Question

In: Statistics and Probability

7. Two children in the same class took the same achievement test at the beginning and...

7. Two children in the same class took the same achievement test at the beginning and the end of the school year. The scores of this standard achievement test form a normal distribution. Both children improved on their scores. The score of Child A increased from the 47th to the 57th percentile while the score of Child B increased form the 87th to the 97th percentile.

a. Did both children make improvement of equal magnitude in terms of z scores? Circle one. (2 points)

YES NO
b. If “Yes”, why? If “No”, who made a greater improvement in z scores? (1 point)

Solutions

Expert Solution

Question (a)

For Child A

Given rhe score of Child A increased from the 47th to the 57th percentile

if the Child A has secured 47th percentile, it implies that there are 47% children who scored less than that Child A

So P(Z<z1) = 0.47

The z-score that has an area of 0.47 to its left approximately from the below attached negative z-table = -0.08 and the exact z-score = -0.0753 from online calcuator

if the Child A has secured 57th percentile, it implies that there are 57% children who scored less than that Child A

So P(Z<z2) = 0.57

The z-score that has an area of 0.57 to its left approximately from the below attached positive z-table = 0.18 and the exact z-score = 0.1764 from online calcuator

So the improvement in z-scores for Child A = 0.1764 - (-0.0753)

= 0.2517

For Child B

Given rhe score of Child B increased from the 87th to the 97th percentile

if the Child B has secured 87th percentile, it implies that there are 87% children who scored less than that Child B

So P(Z<z3) = 0.87

The z-score that has an area of 0.87 to its left approximately from the below attached positive z-table = 1.13 and the exact z-score = 1.1264 from online calcuator

if the Child B has secured 97th percentile, it implies that there are 97% children who scored less than that Child B

So P(Z<z4) = 0.97

The z-score that has an area of 0.97 to its left approximately from the below attached positive z-table = 1.88 and the exact z-score = 1,8808 from online calcuator

So the improvement in z-scores for Child B = 1.8808 - 1.1264

= 0.7544

Clearly the two childs did not make improvement of equal magnitude in terms of z scores, child B's improvement is 0.7544 and Child A's improvement is 0.2517

So Answer is No

Question (b)

Child B's improvement is 0.7544 and Child A's improvement is 0.2517

So Child B made a greater improvement


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