Problem 3. [Statistics competition] ls there a difference in statistics prowess between STA 130B students and STA 131B students? To settle this debate, the statistics department organizes a competition between students among the two classes. Each class sends 5 representatives, and all of them take a standardized exam out of 100 points. Their scores are reported below. STA 130B STA 131B 67 59 35 64 46 49 71 60 Based on this sample, is there a difference in performance between STA 130B and STA 131B students? To answer this question, perform a Mann-Whitney test with a 0.01, using the following three methods: (a) Computing the exact p-value by assessing the rank configurations. (b) Computing the test statistic R and comparing it to the critical value in the table. (c) Using the Gaussian approximation to the Mann-Whitney rank sum.

Check the explanation

Step-by-step 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.

> #the null hypothesis H0: same for both population.

> ##the alternative hypothesis H1 : not same.

> ##Hypothesis for given problem:

> ## H0: there is NO difference exist between STA 130B and STA 131B students.

> ## H1: difference exist 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

> ### NO difference in performance between STA 130B and STA 131B students.

x= -3 and

step-by-step explanation:

to find the roots of a cubic, use division or factoring to find the factors then set equal to 0. as this is a difference of cubes, it factors into

where a is the cube root of .

here   this is 27 and the cube root of 27 is 3. so we write using the form:

.

set each factor to 0 and solve for x.

x-3=0

x=3

the quadratic expression does not factor and must be solved using the quadratic formula.

where a = 1, b=3, and c=9

this cubic has two complex roots.

c. y = 1/4 x + 5

step-by-step explanation:

first, put the equation of the line on the graph into slope-intercept form

(y = mx + b):   y = -4x + 8

second, a perpendicular line has to have "an opposite/reciprocal slope" to the original slope. so, find the "m" value in your original equation (it's -4, right? )

original slope: m = -4   perpendicular slope: change the sign to its opposite and find the reciprocal ("flip" the number)

new slope is now positive (b/c the original slope was negative 4), and it's 1/4 (not the original slope of 4 or 4/1)