R Lab Manual
R Lab Manual
R Lab Manual
DEPARTMENT OF MATHEMATICS
R - LANGUAGE LABORATORY RECORD
SCHOOL OF SCIENCE AND HUMANITIES
(Winter Semester)
ACADEMIC YEAR
2021-2022
R-LANGUAGE LABORATORY RECORD
Subject : …………………………………………..
Degree : …………………………………………..
Department : …………………………………………..
Page 1 of 27
Using External Data:
R offers plenty of options for loading external data, including Excel, Minitab, SAS and SPSS
files.
R – Code:-
Data=read.csv("C:\\Users\\Student\\Desktop\\petroleum import data.csv")
# Importing the External data
>Y=Data$Year
>A=Data$PetroleumImports
>B= Data$TotalPetroleumImports
>C= Data$GPImport
> print(Y)
[1] 1975 1976 1977 1978 1979 1980 1981 1982 1983 1984 1985 1986 1987 1988 1989 1990 1991
1992 1993 1994 1995 1996 1997 1998 1999 2000 2001 2002 2003 2004
> print(A)
[1] 6055 7313 8807 8363 8456 6909 5996 5113 5051 5437 5067 6224 6678 7402 8061
8018 7627 7888 8620 8996 8835 9478 10162 10708 10852 11459 11871 11530 12264 13145
> print(B)
[1] 37.1 41.8 47.7 44.3 45.6 40.5 37.3 33.4 33.1 34.5 32.2 38.2 40.0 42.8 46.5 47.1 45.6 46.3 50.0
50.7 49.8 51.7 54.5 56.6 55.5 58.1 60.4 58.3 61.2 63.4
> print(C)
[1] 19.2 25.1 27.8 26.5 24.4 21.9 20.3 13.6 8.7 9.3 6.1 14.6 16.1 20.8 23.0 24.5 24.1 22.5 20.6
19.2 17.8 16.9 17.2 19.9 22.7 21.7 23.2 19.6 20.3 18.9
Page 2 of 27
Box Plot
> boxplot(B,main="Total PetroleumImports",col=c("green"))
Page 3 of 27
Stem and Leaf Diagram
> stem(C)
6|1
8 | 73
10 |
12 | 6
14 | 6
16 | 1928
18 | 92269
20 | 336879
22 | 5702
24 | 1451
26 | 58
> stem(B,scale=2)
32 | 214
34 | 5
36 | 13
38 | 2
40 | 058
42 | 8
44 | 366
46 | 3517
48 | 8
50 | 077
52 |
54 | 55
56 | 6
58 | 13
60 | 42
62 | 4
Page 4 of 27
Probability Plot
>qqnorm(B,col="blue")
Histogram
> hist(rep(Y,B), main="Distribution of Petroleum Imports", xlab="Petroleum Imports in
Year",ylab=" Total Petroleum Imports(b)",col=c("blue","red"))
Page 5 of 27
Time Series Plot
> T=ts(B)
>T
Time Series:
Start = 1
End = 30
Frequency = 1
[1] 37.1 41.8 47.7 44.3 45.6 40.5 37.3 33.4 33.1 34.5 32.2 38.2 40.0 42.8 46.5
[16] 47.1 45.6 46.3 50.0 50.7 49.8 51.7 54.5 56.6 55.5 58.1 60.4 58.3 61.2 63.4
> plot(T)
Page 6 of 27
2. MEASURES OF CENTRAL TENDENCY, MEASURES OF DISPERSION
FOR RAW DATA AND DISCRETE FREQUENCY DISTRIBUTION
Aim: To find the measures of central tendency , dispersion of the given raw data and discrete
frequency distribution data respectively.
Problem: 1 Find the measures of central tendency and dispersion of the given raw data:
2, 2, 2, 3, 1, 1, 1, 4, 5, 6, 8, 8, 9.
R - Code:-.
