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Consider again the two tests A and B described in Exercise2. If a student is chosen at random, and her scoreon test B is 100, what predicted value of her score on test A has the smallest M.S.E., and what is the value of thisminimum M.S.E.?

Short Answer

Expert verified

The M.S.E value is 5.76

\(\)

Step by step solution

01

Given information

Two different testsAandBare to be given to a student chosen at random from a certain population.Suppose also that the mean score on testAis 85, and the

standard deviation is 10; the mean score on testBis 90,and the standard deviation is 16; the scores on the two tests have a bivariate normal distribution, and the correlation of the two scores is 0.8.

02

Denote the random variables

Let A denote the firsttest scores, and let B denote second test scores.

Then,

\(\begin{array}{*{20}{l}}{{\mu _A} = 85\;}\\{{\sigma _A} = 10}\\{{\mu _B} = 90}\\{{\sigma _B} = 16}\\{p = 0.8}\end{array}\)

\(\)\(\)

03

Define the M.S.E.

For a\(BVN{\rm{ }}\left( {85,10,90,16,0.8} \right)\),

Since the score for test B is 100 is known, then the best prediction is the mean\(E\left( {B|A = a} \right)\)of the conditional distribution of B given that A=a and the M.S.E. of this prediction is the variance\(\left( {1 - {p^2}} \right)\sigma _B^2\)of that conditional distribution.

04

Find the M.S.E. value

\(\begin{array}{c}\left( {1 - {p^2}} \right)\sigma _B^2 = \left( {1 - {{0.8}^2}} \right)\left( {16} \right)\\ = 5.76\end{array}\)

The M.S.E. value is 5.76.

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