In: Statistics and Probability
33. A population consists of the following scores: 12, 1, 10, 3, 7, 3
A. Compute m and s for the population.
B. Find the z-score for each raw score in the population.
C. Transform each score into a new standardized value so that the standardized distribution has a mean of m = 100 and a standard deviation of s = 20.
Do parts A, B, and C, and then from part C, choose which new score would correspond to the original score of 12:
a. 115
b. 120
c. 130
d. 140
Let us consider the given data can be written as (12,1,10,3,7,3).
A) m=mean of the given data= 36/6=6
s=standard deviation of the given data= sqrt(var)=sqrt(19.2)= 4.38178
B) Z-Score for each raw score in the population
Z-score of 12 is Z=(12-m)/s = (12-6)/4.38= 1.37
Z-score of 1 is Z=(1-m)/s = (12-6)/4.38= -1.41
Z-score of 10 is Z=(10-m)/s = (12-6)/4.38=0.91
Z-score of 3 is Z=(3-m)/s = (12-6)/4.38= -0.68
Z-score of 7 is Z=(7-m)/s = (12-6)/4.38= 0.23
Z-score of 3 is Z=(3-m)/s = (12-6)/4.38= -0.68
C) Transformation of the given data such that mean of standardized variable is 100 & standard devation is 20.
Transformation of the variable is (Z-score * s) + m
For 12, (1.37*20) + 100= 127.4
For 1, (-1.41*20) + 100= 77.18
For 10, (0.91*20) + 100= 118.26
For 3, (-0.68*20) + 100= 86.31
For 7, (0.23*20) + 100= 104.56
For 3, (-0.68*20) + 100= 86.31
new score would correspond to the original score of 12 is 130 since 127.4~ 130.
I have done this problem into R. So, I am attaching my R-code
> x=c(12,1,10,3,7,3)
> m=mean(x)
> s=sd(x)
> z=(x-m)/s
> z
[1] 1.3693064 -1.1410887 0.9128709 -0.6846532 0.2282177
-0.6846532
> m
[1] 6
> s
[1] 4.38178
> z*20 + 100
[1] 127.38613 77.17823 118.25742 86.30694 104.56435 86.30694