>X=c(2,2,2,3,1,1,1,4,5,6,8,8,9)
> cat("given data =", X,"\n")
given data = 2 2 2 3 1 1 1 4 5 6 8 8 9
>Mean=mean(X)
> Median=median(X)
> mode=function(X)
+ {
+ uni=unique(X)
+ uni[which.max(tabulate(match(X,uni)))]
+}
> Mode=mode(X)
> S.D=sd(X)
> Variance=var(X)
> Range=range(X)
> Quartile=quantile(X,c(0.25,0.50,0.75))
> Deciles=quantile(X,c(0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1))
> Percentile=quantile(X,c(0.26,0.76,0.85))
> IQR=IQR(X)
> QD=IQR/2
> n=length(X)
> CV=(S.D/Mean)*100
> MDMean= (sum(abs(X-Mean)))/n
> MDMean
> MDMedian= (sum(abs(X-Median)))/n
Page 7 of 27
> GM=prod(X)^(1/n)
> HM=n/sum(1/X)
> cat("Mean of the given data =", Mean,"\n")
Mean of the given data = 4
> cat("Median of the given data =", Median,"\n")
Median of the given data = 3
> cat("Mode of the given data=",Mode,"\n")
Mode of the given data= 2
> cat("Standard Deviation of the given data=",S.D,"\n")
Standard Deviation of the given data= 2.915476
> cat("Variance of the given data=",Variance,"\n")
Variance of the given data= 8.5
> cat("Range of the given data=",Range,"\n")
Range of the given data= 1 9
> cat("Quartile of the given data:",Quartile,"\n")
Quartile of the given data: 2 3 6
> cat("Deciles of the given data:",Deciles,"\n")
Deciles of the given data: 1 1.4 2 2 3 4.2 5.4 7.2 8 9
> cat("Percentiles of the given data for 26%,76% and 85%:",Percentile,"\n")
Percentiles of the given data for 26%,76% and 85%: 2 6.24 8
> cat("Inter Quartile Range of the given data =",IQR,"\n")
Inter Quartile Range of the given data = 4
> cat("Quartile Deviation of the given data =",QD,"\n")
Quartile Deviation of the given data = 2
> cat("Coefficient of Variation of the given data =",CV,"\n")
Coefficient of Variation of the given data = 72.8869
> cat("Mean Deviation about Mean of the given data =",MDMean,"\n")
Mean Deviation about Mean of the given data = 2.461538
> cat("Mean Deviation about Median of the given data =",MDMedian,"\n")
Mean Deviation about Median of the given data = 2.384615
> cat("Geometric Mean of the given data =",GM,"\n")
Geometric Mean of the given data = 3.009174
> cat("Harmonic Mean of the given data =",HM,"\n")
Harmonic Mean of the given data = 2.237094
Page 8 of 27
Problem: 2 Find the measures of central tendency and measures dispersion of the following
discrete frequency distribution.
x 34 56 57 87 90
f 3 4 8 17 4
R-Code:-
> x=c(34,56,76,87,90)
> f=c(3,4,8,17,4)
> Mean= mean(rep(x,f))
> Median= median(rep(x,f))
> Mode=mode(rep(x,f))
> S.D= sd(rep(x,f))
> Variance= var(rep(x,f))
> Quartile=quantile(rep(x,f),c(0.25,0.50,0.75))
> Deciles=quantile(rep(x,f),c(0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1))
> Percentile=quantile(rep(x,f),c(0.26,0.76,0.85))
> IQR=IQR(rep(x,f))
> QD=IQR/2
> N=sum(f)
> CV=(S.D/Mean)*100
> MDMean=sum(abs(x-Mean))/N
> MDMedian=sum(abs(x-Median))/N
> GM=prod(x*f)^(1/N)
> HM=N/sum(x*f)
> cat("Mean of the given data =", Mean,"\n")
Mean of the given data = 77.02778
> cat("Median of the given data =", Median,"\n")
Median of the given data = 87
> cat("Mode of the given data=",Mode,"\n")
Mode of the given data= 87
> cat("Standard Deviation of the given data=",S.D,"\n")
Standard Deviation of the given data= 16.64329
> cat("Variance of the given data=",Variance,"\n")
Variance of the given data= 276.9992
> cat("Quartile of the given data:",Quartile,"\n")
Quartile of the given data: 76 87 87
> cat("Deciles of the given data:",Deciles,"\n")
Deciles of the given data: 56 76 76 76 87 87 87 87 88.5 90
> cat("Percentiles of the given data for 26%,76% and 85%:",Percentile,"\n")
Page 9 of 27
Percentiles of the given data for 26%,76% and 85%: 76 87 87
> cat("Inter Quartile Range of the given data =",IQR,"\n")
Inter Quartile Range of the given data = 11
> cat("Quartile Deviation of the given data =",QD,"\n")
Quartile Deviation of the given data = 5.5
> cat("Coefficient of Variation of the given data =",CV,"\n")
Coefficient of Variation of the given data = 21.60687
> cat("Mean Deviation about Mean of the given data =",MDMean,"\n")
Mean Deviation about Mean of the given data = 2.445216
> cat("Mean Deviation about Median of the given data =",MDMedian,"\n")
Mean Deviation about Median of the given data = 2.722222
> cat("Geometric Mean of the given data =",GM,"\n")
Geometric Mean of the given data = 2.277576
> cat("Harmonic Mean of the given data =",HM,"\n")
Harmonic Mean of the given data = 0.01298233
Page 10 of 27
3. CORRELATION AND REGRESSION ANALYSIS
Problem: 1 Find the correlation of the following data and also draw the scatter diagram.
