Question

For questions 6 through 11, use the following: A researcher wonders if students will spend more time studying for a quiz is they are told that a high mark on that quiz will excuse them from writing a term paper. Two groups of college students were randomly selected from a large lecture course in Art History. The students in Group A were simply told that they would have their first quiz in two days. In Group B, students were told the same thing, except that the promise of "no term paper" for high quiz performance was added. Just before taking the quiz, all students in both groups reported in hours) the time they had spent studying for the quiz. In Group A, the times were 6,4,9,3,2,5,7,4,7,6. In Group B, the times were 12,5, 13, 5,5,7,9,6,9,6. 6. Calculate the means for each group. Group A: Group B: 7. Calculate the standard deviations for each group. Group A: Group B: 8. Calculate the standard error of the means for each group. Group Group B: 9. Calculate the standard error of difference for both groups combined. 10 Calculate the independent t ratio. 11. Do you accept or reject the null hypothesis. Why?

Please help, I would greatly appreciate it

Answer 1

Solution-:

> Gr.A=c(6,4,9,3,2,5,7,4,7,6);Gr.A
[1] 6 4 9 3 2 5 7 4 7 6
> nA=length(Gr.A);nA
[1] 10
> Gr.B=c(12,5,13,5,5,7,9,6,9,6);Gr.B
[1] 12 5 13 5 5 7 9 6 9 6
> nB=length(Gr.B);nB
[1] 10
> #(6)
> MeanA=mean(Gr.A);MeanA
[1] 5.3
> MeanB=mean(Gr.B);MeanB
[1] 7.7
> #(7)
> vA=var(Gr.A);vA
[1] 4.455556
> sdA=sqrt(vA);sdA
[1] 2.110819
> round(sdA,2)
[1] 2.11
> vB=var(Gr.B);vB
[1] 8.677778
> sdB=sqrt(vB);sdB
[1] 2.945807
> round(sdB,2)
[1] 2.95
> #(8)
> SEA=sdA/sqrt(nA);SEA
[1] 0.6674995
> round(SEA,2)
[1] 0.67
> SEB=sdB/sqrt(nB);SEB
[1] 0.9315459
> round(SEB,2)
[1] 0.93
> #(9) Stndarad error for both groups combined
> SE=sqrt((vA/nA)+(vB/nB));SE
[1] 1.146008
> round(SE,2)
[1] 1.15
> #(10)
> t.test(Gr.A,Gr.B,var.equal=T)

Two Sample t-test

data: Gr.A and Gr.B
t = -2.0942, df = 18, p-value = 0.05066
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-4.807672546 0.007672546
sample estimates:
mean of x mean of y
5.3 7.7

> #(11)
> #Here, p-value=0.0506 > LOS=0.05 so we accept null hypothesis.

R-Code:

Gr.A=c(6,4,9,3,2,5,7,4,7,6);Gr.A
nA=length(Gr.A);nA
Gr.B=c(12,5,13,5,5,7,9,6,9,6);Gr.B
nB=length(Gr.B);nB
#(6)
MeanA=mean(Gr.A);MeanA
MeanB=mean(Gr.B);MeanB
#(7)
vA=var(Gr.A);vA
sdA=sqrt(vA);sdA
round(sdA,2)
vB=var(Gr.B);vB
sdB=sqrt(vB);sdB
round(sdB,2)
#(8)
SEA=sdA/sqrt(nA);SEA
round(SEA,2)
SEB=sdB/sqrt(nB);SEB
round(SEB,2)
#(9) Stndarad error for both groups combined
SE=sqrt((vA/nA)+(vB/nB));SE
round(SE,2)
#(10)
t.test(Gr.A,Gr.B,var.equal=T)
#(11)
#Here, p-value=0.0506 > LOS=0.05 so we accept null hypothesis.