In: Statistics and Probability
A company is considering purchasing wine glasses from 1 of 4 possible manufacturers (identified as Manufacturer A, Manufacturer B, Manufacturer C, and Manufacturer D). The cost of the wine glasses is the same for each manufacturer, so the company is conducting a test to determine if there is a difference in the average life of glasses produced by the manufacturers. Four (4) glasses are randomly sampled from each manufacturer and the glasses are tested until they break. A partially completed ANOVA table of the results is shown below. The F critical value is shown for α = 0.05.
ANOVA |
|||||
Source of Variation |
SS |
df |
MS |
F |
F crit |
Between Groups |
204993 |
3 |
3.4903 |
||
Within Groups |
78218 |
||||
Total |
283211 |
15 |
1. State the Null and alternative hypotheses of interest.
Ho: __________________________________________
HA: __________________________________________
Explanation:-
Given the partially data ANOVA table
source of Variation | sum of squares | degrees of freedom | mean squares | tabulated value F |
Between groups | 204993 | 3 | - | 3.49 |
within groups | 78213 | - | - | |
Total | 283211 | 15 | - |
Step(i):-
Mean of squares between groups
=
=
Mean of squares between within groups
=
=
Step(ii):-
Now Complete ANOVA table
source of Variation | sum of squares | degrees of freedom | mean squares | critical value F |
Between groups | 204993 | 3 | 68,331 | 3.49 |
within groups | 78213 | 12 | 6518.16 | |
Total | 283211 | 15 | 74,849.16 |
a)
Null hypothesis:- H0: The manufacturer has do not effect on the average life of the wine glasses
Alternative hypothesis:H1: The manufacturer has effect on the average life of the wine glasses
b) Mean of squares between groups = 68,331
c) Mean of squares within groups = 6518.16
d) F-statistic
the critical value F(3 , 12 ) = 3.49
Conclusion:-
The calculated value F = 10.48 > the critical value 3.49
Null hypothesis is rejected
Alternative hypothesis is accepted
The manufacturer has effect on the average life of the wine glasses