X 10 12 13 16 17 20 25
Y 19 22 26 27 29 33 37
R-Code:-
> X=c(10,12,13,16,17,20,25)
> Y=c(19,22,26,27,29,33,37)
> r=cor(X,Y, method="pearson")
> cat("Correlation coefficient of the given data =",r,"\n")
Correlation coefficient of the given data = 0.9801983
> Rho=cor(X,Y, method="spearman")
> cat("Rank Correlation coefficient of the given data =",Rho,"\n")
Rank Correlation coefficient of the given data = 1
> plot(X,Y, main="Scatter Plot",xlab="X Data", ylab="Y Data", col="blue",pch="*")
Page 11 of 27
Problem: 2 Find the regression of the following data
R-Code:-
> X=c(151,174,138,186,128,136,179,163,152,131)
> Y=c(3,81,56,91,47,57,7,72,62,48)
> reg1=lm(X~Y)
> reg1
Call:
lm(formula = X ~ Y)
Coefficients:
(Intercept) Y
146.1282 0.1464
> summary(reg1)
Call:
lm(formula = X ~ Y)
Residuals:
Min 1Q Median 3Q Max
-25.009 -17.937 0.614 13.592 31.847
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 146.1282 14.9235 9.792 9.93e-06 ***
Y 0.1464 0.2529 0.579 0.579
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
> reg2=lm(Y~X)
> reg2
Call:
lm(formula = Y ~ X)
Coefficients:
(Intercept) X
10.1555 0.2747
> summary(reg2)
Page 12 of 27
Call:
lm(formula = Y ~ X)
Residuals:
Min 1Q Median 3Q Max
-52.322 1.731 8.714 15.328 29.756
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 10.1555 73.5694 0.138 0.894
X 0.2747 0.4744 0.579 0.579
Page 13 of 27
4. CONFIDENCE INTERVAL FOR THE POPULATION MEAN
R-Code:-
>n=64
>sigma=1.6
>xbar=90
>sem=sigma / sqrt(n);
>sem
[1] 0.2
>E=qnorm(0.975)*sem;
>E
[1] 0.3919928
>xbar+c(-E,E)
[1] 89.081 90.39199
Inference:-
This implies that the true value of the population mean length of the components will fail in
this interval hence we infer that the 95% confidence interval ensure confidence by the customer.
Page 14 of 27
Problem:2
A Stock market analyst wants to estimate the average return on a certain stock. A random
sample of 15 days yields an average return of mean is 10.37% and a standard deviation of s = 3.5
%. Assuming a normal population of returns, give a 95% confidence interval for the average return
on this stock.
R-Code:-
> n=15
> xbar=10.37
> s=3.5
> sem=s/sqrt(n);
>sem
[1] 0.9036961
> E=qnorm(0.975)*sem;
>E
[1] 1.771212
> xbar+c(-E,E)
Inference:-
This implies that the analyst may be 95% sure that the average annualized return on the
stock is anywhere from 8.60 % to 12.14%.
