Question

Suppose that you are in charge of evaluating teacher performance at a large elementary school. One tool you have for this evaluation is reports of the average student reading test score in each classroom. You also know that across the whole school, the average student reading score was 80 points and the standard deviation in scores was 10 points.
Determine:
(a) If each class has 25 students in it, what is the standard error of the classroom average score?
(b) In what range do you expect the average classroom test score to fall 95% of the time?
(c) What is the approximate probability that a classroom will have an average test score of 79 or higher?
(d) Do you think the probability that a classroom has an average test score of 79 or higher would be greater or smaller if there were only 15 students in a class? Explain your answer in 2-3 sentences.
(e) Do you think the probability that a classroom has an average test score of 79 or higher would be greater or smaller if the standard deviation of individual student reading scores was only 5 points (instead of 10)?

Answer 1

Answer:

a) Standard error = 2

b) Range = (76.08, 83.92)

c) P=0.69

d) Smaller

e) Greater

Step-by-step explanation:

a) When we have a sample taken out of the population, the standard error of the mean is calculated as:

$\sigma_m=\dfrac{\sigma}{\sqrt{n}}=\dfrac{10}{\sqrt{25}}=\dfrac{10}{5}=2$

where n is te sample size (n=25) and σ is the population standard deviation (σ=10).

Then, the standard error of the classroom average score is 2.

b) The calculations for this range are the same that for the confidence interval, with the difference that we know the population mean.

The population standard deviation is know and is σ=10.

The population mean is M=80.

The sample size is N=25.

The standard error of the mean is σM=2.

The z-value for a 95% confidence interval is z=1.96.

The margin of error (MOE) can be calculated as:

$MOE=z\cdot \sigma_M=1.96 \cdot 2=3.92$

Then, the lower and upper bounds of the confidence interval are:

$LL=M-t \cdot s_M = 80-3.92=76.08\\\\UL=M+t \cdot s_M = 80+3.92=83.92$

The range that we expect the average classroom test score to fall 95% of the time is (76.08, 83.92).

c) We can calculate this by calculating the z-score of X=79.

$z=\dfrac{X-\mu}{\sigma}=\dfrac{79-80}{2}=\dfrac{-1}{2}=-0.5$

Then, the probability of getting a average score of 79 or higher is:

$P(X>79)=P(z>-0.5)=0.69146$

The approximate probability that a classroom will have an average test score of 79 or higher is 0.69.

d) If the sample is smaller, the standard error is bigger (as the square root of the sample size is in the denominator), so the spread of the probability distribution is more. This results then in a smaller probability for any range.

e) If the population standard deviation is smaller, the standard error for the sample (the classroom) become smaller too. This means that the values are more concentrated around the mean (less spread). This results in a higher probability for every range that include the mean.