- #1

chwala

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- 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.

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