Solve the problem involving Anova -Statistics

[chwala]

Homework Statement

See attached

Relevant Equations

stats

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.

 

[BvU]

Oops !

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

I have 4.553 there ...

$\$

 

[chwala]

Oops !

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

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.

 

[nuuskur]

Explain why your alternative hypothesis is as given. That is not the opposite of $\mu_1=\mu _2=\mu _3$.

 

[chwala]

I may need your input there...i tried to follow literature on such like questions, should it be

Let $H_0: μ_1=μ_2=μ_3$
$H_A: μ_1≠μ_2≠μ_3$

 

[nuuskur]

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
$$\mu _1\neq \mu _2 \quad\mathrm{or}\quad \mu _2\neq\mu _3\quad\mathrm{or}\quad \mu_1\neq \mu _3.$$

 

[chwala]

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
$$\mu _1\neq \mu _2 \quad\mathrm{or}\quad \mu _2\neq\mu _3\quad\mathrm{or}\quad \mu_1\neq \mu _3.$$

but how different is this;

\mu _1 = \mu _2$and$\mu _2=\mu _3$and$\mu _1=\mu _3## $$\mu _1\neq \mu _2 \quad\mathrm{or}\quad \mu _2\neq\mu _3\quad\mathrm{or}\quad \mu_1\neq \mu _3.$$

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...

 

[nuuskur]

$\mu _1 \neq \mu _2 \neq \mu _3$ is read as
$$\mu _1\neq \mu _2 \quad\mathrm{and}\quad \mu _2\neq \mu _3,$$
which is not equivalent to what I wrote. This is not debatable. If your work is sloppy, your results are sloppy/incorrect.

 

[chwala]

$\mu _1 \neq \mu _2 \neq \mu _3$ is read as
$$\mu _1\neq \mu _2 \quad\mathrm{and}\quad \mu _2\neq \mu _3,$$
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.