R Notebook

fat <- c(64,    72, 68,77,56,95,
         78, 91, 97, 82, 85, 77,        
         75,93,78,71,63,76,        
         55,66,49,64,70,68)
Fat_type= c(rep("Fat1",6), rep("Fat2",6), rep("Fat3",6), rep("Fat4",6))
dataset<- data.frame(fat,Fat_type)
plot(fat ~ Fat_type, data=dataset)

comments

results  = aov(fat~ Fat_type, data=dataset) 
summary(results)

            Df Sum Sq Mean Sq F value  Pr(>F)   
Fat_type     3   1636   545.5   5.406 0.00688 **
Residuals   20   2018   100.9                   
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

I reject the Null Hypothesis that the means are the same. The value is very small.

pairwise.t.test(fat, Fat_type, p.adjust="bonferroni")

    Pairwise comparisons using t tests with pooled SD 

data:  fat and Fat_type 

     Fat1   Fat2   Fat3  
Fat2 0.2189 -      -     
Fat3 1.0000 0.8182 -     
Fat4 0.6005 0.0046 0.1529

P value adjustment method: bonferroni 

There is a significant difference between Fat 2 at Fat 3 and Fat 2 at Fat 4.

TukeyHSD(results, conf.level = 0.95)

  Tukey multiple comparisons of means
    95% family-wise confidence level

Fit: aov(formula = fat ~ Fat_type, data = dataset)

$Fat_type
          diff        lwr       upr     p adj
Fat2-Fat1   13  -3.232221 29.232221 0.1461929
Fat3-Fat1    4 -12.232221 20.232221 0.8998057
Fat4-Fat1  -10 -26.232221  6.232221 0.3378150
Fat3-Fat2   -9 -25.232221  7.232221 0.4270717
Fat4-Fat2  -23 -39.232221 -6.767779 0.0039064
Fat4-Fat3  -14 -30.232221  2.232221 0.1065573

There is a significant difference between Fat 2 and Fat 4. That P Value is much smaller than 0.05. Thus, we reject the null.