Question
In: Economics
Three suppliers provide parts in shipments of 500 units. Random samples of six shipments from each...
- Three suppliers provide parts in shipments of 500 units. Random samples of six shipments from each of the three suppliers were carefully checked and the numbers of numbers not conforming to standards were recorded. These numbers are listed in the following table:
|
Supplier A |
Supplier B |
Supplier C |
|
28 37 34 29 31 33 |
22 27 29 20 18 30 |
33 29 39 33 37 38 |
- Prepare the analysis of variance table for these data
- Test the null hypothesis at the 5% of that the population mean numbers of parts per shipment not conforming to standards are the same for all three suppliers.
Solutions
Expert Solution
i. The degree of freedom for the groups would be m-1 or 2, for m is number of groups. The degree of freedom for the error would be n-m or 15, for n is the total number of observations.
The sum of square total would be 
or 
or 
. The sum of square of error would be 
or 
or 
or 
. The sum of square between would be 
or 
or 
.
The anova table would be as below.
| DF | SS | MS | F | P-val | |
| Group | 2 | 354.1111 | 354.1111/2=177.0556 |  |
0.0014 |
| Error | 15 | 254.1667 | 254.1667/15=16.9445 | ||
| Total | 17 | 608.2778 |
ii. The null hypothesis is that the mean of the groups is the same for all, while the alternate would be that the mean of the groups is not same for all.
The F-statistic obtained is 
. The critical F would be 
, and since 
, we may reject the null that the mean is the same for all. This
means that the variation between groups is quite high. This can
also be confirmed by the p-value, which is less than 0.05 alpha
level.
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