R Homework 6

# Chicken Weights by Feed Type ####

# An experiment was conducted to measure and compare the effectiveness
# of various feed supplements on the growth rate of chickens. 

data("chickwts")
summary(chickwts)

##      weight             feed   
##  Min.   :108.0   casein   :12  
##  1st Qu.:204.5   horsebean:10  
##  Median :258.0   linseed  :12  
##  Mean   :261.3   meatmeal :11  
##  3rd Qu.:323.5   soybean  :14  
##  Max.   :423.0   sunflower:12

Plot your data first (as always). Be sure to include an informative figure caption.

plot(chickwts$weight~chickwts$feed)

Figure 1: Chicken weights based on different foods eaten

Are your data normal and have homogeneity of variance? Provide three pieces of graphical evidence that argues your case. Be sure to include informative figure captions. Explain your answer.

I’d say that this data has homogeneity of variance because the spread of the area on the box plots are around the same. The Q-Q plot shows that the normality because the points lie on the best fit line, towards the ends of the data there seem to be some stragglers.

hist(chickwts$weight) #This shows that there is a normal distribution

Figure 2: Chicken Weights plotted on a histogram

boxplot(chickwts$weight~chickwts$feed) #This shows equal variance

Figure 3: Chicken weights data plotted on a box plot for equal variance

qqnorm(chickwts$weight) #This also shows for normality
qqline(chickwts$weight)

Figure 4: Chicken weights Q-Q plot to check for normality

Did you transform your data? If so, state which transformation you used. Provide three pieces of evidence that your data more closely approximates a normal distribution and homogeneity of variance. If not, state why you did not transform the data.

I didn’t transform the data. As we see in figures 3-5 there these functions don’t normalize the data and the original function is more normal before the fact.

#The following codes are to check to see if we can transform the data to make it more normal.
chickwts$weightLog<-log10(chickwts$weight) 
qqnorm(chickwts$weightLog)
qqline(chickwts$weightLog)

Figure 5: Chicken weights Q-Q plot transformed using log base 10

chickwts$weightSqrt<-sqrt(chickwts$weight)
qqnorm(chickwts$weightSqrt)
qqline(chickwts$weightSqrt)

Figure 6:Chicken weights Q-Q plot transformed using the square root function

chickwts$weightSqrd<-(chickwts$weight)^2
qqnorm(chickwts$weightSqrd)
qqline(chickwts$weightSqrd)

figure 7: Chicken weights Q-Q plot transformed using the square function

Create an ANOVA object with the appropriate data (raw or transformed). Present your results in the R Markdown file. Did you accept or reject the null hypothesis? Are the results statistically significant? Provide and interpret two evidence graphs that the residuals meet the assumption of the ANOVA.

The p-value was a really small number so we can reject the null hypothesis and say that there is a significant difference.

weight.aov<-aov(weight~feed, data=chickwts)
plot(weight.aov)

Figure 8: ANOVA for chicken weight by feed.

summary(weight.aov)

##             Df Sum Sq Mean Sq F value   Pr(>F)    
## feed         5 231129   46226   15.37 5.94e-10 ***
## Residuals   65 195556    3009                     
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Did you perform a multiple comparison test? Why or why not? Explain. Present your code if appropriate.

Since we rejected the null hypothesis in the previous question, we can run a multiple comparison test because there is a significant difference. The alternate hypothesis says there is a difference, but we can’t know where it is until we run this test and see that there is a significant difference between Horsebean and Casein, Linseed and Casein, Soybean and Casein, Meatmeal and Horsebean, Soybean and Horsebean, Sunflower and Horsebean, Sunflower and Linseed, Sunflower and Soybean.

TukeyHSD(weight.aov)

##   Tukey multiple comparisons of means
##     95% family-wise confidence level
## 
## Fit: aov(formula = weight ~ feed, data = chickwts)
## 
## $feed
##                            diff         lwr       upr     p adj
## horsebean-casein    -163.383333 -232.346876 -94.41979 0.0000000
## linseed-casein      -104.833333 -170.587491 -39.07918 0.0002100
## meatmeal-casein      -46.674242 -113.906207  20.55772 0.3324584
## soybean-casein       -77.154762 -140.517054 -13.79247 0.0083653
## sunflower-casein       5.333333  -60.420825  71.08749 0.9998902
## linseed-horsebean     58.550000  -10.413543 127.51354 0.1413329
## meatmeal-horsebean   116.709091   46.335105 187.08308 0.0001062
## soybean-horsebean     86.228571   19.541684 152.91546 0.0042167
## sunflower-horsebean  168.716667   99.753124 237.68021 0.0000000
## meatmeal-linseed      58.159091   -9.072873 125.39106 0.1276965
## soybean-linseed       27.678571  -35.683721  91.04086 0.7932853
## sunflower-linseed    110.166667   44.412509 175.92082 0.0000884
## soybean-meatmeal     -30.480519  -95.375109  34.41407 0.7391356
## sunflower-meatmeal    52.007576  -15.224388 119.23954 0.2206962
## sunflower-soybean     82.488095   19.125803 145.85039 0.0038845

Summarize your results in a paragraph similar to the example in the ``Reporting Your Results’’ section.

There is a significant difference between the weights of the chickens and the feed type used (ANOVA p-value=0.0000). The data was not transformed and is the raw data obtained from the field. A Tukey HSD multiple comparison test was used to determine which of the feed was significantly different from the other. The weights of the chicken who ate Horsebean versus Casein(p=0.000) are significantly lower and Linseed versus Casein(p=0.000) are also significantly lower. However, chickens who ate Soybean versus Casein(p=0.008), Meatmeal versus Horsebean(p=0.000), Soybean versus Horsebean(p=0.004), Sunflower versus Horsebean(p=0.000), Sunflower versus Linseed(p=0.000), Sunflower versus Soybean(p=0.004) are significantly higher than it’s counterpart.

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