For each problem provide the following: . Hypotheses . Pre-test results and tran
ID: 3360775 • Letter: F
Question
For each problem provide the following: . Hypotheses . Pre-test results and transformations (if necessary Appropriate ANOVA table . Post-test (if applicable) .Conclusion: statistical & biological 1. A botanist is studying the effects of different fertilizers on wheat growth. The wheat berry production for each type of fertilizer is found below. Does there appear to be a difference in the number of wheat berries produced depending on fertilizer type? 3. Cow 4. Chicken 1. SuperGro2. CrazyPlant 30 32 38 26 25 28 38 38 32 30 37 32 27 31 27 26 29 27 25 25 32 31 29 27 32 30 30Explanation / Answer
Hypothesis of the data,
we are given above the table of ANOVA,
The hypotheses of interest in an ANOVA are as follows:
where k = the number of independent comparison groups.
the formulas for calculation for ANOVA,
df
ss
mss
f
p
Between
k-1
SSb
MSB=SSB/k-1
F=MSB/MSE
within
n-k
SSE
MSE=SSE/n-k
total
n-1
SST
the analysis table for anova is
Source of Variation
SS
df
MS
F
P-value
F crit
Between Groups
160.1071
3
53.36905
4.261407
0.015099
3.008787
Within Groups
300.5714
24
12.52381
Total
460.6786
27
conclusion:
p-value is =0.01 , p<0.5 therefor we reject the null hypothe sis and we can conclud that the there is significant diffreance between the all groups.
Post test(Tukeys test)
The p-value corrresponing to the F-statistic of one-way ANOVA is lower than 0.05 which strongly suggests that
one or more pairs of treatments are significantly different. You have k=4k=4 treatments, for which we shall apply
Tukey's HSD test to each of the 6 pairs to pinpoint which of them exhibits statistially significant difference.
We first establish the critical value of the Tukey-Kramer HSD QQ statistic based on the k=4 treatments
and =24 degrees of freedom for the error term, for significance level =0.05 (p-values) in
the Studentized Range distribution. We obtain these ctitical values forQ for of 0.05,
critical value of Q at Q =0.05,k=4,=24 Q= 3.9015.
so in the table group B and Group C has significant diffreance from each other.
thanks.
df
ss
mss
f
p
Between
k-1
SSb
MSB=SSB/k-1
F=MSB/MSE
within
n-k
SSE
MSE=SSE/n-k
total
n-1
SST
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