Interpretation of tw-way ANOVA result
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Entering edit mode
4.0 years ago

Dear all,

does anybody knows how to interpret two way ANOVA result. My data looks like:

Protein_name    holo_1      holo_2
A1              82.3965243  70.91176151
B1              27.26637961 47.63355456
C1              97.75786493 64.92764661
D1              115.9354513 127.4018061
E1              130.4860545 163.4261778
F1              57.13565305 142.0628876
G1              88.66907173 87.42791862
H1              184.2934171 150.3209662
I1              95.70968618 68.99684474
J1              53.80736258 79.40920466
K1              166.5425346 97.48123164


I have run two ANOVA test in R studio with this commands:

x = c(82.3965243, 27.26637961, 97.75786493, 115.9354513, 130.4860545, 57.13565305, 88.66907173, 184.2934171, 95.70968618, 53.80736258, 166.5425346, 70.91176151, 47.63355456, 64.92764661, 127.4018061, 163.4261778, 142.0628876, 87.42791862, 150.3209662, 68.99684474, 79.40920466, 97.48123164)

h = as.factor(c(1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,2))
p = as.factor(c(1,2,3,4,5,6,7,8,9,10,11,1,2,3,4,5,6,7,8,9,10,11))

cbind(x,h,p)

AOV.OUT = aov(x~h+p)
summary(AOV.OUT)

shapiro.test(AOV.OUT$resi)  The output is : > cbind(x,h,p) x h p [1,] 82.39652 1 1 [2,] 27.26638 1 2 [3,] 97.75786 1 3 [4,] 115.93545 1 4 [5,] 130.48605 1 5 [6,] 57.13565 1 6 [7,] 88.66907 1 7 [8,] 184.29342 1 8 [9,] 95.70969 1 9 [10,] 53.80736 1 10 [11,] 166.54253 1 11 [12,] 70.91176 2 1 [13,] 47.63355 2 2 [14,] 64.92765 2 3 [15,] 127.40181 2 4 [16,] 163.42618 2 5 [17,] 142.06289 2 6 [18,] 87.42792 2 7 [19,] 150.32097 2 8 [20,] 68.99684 2 9 [21,] 79.40920 2 10 [22,] 97.48123 2 11 > AOV.OUT = aov(x~h+p) > summary(AOV.OUT) Df Sum Sq Mean Sq F value Pr(>F) h 1 0 0.0 0.000 1.0000 p 10 29209 2920.9 3.367 0.0343 * Residuals 10 8674 867.4 --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 > shapiro.test(AOV.OUT$resi)


I am trying to compare my sample two way : (1) holo1 vs holo2 and (2) across the proteins as well.

Does anybody know how to interpret this result? Shapiro-Wilk normality test

data:  AOV.OUT\$resi
W = 0.97786, p-value = 0.8795

ANOVA • 823 views
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Entering edit mode
4.0 years ago
ashish ▴ 680

You need to do a post hoc test to get the results you are looking for. Using the data you have provided I did HSD test.

anova <- aov(x ~ h + p, data = data)

library(agricolae)
HSD.test(anova, trt = c("h", "p"), console = TRUE)

Treatments with the same letter are not significantly different.

x        groups
holo_1:H1 184.29342      a
holo_1:K1 166.54253      a
holo_2:E1 163.42618      a
holo_2:H1 150.32097      a
holo_2:F1 142.06289      a
holo_1:E1 130.48605      a
holo_2:D1 127.40181      a
holo_1:D1 115.93545      a
holo_1:C1  97.75786      a
holo_2:K1  97.48123      a
holo_1:I1  95.70969      a
holo_1:G1  88.66907      a
holo_2:G1  87.42792      a
holo_1:A1  82.39652      a
holo_2:J1  79.40920      a
holo_2:A1  70.91176      a
holo_2:I1  68.99684      a
holo_2:C1  64.92765      a
holo_1:F1  57.13565      a
holo_1:J1  53.80736      a
holo_2:B1  47.63355      a
holo_1:B1  27.26638      a


It seems there are no significant differences in any of the combinations here. Besides that

1. Why aren't you checking the interaction effects? You can do that by using x ~ h + p + h:p instead of x ~ h + p while doing anova.
2. Do you have replicates for each of these readings ?