using Excel
data -> data analysis -> Anova Single Factor
Anova: Single Factor |
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SUMMARY |
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Groups |
Count |
Sum |
Average |
Variance |
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Distilled
H2O |
20 |
400.15 |
20.0075 |
0.0055 |
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Petro |
20 |
399.32 |
19.9660 |
0.0086 |
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NaCl |
20 |
398.62 |
19.9310 |
0.0081 |
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MgCl |
20 |
398.53 |
19.9265 |
0.0103 |
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NaCl + MgCl |
20 |
396.74 |
19.8370 |
0.0073 |
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ANOVA |
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Source of Variation |
SS |
df |
MS |
F |
P-value |
F crit |
Between
Groups |
0.3180 |
4 |
0.0795 |
9.9463 |
0.0000 |
2.4675 |
Within
Groups |
0.7593 |
95 |
0.0080 |
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Total |
1.0773 |
99 |
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since p-value = 0.0000 < alpha
we reject the null hypothesis
we conclude that there is significant difference in means
2)
option C),D) and E) are correct
it wrong then remove option C)
3)
There is a significant difference in growth between the five
groups of Phragmites plants.
I would reject the null hypothesis.
option )A and C) are correct