In: Statistics and Probability
Fenthion is a pesticide used against the olive fruit fly in olive groves. It is toxic to humans, so it is important that there be no residue left on the fruit or inolive oil that will be consumed. One theory was that, if there is residue of the pestice left in the olive oil, it would disspate over time. Chemists set out to test that theory by taking a random sample of small amounts of olive oil with fenthion residue and measuring the amount of fenthion in the oil at 3 different times over the year0day 0, day 281, and day 365.
SampleNumber | Group | Day | Fenthion | FenthionSulphoxide | FenthionSulphone | Time |
1 | 1 | 0 | 2.02 | 0 | 1.66 | 0 |
2 | 1 | 0 | 2.05 | 0 | 1.83 | 0 |
3 | 1 | 0 | 1.98 | 0 | 1.48 | 0 |
4 | 2 | 0 | 2.01 | 0 | 1.58 | 0 |
5 | 2 | 0 | 1.87 | 0 | 1.54 | 0 |
6 | 2 | 0 | 1.92 | 0 | 1.5 | 0 |
1 | 1 | 281 | 1.48 | 0.61 | 1.4 | 3 |
2 | 1 | 281 | 1.36 | 0.7 | 1.54 | 3 |
3 | 1 | 281 | 1.39 | 0.78 | 1.59 | 3 |
4 | 2 | 281 | 1.23 | 0.94 | 1.63 | 3 |
5 | 2 | 281 | 1.31 | 0.8 | 1.61 | 3 |
6 | 2 | 281 | 1.14 | 0.99 | 1.54 | 3 |
1 | 1 | 365 | 1.15 | 0.8 | 1.72 | 4 |
2 | 1 | 365 | 1.13 | 0.78 | 1.61 | 4 |
3 | 1 | 365 | 1.1 | 0.88 | 1.52 | 4 |
4 | 2 | 365 | 1.08 | 0.92 | 1.55 | 4 |
5 | 2 | 365 | 0.71 | 1.27 | 1.67 | 4 |
6 | 2 | 365 | 0.9 | 1.05 | 1.79 | 4 |
a. Two variables given in the data are fenthion and time. Which variable is the response variable and which variable is the explanatory variable? Explain.
b. Check the conditions necessary for conducting an ANOVA to analyze the amount of fenthion present in the samples. If the conditions are met, report the results of the analysis.
c. Transform the amount of fenthion using the exponential. Check the conditions necessary for conducting an ANOVA to analyze the exponential of the amount of fenthion present in the samples. If the conditions are met, report the results of the analysis
a)
b)
x=read.csv("book1cvs")
names(x)
Fenthion=x[1]
Time=(rep("z",6),rep("t",6),rep("f",6))
result=aov(Fenthiontime)
summery(result)
summery(result)
Df sum sQ mean sq F valvue pr(>F)
time 2 2 9066514533 89.80 4.75e-0.9***
residual 15 0.2447 0.0163
---
signif.codes:0'***"0.001'**'0.01'*' 0.05' '0.1' '1'
conditions
plot(result,1)#Homogeneity of variances lobrary(car)
>levene test(fenthiontime)
levene's test for Homogeneity of variance (center +median)
Df F value pr(>F)
group 2 0.6604 0.531
15
warning message:
in levene test.default(y=y,group=group,....):group coerced to factor.
plot(result,2) # normality
>aov_residual=residuals(object=result)
>shapioro.test(x=aov_residuals)
shapiro-wilk normality test
data:aov_residuals
W=0.92931,
p-value =0.1889