R Notebook

Krister Martinez 3

Four different designs for a digital computer circuit are being studied to compare the amount of noise present. The following data have been obtained:

Dependent variable. Noise Present amount

The factor. Circuit Design

List the factor levels. Design 1, Design 2, Design 3, Design 4.

If you decided to conduct multiple t tests instead of ANOVA, list all the t tests you would have to conduct. In other words, list each pair of designs that you would need to compare using t tests. D1 vs D2, D1 vs D3, D1 vs D4. D2 vs D3, D2 vs D4. D3 vs D4.

20 pigs are assigned at random among 4 experimental groups. Each group is fed a different diet. The data are the pig’s weight, in kilograms, after being raised on these diets for 10 months. We wish to determine whether the mean pig weights are the same for all 4 diets or not.

Download the CSV file “Pigs_Weights”. Then, read it into R.

Pigs_Weights_DF

NA

The dependent variable. The pigs Weight

The factor. The Diet Groups

The factor levels. Diet1, Diet2, Diet3, Diet4

levels(Pigs_Weights_DF$Diet)

[1] "D1" "D2" "D3" "D4"

str (Pigs_Weights_DF)

'data.frame':   20 obs. of  2 variables:
 $ Diet  : Factor w/ 4 levels "D1","D2","D3",..: 1 1 1 1 1 2 2 2 2 2 ...
 $ Weight: num  80.8 77.1 85 78.7 81.8 88.3 87.7 79 86.3 79.9 ...

State the hypotheses (Ho and Ha).

Ho: All diets result in the same mean weight

Ha: At least 2 of the diets result in different mean weights

Compute the average weight for each diet.

sort(tapply(Pigs_Weights_DF$Weight, Pigs_Weights_DF$Diet, mean))

   D1    D2    D4    D3 
80.68 84.24 86.38 87.34 

Run ANOVA and make the appropriate conclusion. Use a significance level of 0.05.

anova_Pigs_weights = aov(Pigs_Weights_DF$Weight ~ Pigs_Weights_DF$Diet)

summary(anova_Pigs_weights)

                     Df Sum Sq Mean Sq F value Pr(>F)  
Pigs_Weights_DF$Diet  3  130.8   43.60   3.134 0.0547 .
Residuals            16  222.5   13.91                 
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Because the P value is bigger than 0.05 we would fail to reject the null hypothesis. We cannot conclude that there are significant differences between group means. Support Ho, reject Ha.

5. (20 points) For some extra credit, I plotted the data and the results using boxplots to get an idea of the group distributions and potential differences. :)

# Creating a boxplot
boxplot(Weight ~ Diet, data = Pigs_Weights_DF,
        xlab = "Diet Type", ylab = "Weight (kg)",
        main = "Boxplot of Weights Across Different Diets",
        col = c("lightblue", "lightpink", "lightgreen", "lightyellow"),
        notch = FALSE)

# Adding labels
text(x = 1:4, y = tapply(data$Weight, data$Diet, max) + 1, 
     labels = c("D1", "D2", "D3", "D4"),
     col = "red", cex = 1)

Error in data$Diet : object of type 'closure' is not subsettable

Consider the data shown in the table from question 1. Enter this data in R by following these steps:

Create a vector called “design” by using this code: design= rep (c(“1”,“2”,“3”,“4”), each= 5)

design= rep (c("1","2","3","4"), each= 5)
design

 [1] "1" "1" "1" "1" "1" "2" "2" "2" "2" "2" "3" "3" "3" "3" "3" "4" "4" "4" "4" "4"

Create a vector called “noise” and just enter all the values from question 1’s table. Enter the values by row (i.e., all the values from row 1 followed by all the values from row 2,…)

noise = c(21,20,29,30,24,30,31,33,26,30,27,26,25,35,39,35,41,33,28,38)
noise 

 [1] 21 20 29 30 24 30 31 33 26 30 27 26 25 35 39 35 41 33 28 38

ANOVA test:

State the hypotheses (Ho and Ha) Ho: states that there are no significant differences among the group means Ha: posits that at least one group mean is different from the others.

Run ANOVA and make the appropriate conclusion. Use a significance level of 0.05.

anova_design_noise = aov(noise ~ design)
summary (anova_design_noise)

            Df Sum Sq Mean Sq F value Pr(>F)  
design       3    261   86.98   3.845 0.0301 *
Residuals   16    362   22.63                 
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Since the P value 0.0301 is smaller than our given significance level of 0.05 we will reject the the null hypothesis Ho. At least 2 design means make a different impact in the noise levels.

3. (15 points) Conduct a post-hoc test and discuss which designs lead to different average noise levels.

sort (tapply(noise,design, mean))

   1    2    3    4 
24.8 30.0 30.4 35.0 

pairwise.t.test (noise, design, p.adjust.method = "bonfe")

    Pairwise comparisons using t tests with pooled SD 

data:  noise and design 

  1     2     3    
2 0.619 -     -    
3 0.487 1.000 -    
4 0.022 0.696 0.875

P value adjustment method: bonferroni 

post_hoc <- TukeyHSD(anova_design_noise)
print(post_hoc)

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

Fit: aov(formula = noise ~ design)

$design
    diff       lwr       upr     p adj
2-1  5.2 -3.406868 13.806868 0.3419780
3-1  5.6 -3.006868 14.206868 0.2826091
4-1 10.2  1.593132 18.806868 0.0176270
3-2  0.4 -8.206868  9.006868 0.9991237
4-2  5.0 -3.606868 13.606868 0.3744075
4-3  4.6 -4.006868 13.206868 0.4442230

Both of our Post-Hoc test, Bonfe and Tukey, plus our box plot graph, tell us that the 4-1 have a P-Value of less than our given significance level of 0.05

I plotted the data and the results using box plots to get an idea of the group distributions and potential differences.

# Creating a boxplot with labels
boxplot(noise ~ design,
        main = "Boxplot of Noise Levels Across Different Designs",
        xlab = "Design",
        ylab = "Noise Level",
        col = c("lightblue", "lightgreen", "lightpink", "lightyellow"),
        names = c("Design 1", "Design 2", "Design 3", "Design 4"))

I was reading more details about the 2 different Post-Hoc tests, I have learned that both test serve to investigate where the significant differences lie following a significant ANOVA test. TukeyHSD is utilized to identify pairs of means that are significant from each other, while, the Bonferroni correction corrects the problem of inflated Type I errors during multiple compararisons but that corrections could also create more Type II errors. I like the TukeyHSD Post_Hoc test better. It is in its name “Honest”.

The End