In: Statistics and Probability
ANOVA |
df |
MS |
F |
|
SSTR |
134 |
2 |
67 |
4.9 |
SSE |
164 |
12 |
13.67 |
|
SST |
298 |
14 |
Paper Towel Absorbency
A |
B |
C |
|
1 |
91 |
99 |
83 |
2 |
100 |
96 |
88 |
3 |
88 |
94 |
89 |
4 |
89 |
89 |
86 |
(OR) TRY THIS ANSWER IN TYPING
From the ANOVA table, we infer that the Critical value of F is 3.8853 and since the calculated value of F is 4.9>the critical value, there is at least a pair of treatments whose difference is significant. Therefore, we use the LSD to find out the pair.
a).
Fisher's LSD is given by the formula where t is the tabulated value of t at 10% level with df as the error df, is the means sum of squares for error, are the replications for treatment 1 and 2. In our case, the number of replications for all the three treatments are the same.
For our case, the , , . Substituting teh values, we have,
We now calculate the means of each treatment.
A | B | C | |
1 | 91 | 99 | 83 |
2 | 100 | 96 | 88 |
3 | 88 | 94 | 89 |
4 | 89 | 89 | 86 |
Mean | 92 | 94.5 | 86.5 |
Whenever the absolute value difference between any pair of tretments is more tahn the LSD, we then conclude that the pair is statistically significant.
We then calculate the absolute difference for comparisions as follows:
Treatments | Mean | Comparision | Difference | ABS(Diff) | LSD | Significant? |
A | 92 | A Vs. B | -2.5 | 2.5 | 4.6596 | No |
B | 94.5 | A Vs. C | 5.5 | 5.5 | 4.6596 | Yes |
C | 86.5 | B Vs. C | 8 | 8 | 4.6596 | Yes |
b).
We see that the Differences A and B are not significant but A and C and B and C are significant. Hence we conclude that the among the three, C is the least. Hence, I don't agree with the Director's arguet that the ability to absorb water is the same for all the three brands since A and B are statistically difference from C.
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