Question

Students in a school were asked to participate in a study of the effects of a new teaching method on reading skills of 10th graders. To determine the effectiveness of the new method, a reading test was given to each student before applying the new method (pre-test). Another test was given to the same students after applying the new method (post-test). Experts consider that a good measure of improvement is given by the following formula: Improvement=20(Post-test score)- 10(Pre-test score). That is, improvement is 20 times the post-test score minus 10 times the pre-test score. An improvement ? 300 is considered good. It is known that the average score in the pre-test is 40, average score in the post-test is 40, standard deviation in the post-test is 6, standard deviation in the pre-test is 5, and correlation between the scores in the two tests is 0.5

What is the standard deviation of the improvement?

Write the Variance covariance matrix for the pre and post-test scores vector random variable.

Answer 1

define :

X : pre test score

Y: post test score

improvement =( 20* post test score) - (10*pre test score)

average improvement = (20*40) -(10*40) =400 , which is greater than 300

so there is improvement on an average.

variance of improvement

= {(20^2)*variance of post test score} + {(10^2)* variance of pre test score} - {2*covariance(X,Y)} ......... (1)

[ since V(aX-bY)= (a^2) V(X)+ (b^2)V(Y) - (2*cov(X,Y)) ]

correlation coefficient between X and Y is defined by

p= cov(X,Y) / [{SD(X)*SD(Y)}]

[SD= standard deviation

V= variance

cov = covariance ]

given that p= 0.5,SD(X)= 5, SD(Y)=6

So, Cov(X,Y)= 0.5 * 5*6= 15

from relation (1) we get

variance of improvement

={400*(6^2)} +{100*(5^2)} - (2*15) = 16870

standard deviation of improvement = sqrt ( 16870)= 129.88

[since SD(Z)= sqrt ( V(Z))]

variance covaraince matrix of pre and post test score vector (X Y) is given by

sigma =

[V(X) Cov(X,Y)

Cov(X,Y) V(Y) ]

=[ 25 15

15 36 ]

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