# 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.

boxplot(chickwts) #boxplot of data

qqnorm(chickwts$weight) #qq plots of data
qqline(chickwts$weight)

plot(chickwts$weight~chickwts$feed) #more detailed boxplot of data

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.

The data is mostly normal based on what I saw from the qqplots and how close the data was to 1:1 line, and it does have homogenity of variance because the data in the Residuals vs Fitted data is good and almost has the shape of the rectangle.

qqnorm(chickwts$weight) #qq plot of our data
qqline(chickwts$weight)

chickwts.aov <- aov(weight ~ feed, data = chickwts)
plot(chickwts.aov)  #used this plot to determine homogenity

plot(chickwts$weight~chickwts$feed) #detailed boxplot of our data

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.

No, I did not need to transform my data because it was normal and had equal variance so it was ready to run in ANOVA.

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.

I reject my null hypothesis because my results were statistically different. My null hypothesis was that the means were equal, which they were not.

summary(chickwts.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.

Yes I performed a multiple comparison test to see which means specifally differenent. I did this because my pvalue was significantly different.

TukeyHSD(chickwts.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.

The chicken feed weights are significantly different when predicted by the ANOVA test because the p-value was 5.94e-10 which means that I rejected the null hypothesis stating that all means are equal. A Tukey HSD multiple comparision test assessed whether statistical differences occur between different weights of feed. The compared means that were significantly different were horsebean-casein(p=0.001), linseed-casein(p=0.001), soybean-casein(p=0.008), lineseed-horsebean(p=0.141), meatmeal-horsebean(p=0.001), soybean-horsebean, sunflower-horsebean, meatmeal-linseed, sunflower-horsebean(p=0.004), meatmeal-linseed(p=0.127), sunflower-linseed(p=0.001), sunflower-meatmeal(p=0.220) sunflower-soybean(p=0.003).

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