Question

D5 Nine teenfortyfour to Nineteensixtynine Nineteenseventy to Twentythirteen 10 12 13 14 16 17 18 19 20 21 0 24 25 26 27 28 29 30 31 32 0 34 35 36

Show R work and conclusions thanks.

Answer 1

#1. & 2.

> #variable A for 1944-1960
> #variable B for 1970-2013
> A=c(3,2,1,2,4,3,7,2,3,3,2,5,2,2,4,2,2,6,0,2,5,1,3,1,0,3)
> B=c(2,1,0,1,2,3,2,1,2,2,2,3,1,1,1,3,0,1,3,2,1,2,1,1,0,5,6,1,3,5,3,3,2,3,6,7,2,2,5,2,5,4,2,0)
#3.

> par(mfrow=c(1,2))
> boxplot(A,main="A")
> boxplot(B,main="B")

#BOXPLOTS indicate that means are same.

#4.

#Null hypothesis $\mu _{A}=\mu _{B}$

#Alternative hypothesis $\mu _{A}\neq \mu _{B}$

#5.

> var.test(A,B)

F test to compare two variances

data: A and B
F = 0.97515, num df = 25, denom df = 43, p-value = 0.9692
alternative hypothesis: true ratio of variances is not equal to 1
95 percent confidence interval:
0.4955629 2.0523066
sample estimates:
ratio of variances
0.9751496

#p-value=0.9692 which is greater than 0.05. Therefore, variances of A and B are homogeneous.

#Student's two sample t-test is to be carried out.

> t.test(A,B,var.equal=TRUE)

Two Sample t-test

data: A and B
t = 0.77921, df = 68, p-value = 0.4386
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0.5130186 1.1703613
sample estimates:
mean of x mean of y
2.692308 2.363636

#p-value=0.4386 which is greater than 0.05. Therefore, means of A and B are equal.

#6.

#Conclusion: Null Hypothesis is accepted. Means for A (1944-1960) and B (1970-2013) are equal.