Certain types of nerve cells have the ability to regenerate a part of the cell t
ID: 3311753 • Letter: C
Question
Certain types of nerve cells have the ability to regenerate a part of the cell that has been amputated. In an early study of this process, measurements were made on the nerves in the spinal cord in rhesus monkeys. Nerves emanating from the left side of the cord were cut, while nerves from the right side were kept intact. During the regeneration process, the content of creatine phosphate (CP) was measured in the left and the right portion of the spinal cord.
Data for the right (control) side (X) and for the left (regenerating) side (Y ) is shown below. The units of measurement are mg CP per 100 gm tissue.
Right side (X) control: 11.8, 33, 22.1, 18.9, 15.2, 26.6, 16.8, 12.3, 16.5, 11.4, 17.1, 19.9, 13.4, 9.5, 8.7, 24, 29.1, 25.6
Left side (Y) regenerating: 13.7, 3.4, 6.2, 8.8, 7.1, 5.9, 7.9, 12.2, 10.5, 6.1, 14, 3.3, 11.7, 4.6, 13.1, 7.6, 5.1, 6.8
Consider a t-test to compare the two sides at = 0.01 using a two-sided alternative.
(a) Using R and other reasoning, check the assumptions and comment on the validity of the study. Regardless of your conclusion, do the remainder of the problem.
(b) Using R, compute and interpret a 99% confidence interval for the mean difference in creatine phosphate between the cut and control sides.
(c) If you were to conduct a hypothesis test, what would be the conclusion in the context of this setting?
Explanation / Answer
> x=c(11.8,33,22.1,18.9,15.2,26.6,16.8,12.3,16.5,11.4,17.1,19.9,13.4,9.5,8.7,24,29.1,25.6)
> y=c(13.7,3.4,6.2,8.8,7.1,5.9,7.9,12.2,10.5,6.1,14,3.3,11.7,4.6,13.1,7.6,5.1,6.8)
> alpha= 0.01
> x=as.matrix(x)
> y=as.matrix(y)
> d=x-y
> d=as.matrix(d)
> d
[,1]
[1,] -1.9
[2,] 29.6
[3,] 15.9
[4,] 10.1
[5,] 8.1
[6,] 20.7
[7,] 8.9
[8,] 0.1
[9,] 6.0
[10,] 5.3
[11,] 3.1
[12,] 16.6
[13,] 1.7
[14,] 4.9
[15,] -4.4
[16,] 16.4
[17,] 24.0
[18,] 18.8
> #H0: mu_d = 0 vs H1: mu_d is not equal to 0
> n=nrow(d)
> n
[1] 18
> dbar=mean(d)
> vard =var(d)
> SD=sqrt(vard)
> SEd =SD/sqrt(n)
> ttable = qt(0.995,df=17)
> dbar
[1] 10.21667
> SD
[,1]
[1,] 9.46506
> SEd
[,1]
[1,] 2.230936
> ttable
[1] 2.898231
b]
> #99% confidence interval
> Upperlimit = dbar +(ttable*SEd)
> Lowerlimit = dbar -(ttable*SEd)
> Upperlimit
[,1]
[1,] 16.68243
> Lowerlimit
[,1]
[1,] 3.7509
c]
> #To test hypothesis,
> #t=(dbar-mu_d)/SEd
> t = dbar/SEd
> t
[,1]
[1,] 4.579543
t = 4.5795 > ttable = 2.8982
Reject the null hypothesis i.e H0.
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