Question

You find out that the average 10th grade math test score for section 6 of the local high school is 87 for the 25 students in the class. The average test score for all 10th grade math students across the state is 85 for 1,800 students. The standard deviation for the state is 3.8. What Z score do you calculate, and what is the area between the mean and the Z score (found on the table at the end of the course text)? What does this mean about the probability of this test score difference occurring by chance? Is it less than 0.05? Explain.

Answer 1

Mean score across state, _{$\small \mu$} =85

Standard Deviation, $\small \sigma$ = 3.8

Mean score of a high school , $\small \bar x$ =87

Sample size, n=25

$\small z-score,\ z =\frac{\bar x-\mu}{\sigma/\sqrt{n}}$

$=\frac{87-85}{3.8/ \sqrt{25}}=2.63$

z-score for the mean =0

Area between the mean and the z-score =

P(0 < z < 2.63)

= P(z < 2.63) - P(z < 0)

= 0.9957 - 0.5000

= 0.4957

This mean that the probability of this test score difference occurring by chance is 0.4957.

It is not less than 0.05.