Page 15 of 27
5. Z-TEST FOR SINGLE MEAN AND DIFFERENCE OF TWO SAMPLE
MEANS
Aim: To test whether there is significant difference between
(i) sample mean and population mean,
(ii) two sample means
Problem:1
Suppose the manufacturer claims that the mean lifetime of a light bulb is more than
10,000 hours. In a sample of 30 light bulbs, it was found that they only last 9,900 hours on
average. Assume the population standard deviation is 120 hours. At 0.05 significance level, can
we reject the claim by the manufacturer?
R-Code:-
>xbar=9900 #sample mean
> mu0=10000 #hypothesized value
>sigma=120 #population standard deviation
> n=30 #sample size
> z=abs((xbar-mu0)/(sigma/sqrt(n)))
>z # test statistic
[1] 4.564355
>alpha=0.05
>z.alpha=qnorm(1-alpha)
>z.alpha
[1]1.644854
Inference:-
The test statistic 4.5644 is less than the critical value of 1.6449. Hence, at .05 significance
level, we reject the claims that mean lifetime of a light bulb is above 10,000 hours.
Page 16 of 27
Problem:2
A college conduct both day and night classes intended to be identical. A sample of 100
day students yields examination results as X 72.4, 14.8 and a sample of 200 night students
as X 73.9, 17.9 . Are the two means statistically equal at 5% level?
R code:-
[1] 0.7702493
>alpha=0.05
>z.alpha=qnorm(1-alpha/2)
> z.alpha
[1] 1.959964
Inference:-
The test statistic 0.7702493 is less than the critical value of 1.959964 at .05 significance
level. Hence, we accept H0. This implies that the two means are statistically equal.
Page 17 of 27
6. Z-TEST FOR SINGLE PROPORTION AND DIFFERENCE OF TWO
SAMPLE PROPORTIONS
Aim: To test whether there is significant difference between
(i) sample proportion and population proportion
(ii) two sample proportion.
Problem:1
Suppose that 12% of apples harvested in an orchard last year was rotten. 30 out of 214
apples in a harvest sample this year turns out to be rotten. At .05 significance level, can we reject
the null hypothesis that the proportion of rotten apples in harvest stays below 12% this year?
Null hypothesis (H0) : p ≤ 0.12
Alternative hypothesis (H1) : p > 0.12
We begin with computing the test statistic,
R-Code:-
>pbar=30/214 # sample proportion
> p0=0.12 # hypothesized value
> n=214 # sample size
> z=abs((pbar-p0)/sqrt(p0*(1-p0)/n))
>z # test statistic
[1] 0.908751
>alpha=0.05
>z.alpha=qnorm(1-alpha)
>z.alpha # critical value
[1] 1.644854
Inference:-
The test statistic 0.90875 is not greater than the critical value of 1.6449. Hence, at .05
significance level, we do not reject the null hypothesis that the proportion of rotten apples in harvest
stays below 12% this year.
Page 18 of 27
Problem:2
Random samples of 400 men and 600 women asked whether they would like to have a fly
over near their residence. 200 men and 325 women were in favour of it. Test the equality of men
and women in the proposal.
R – Code:-
Null hypothesis (H0) : p1=p2
Alternative hypothesis (H1) : p1 ≠ p2
We begin with computing the test statistic,
>alpha=0.05
>z.alpha=qnorm(1-alpha/2)
>z.alpha # critical value
[1] 1.959964
Inference:-
The test statistic 1.292611 is lesser than the critical value of 1.959964 at .05 significance level.
Hence, we accept the null hypothesis. Thus, the men and women are equally favour in the proposal.
Page 19 of 27
7. STUDENT’S t-TEST FOR SINGLE MEAN AND DIFFERENCE OF TWO SAMPLE
MEANS
Problem 1:- A random sample of 10 boys has the following IQ’s: 70, 120, 110, 101, 88, 83,
95, 98, 107, 100. Do these data supports the assumption of population mean IQ of 100.
Null hypothesis (H0) : µ = 100
Alternative hypothesis (H1) : µ ≠ 100
We begin with computing the test statistic,
R – Code:-
> x=c(70,120,110,101,88,83,95,98,107,100)
>t.test(x,mu=100)
One Sample t-test
data: x
t = -0.62034, df = 9, p-value = 0.5504
Alternative hypothesis: true mean is not equal to 100
95 percent confidence interval:
86.98934 107.41066
sample estimates:
mean of x
97.2
Compute the critical value at .05 significance level:
>qt(0.975,9)
[1] 2.262157
Inference:-
Since calculated value of t (0.62034) is lesser than the table value of t (2.262157) at 5%
LOS with 9 dof. Accept H0, ie. The mean IQ of the population can be assumed as 100.
