Answer:
Check the explanation
Explanation:
[USED R-Software]
> ##General Hypothesis:
> ##the null hypothesis H0 : the distributions of both populations are same.
> ##the alternative hypothesis H1 : the distributions are not same.
> ##Hypothesis for given problem:
> ## H0: there is NO difference in performance between STA 130B and STA 131B students.
> ## H1: there is difference in performance between STA 130B and STA 131B students.
> STA_130B=c(67,88,71,60,77)
> STA_131B=c(59,35,64,46,49)
> Score=c(STA_130B,STA_131B)
> Score #Combined (n1+n2) scores
[1] 67 88 71 60 77 59 35 64 46 49
> R=rank(Score) #Ranks assigned to combined scores
> R
[1] 7 10 8 5 9 4 1 6 2 3
> W1=sum(R[1:5]) ##the total Sum of ranks of scores of STA_130B
> W1
[1] 39
> W2=sum(R[6:10]) ##the total Sum of ranks of scores of STA_131B
> W2
[1] 16
> n1=5
> n2=5
> U1=W1-((n1)*(n1+1)/2)
> U1
[1] 24
> U2=W2-((n2)*(n2+1)/2)
> U2
[1] 1
> U=min(U1,U2)
> U
[1] 1
> ##independent 2-group Mann-Whitney U Test in R is wilcoxon rank sum test.
> ##wilcox.test(y,x) ;where y and x are numeric
> wilcox.test(STA_131B,STA_130B,exact=TRUE,conf.level=0.99,paired=FALSE)
Wilcoxon rank sum test
data: STA_131B and STA_130B
W = 1, p-value = 0.01587
the substitute hypothesis: true location shift is not equal to 0
> ##p-value=0.01587 > aplha=0.01, Accept H0.
> ##Calculated value of U=1
> ##From Statistical Tables,Critical value of Mann-Whitney U test at alpha=0.01,n1=n2=5 is 0.
> ### Calculated U=1 > Tabulated U=0, Accept H0.
> ##Normal/ GAUSSIAN approximation:
> EU=(n1*n2)/2 ##Expected value of U
> EU
[1] 12.5
> VarU=(n1*n2*(n1+n2+1))/12 ##Variance value of U
> VarU
[1] 22.91667
> Z=(U-EU)/sqrt(VarU)
> Z
[1] -2.402272
> qnorm(1-0.01/2)
[1] 2.575829
> ## qnorm(1-0.01/2)= 2.575829
> ##Z follows Normal(0,1)
> ##If calculated |Z| > tabulatedZ, then Reject H0 at alpha % l.o.s.
> ##Here, calculated |Z|= 2.402272< tabulatedZ= 2.575829, thus Accept H0 at 1% l.o.s
> ### There is NO difference in performance between STA 130B and STA 131B students.