mu _3\quad\mathrm{or}\quad \mu_1\neq \mu _3.<br />

from this;

##H_0: μ_1=μ_2=μ_3##
##H_A: μ_1≠μ_2≠μ_3##

I think this is also fine mate...
 
##\mu _1 \neq \mu _2 \neq \mu _3## is read as
<br /> \mu _1\neq \mu _2 \quad\mathrm{and}\quad \mu _2\neq \mu _3,<br />
which is not equivalent to what I wrote. This is not debatable. If your work is sloppy, your results are sloppy/incorrect.
 
nuuskur said:
##\mu _1 \neq \mu _2 \neq \mu _3## is read as
<br /> \mu _1\neq \mu _2 \quad\mathrm{and}\quad \mu _2\neq \mu _3,<br />
which is not equivalent to what I wrote. This is not debatable. If your work is sloppy, your results are sloppy/incorrect.
Ok noted...learning point for me, thanks nuuskur.
 
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