Question

Please use R to solve part e and f

The data file data2.txt gives a data set with two variables x and y. The first column in the data set is just row numbers not useful for this question.

(e) Use the Shapiro-Wilks test to test for Normality of the data. State your null and alternative hypotheses, p-value and conclusion. Use α = 0.05

(f) Apply the transformation y 0 = log(y) and run the regression on y 0 on x. Now repeat parts (c), (d) and (e) with the residuals from this transformed model.

Data file:

Row x y
1 60.26 63.95
2 64.64 67.42
3 69.17 73.36
4 61.49 65.30
5 65.10 68.74
6 61.34 65.22
7 84.12 86.78
8 73.58 75.01
9 69.51 71.46
10 51.94 56.18
11 54.39 58.07
12 69.25 72.53
13 76.64 78.53
14 73.16 75.50
15 67.99 70.37
16 42.23 47.08
17 62.95 67.80
18 70.12 74.66
19 63.96 66.32
20 60.32 64.22
21 60.33 64.56
22 55.28 58.95
23 51.48 58.13
24 76.90 84.55
25 69.79 71.75
26 79.31 80.64
27 68.12 71.40
28 65.70 68.22
29 50.85 54.99
30 59.47 63.75

Answer 1

I am sharing R markdown output file. R commands are in blue , output in bold black and comments/conclusions in italic bold brown.

library(xlsx)
importing dataframe
xy_data = read.xlsx("C:\\Users\\ADMIN\\Desktop\\xydata.xlsx",1)

Accessing x and y variables from the data

x = xy_data$x
y = xy_data$y

e) Checking normality of both the variables.
level of significance (alpha) = 0.05.

Hypothesis –

H0: Data comes from normally distributed population. VS

H1: Data is not from normally distributed population.

For variable x -

shapiro.test(x)

## Shapiro-Wilk normality test
##
## data: x
## W = 0.98819, p-value = 0.9786

p - value = 0.9786
p - value > alpha i.e 0.9786>0.05
Decision - Accept H0 ( Null hypothesis is rejected if p-value < alpha)
Conclusion - Variable x is from normally distributed population.

For variable y -
H0: Data comes from normally distributed population.   VS

H1: Data is not from normally distributed population.


shapiro.test(y)

##
## Shapiro-Wilk normality test
##
## data: y
## W = 0.98766, p-value = 0.9735

p - value = 0.9735
p - value > alpha i.e 0.9735>0.05
Decision - Accept H0
conclusion - Variable y is from normally distributed population.

Hence , data comes from normally distributed population.

f) Transformation on y - y0 = log(y)

Using log to the base e function to transform y

y0 = log(y)

Fitting regression line on y0 on x taking y0 as response and x as predictor.

fit = lm(y0~x)
summary(fit)

##
## Call:
## lm(formula = y0 ~ x)
##
## Residuals:
##       Min        1Q    Median        3Q       Max
## -0.046893 -0.012009 -0.004173 0.010305 0.051214
##
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)   
## (Intercept) 3.3050759 0.0265154 124.65   <2e-16 ***
## x           0.0140579 0.0004061   34.62   <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.02037 on 28 degrees of freedom
## Multiple R-squared: 0.9772, Adjusted R-squared: 0.9764
## F-statistic: 1198 on 1 and 28 DF, p-value: < 2.2e-16

Values of Rsquare are close to 1. Hence model is strong.

Accessing residuals from the fitted model

residuals = fit$residuals

Checking normaltiy of the residuals

H0: Data comes from normally distributed population. VS

H1: Data is not from normally distributed population.

shapiro.test(residuals)

##
## Shapiro-Wilk normality test
##
## data: residuals
## W = 0.98349, p-value = 0.9088

p - value = 0.9088
p - value > alpha i.e 0.9088>0.05
Decision - Accept H0
Conclusion - residuals obtained from the regression model are from normally distributed population.