Solve the problem involving Anova -Statistics

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In summary, the conversation discusses a statistical analysis of three different suppliers' parachutes. The critical value and null and alternative hypotheses are provided, along with calculations for within and between group variances and the resulting F value. The conclusion is that the null hypothesis is rejected and the alternative hypothesis, that there is a difference between the suppliers' parachutes, is accepted. The correct alternative hypothesis should be that at least two of the means are not equal, rather than all three means being not equal.
  • #1
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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  • #2
Oops ! :wink:

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

I have 4.553 there ...

##\ ##
 
  • #3
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.
 
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  • #4
Explain why your alternative hypothesis is as given. That is not the opposite of ##\mu_1=\mu _2=\mu _3##.
 
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  • #5
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##
 
  • #6
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
[tex]
\mu _1\neq \mu _2 \quad\mathrm{or}\quad \mu _2\neq\mu _3\quad\mathrm{or}\quad \mu_1\neq \mu _3.
[/tex]
 
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  • #7
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
[tex]
\mu _1\neq \mu _2 \quad\mathrm{or}\quad \mu _2\neq\mu _3\quad\mathrm{or}\quad \mu_1\neq \mu _3.
[/tex]
but how different is this;

\mu _1 = \mu _2## and ##\mu _2=\mu _3## and ##\mu _1=\mu _3## [tex]
\mu _1\neq \mu _2 \quad\mathrm{or}\quad \mu _2\neq\mu _3\quad\mathrm{or}\quad \mu_1\neq \mu _3.
[/tex]

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...
 
  • #8
##\mu _1 \neq \mu _2 \neq \mu _3## is read as
[tex]
\mu _1\neq \mu _2 \quad\mathrm{and}\quad \mu _2\neq \mu _3,
[/tex]
which is not equivalent to what I wrote. This is not debatable. If your work is sloppy, your results are sloppy/incorrect.
 
  • #9
nuuskur said:
##\mu _1 \neq \mu _2 \neq \mu _3## is read as
[tex]
\mu _1\neq \mu _2 \quad\mathrm{and}\quad \mu _2\neq \mu _3,
[/tex]
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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Related to Solve the problem involving Anova -Statistics

1. What is ANOVA and when is it used?

ANOVA, or Analysis of Variance, is a statistical method used to compare the means of three or more groups. It is used when there are multiple groups being compared and the researcher wants to determine if there is a significant difference between the groups.

2. How does ANOVA differ from t-tests?

ANOVA is used to compare the means of three or more groups, while t-tests are used to compare the means of only two groups. ANOVA also takes into account the variance within each group, while t-tests only consider the difference between the means.

3. What is the difference between one-way and two-way ANOVA?

One-way ANOVA is used when there is only one independent variable being studied, while two-way ANOVA is used when there are two independent variables being studied. Two-way ANOVA also allows for the examination of interactions between the two variables.

4. How do you interpret the results of an ANOVA test?

The results of an ANOVA test will provide a p-value, which indicates the probability of obtaining the observed results by chance. If the p-value is less than the chosen significance level (usually 0.05), then there is a significant difference between at least two of the groups. Additionally, the ANOVA test will provide an F-statistic, which is used to determine the significance of the results.

5. What are the assumptions of ANOVA?

The assumptions of ANOVA include: normality of the data, equal variances between groups, independence of observations, and random sampling from the population. Violation of these assumptions can affect the accuracy of the ANOVA results and may require alternative statistical methods.

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