Hi Please , l want to you to answer the All questions ( a , b , c ) in the Quiz

Question

Hi

Please , l want to you to answer the All questions ( a , b , c ) in the Quiz page

Thanks

Quiz 4 -Number 8.12 (page 437) This problem refers to example 8.1 on page 409 a. Calculate the 15 residuals...you can get this out of SPSS. b. Use Levene's test to test the equality of variance assumption. Make sure you set up the null and alternate hypothesis, and report the p-value in your conclusion. Use a S.05. c. With the residuals use Shapiro-Wilk's test to test the normal distribution assumption. Make sure you set up the null and alternate hypothesis, and report the p-value in your conclusion. Use a .05.

Explanation / Answer

we shall answer this usign the open source statistical package R

The complete R snippet is as follows

data.df<- read.csv("C:\Users\586645\Downloads\Chegg\contentmg.csv",header=TRUE)
str(data.df)

# H0 : The samples comes from a normal distributio
# H1 : The samples do not come from a normal distribution
##

shapiro.test(data.df$one)
shapiro.test(data.df$two)
shapiro.test(data.df$three)

## as the p values of all the results are greater than 0.05 , hence we fail to
## reject the null hypothesis and conclude that they come from normal distribution

#perform the omnibus 1 way anova analysis

library(reshape2)
data.df<- melt(data.df)

# perform anova analysis
a<- aov(lm(value~ variable,data=data.df))

#summarise the results
summary(a)
a$residuals

# levene test

library(car)
attach(data.df)
leveneTest(value,variable)

## H0 : There is an equality of variance in the variables
## H1 :There is no equality of variance in the variables

## as the p value is not less than 0.05 hence , we fail to reject the null hypotehsis

The results are

> summary(a)
Df Sum Sq Mean Sq F value Pr(>F)
variable 2 60.4 30.20 0.828 0.46
Residuals 12 437.6 36.47   
> a$residuals
1 2 3 4 5 6
-6.200000e+00 7.800000e+00 2.800000e+00 8.000000e-01 -5.200000e+00 9.000000e+00
7 8 9 10 11 12
-1.554312e-15 -1.000000e+00 -3.000000e+00 -5.000000e+00 1.120000e+01 1.200000e+00
13 14 15
-1.800000e+00 -4.800000e+00 -5.800000e+00

> shapiro.test(data.df$one)

   Shapiro-Wilk normality test

data: data.df$one
W = 0.93653, p-value = 0.6415

> shapiro.test(data.df$two)

   Shapiro-Wilk normality test

data: data.df$two
W = 0.86708, p-value = 0.2548

> shapiro.test(data.df$three)

   Shapiro-Wilk normality test

data: data.df$three
W = 0.86984, p-value = 0.2658

> leveneTest(value,variable)
Levene's Test for Homogeneity of Variance (center = median)
Df F value Pr(>F)
group 2 0.123 0.8854
12

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.