Solve the problem involving Anova -Statistics

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SUMMARY

The discussion centers on solving an ANOVA problem with three groups (k=3) and a total of 15 observations (N=15). The analysis reveals that the critical value at α=0.05 is 3.89, leading to the rejection of the null hypothesis (H0: μ1=μ2=μ3) in favor of the alternative hypothesis (HA: μ1≠μ2≠μ3), indicating a significant difference between the groups. The calculated F-value of 5.73681 exceeds the critical value, confirming the presence of variance among the groups. The discussion also highlights the importance of precise hypothesis formulation and the implications of sloppy work on statistical results.

PREREQUISITES
  • Understanding of ANOVA (Analysis of Variance) concepts
  • Familiarity with hypothesis testing and critical values
  • Knowledge of variance calculations (between-group and within-group variance)
  • Basic statistical literacy, including interpretation of F-values
NEXT STEPS
  • Study the derivation and application of ANOVA formulas in statistical software like R or Python's SciPy library.
  • Learn about post-hoc tests following ANOVA to determine specific group differences.
  • Explore the assumptions of ANOVA, including normality and homogeneity of variances.
  • Investigate the implications of Type I and Type II errors in hypothesis testing.
USEFUL FOR

Statisticians, data analysts, researchers conducting experiments, and students studying statistics who want to deepen their understanding of ANOVA and hypothesis testing.

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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