Page 20 of 27
Problem 2 :- Comparing two independent sample means, taken from two populations with
unknown variance. The following data shows the heights of individuals of two different
countries with unknown population variances. Is there any significant difference between the
average heights of two groups?
A: 175 168 168 190 156 181 182 175 174 179
B: 120 180 125 188 130 190 110 185 112 188
R - Code:-
> A=c(175,168,168,190,156,181,182,175,174,179)
>A
[1] 175 168 168 190 156 181 182 175 174 179
> B=c(120,180,125,188,130,190,110,185,112,188)
>B
[1] 120 180 125 188 130 190 110 185 112 188
>t.test(A,B)
data: A and B
t = 1.8827, df = 10.224, p-value = 0.08848
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-3.95955 47.95955
sample estimates:
mean of x mean of y
174.8 152.8
> qt(0.975,10+10-2)
[1] 2.100922
Inference:-
Since calculated value of t (1.8827) is lesser than the table value of t (2.100922) at 5%
LOS with 18 dof. Accept H0, ie. There is no significant difference between the average heights
of two groups.
Page 21 of 27
8. STUDENT t-TEST FOR PAIRED OBSERVATION
Aim: To test whether there is any significant difference paired observations based on one tailed and
two tailed test.
Problem: 1 An IQ test was administrated to five persons before and after they were trained. The
results are given below:
Candidates : I II III IV V
IQ before training 110 120 123 132 125
IQ after training 120 118 125 136 121
Test whether there is any change in IQ after the training programme.
R – Code:-
> Before_Training=c(110,120,123,132,125)
> Before_Training
[1] 110 120 123 132 125
> After_Training=c(120,118,125,136,121)
> After_Training
[1] 120 118 125 136 121
> t.test(Before_Training,After_Training,paired=TRUE)
Paired t-test
Inference:-
Since calculated value of t (0.8165) is lesser than the table value of t (2.776445) at 5%
LOS with 4 dof. Accept H0, ie. There is no significant change in IQ due to the training
programme.
Page 22 of 27
Problem: 2 Consider the paired data below that represents cholesterol levels on 10men before and
after a certain medication
Before(x) 237 289 257 228 303 275 262 304 244 233
After(y) 194 240 230 186 265 222 242 281 240 212
Test the claim that, on average, the drug lowers cholesterol in all men. i.e., test the claim
that d 0 . Test this at the 0.05 significance level.
R - Code:-
> before=c(237,289,257,228,303,275,262,304,244,233)
> before
[1] 237 289 257 228 303 275 262 304 244 233
> after=c(194,240,230,186,265,222,252,281,240,212)
> before
[1] 237 289 257 228 303 275 262 304 244 233
> t.test(before,after, paired=TRUE,alternative="greater",mu=0)
Paired t-test
Inference:-
Since calculated value of t (5.9151) is greater than the table value of t (1.833113) at 5% LOS
with 9 dof.We can reject the null hypothesis and support the claim.
Page 23 of 27
9. F- TEST OF EQUALITY OF VARIANCES
Aim: To test whether there is significant difference between population variances.
Problem: Five Measurements of the output of two units have given the following results (in
kilograms of material per one hour of operation). Assume that both samples have been obtained
from normal population, test at 5% significance level if two samples have the same variance.
H0:S12=S22
H1: S12 S22
Level of Significance: 0.05
R-Code:-
> Unit_A=c(14.1,10.1,14.7,13.7,14.0)
> Unit_A
[1] 14.1 10.1 14.7 13.7 14.0
> Unit_B=c(14.0,14.5,13.7,12.7,14.1)
> Unit_B
[1] 14.0 14.5 13.7 12.7 14.1
> var.test(Unit_A,Unit_B)
F test to compare two variances
data: Unit_A and Unit_B
F = 7.3304, num df = 4, denom df = 4, p-value = 0.07954
alternative hypothesis: true ratio of variances is not equal to 1
95 percent confidence interval:
0.7632268 70.4053799
sample estimates:
ratio of variances
7.330435
> qf(0.975,4,4)
[1] 9.60453
Inference:
Since calculated value of F (7.3304) is lessser than the table value of F(9.60453) at 5%
LOS with 9 dof. Hence, we accept the null hypothesis.
Page 24 of 27
10. ANALYSIS OF VARIANCE - ONE WAY CLASSIFICATION
Aim:- To test whether there is significant difference between the row means (column means).
A B C D
38 40 41 39
45 42 49 36
40 38 42 42
Can we infer that the mean lifetime of the four brands of electric bulbs are equal?
Null Hypothesis (H0): 1 2 3 4 (I.e. Average lifetimes of four different bulbs are same)
Alternate Hypothesis (H1): 1 2 3 4 (I.e. Average lifetimes of four different bulbs are
different)
R – Code:-
>data=c(38 ,45, 40, 40 ,42 ,38, 41, 49, 42 ,39 ,36 ,42)
>data
[1] 38 45 40 40 42 38 41 49 42 39 36 42
> brands=c("A","A","A","B","B","B","C","C","C","D","D","D")
>brands
[1] "A" "A" "A" "B" "B" "B" "C" "C" "C" "D" "D" "D"
>One_Way_Anova=aov(data~brands)
>summary(One_Way_Anova)
Df Sum Sq Mean Sq F value Pr(>F)
brands 3 42 14.00 1.244 0.356
Residuals 8 90 11.25
>qf(0.95, df1=3, df2=8)
[1] 4.066181
Inference:-
Since calculated value of F (1.244) is lesser than the table value of F (4.066181) at 5% LOS
with dof (3,8). We accept H0. The mean life times of the four electric bulbs are equal.
Page 25 of 27
11. ANALYSIS OF VARIANCE - TWO WAY CLASSIFICATION
Aim:- To test whether there is significant difference between the row means as well as column
means.
Problem: The following data represent the number of units of production per day turned out by
different workers using 4 different types of machines.
A B C D
1 44 38 47 36
2 46 40 52 43
3 34 36 44 32
4 43 38 46 33
5 38 42 49 39
Test whether the five men differ with respect to mean productivity and test whether the mean
productivity is the same for the four different machine types.
Null Hypothesis (H0) : There is no significant difference between men and machines.
Alternate Hypothesis (H1) : There is significant difference between men and machines.
R – Code:-
> data=c(44,38,47,36,46,40,52,43,34,36,44,32,43,38,46,33,38,42,49,39)
>data
[1] 44 38 47 36 46 40 52 43 34 36 44 32 43 38 46 33 38 42 49 39
>workers=gl(5,4)
>workers
[1] 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 5 5 5 5
Levels: 1 2 3 4 5
>machine=gl(4,1,20)
>machine
[1] 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4
Levels: 1 2 3 4
>Twoway_Anova=aov(data~workers+machine)
>summary(Twoway_Anova)
Df Sum Sq Mean Sq F value Pr(>F)
workers 4 161.5 40.37 6.574 0.00485 **
machine 3 338.8 112.93 18.388 8.78e-05 ***
Residuals 12 73.7 6.14
>qf(.95, df1=4, df2=12)
[1] 3.259167
Inference:-
Since calculated values of F (6.574 & 18.388) are greater than the table value of F
(3.259167&3.490295) at 5% LOS with dof(4,12) & dof(3,12). We reject H0. There is significance
difference between men and machines.
Page 26 of 27
12. ANALYSIS OF VARIANCE - LATIN SQUARE DESIGN
Aim:- To test whether there is significant difference between the row means, column means and
treatments.
Problem: Four varieties A, B, C, D of a crop are tested in a Latin Square design with four
replications. The plot yields are given in kgs as below:
A6 C5 D6 B9
C8 A4 B6 D4
B7 D6 C 10 A6
D7 B4 A8 C9
Analyze the experimental yield and state your conclusion.
Inference:-
Since calculated values of F (2.5304, 1.0000 & 1.6957) are lesser than the table value of F
(8.940645) at 5% LOS with dof (6,3). We accept H0. There is no significant difference between
the row means, column means and treatments.
---*****---
Page 27 of 27