Worked problems for standard scores, percentiles, and stanines

Download Report

Transcript Worked problems for standard scores, percentiles, and stanines

Item 1
Picabo came in at a speed of 100 mph on the downhill. Tommy, on a bad day,
came in at the same speed. The average female speed on the downhill is 80
mph with a SD of 5. The average male speed on the downhill is 110 with a SD
of 10. Just how much better than Tommy is Picabo?
Tommy is coming in at the 16th percentile. He needs to stop that partying.
Picabo is at the 99+ percentile. She needs to watch those steroids.
Picabo is definitely better than Tommy. Whoa! Actually, they are the same—
they both skied at the same speed, 100 mph.
Only when compared to the norms for their gender do we see a difference in their
performance.
In this case, however, there is an absolute standard of comparison. This is a case
of criterion-referenced measurement.
Item 2
On a History test, Mortimer, a 5th grade student, scored 80%
of the items correct, yet his percentile rank was 36. On
further investigation, you learn that Mortimer scored at the
4th Stanine. On the other hand, on a Science test, Mortimer
got 45% of the items correct and a percentile rank of 74,
which placed him in the 6th Stanine.
If you were interpreting/explaining Mortimer’s performance,
in History and Science, to his parent, what would you say?
So, what’s going on here? 80% correct gets Mort to the 36th %tile in History,
while only 45% correct gets him to the 74th %tile in Science.
The answer, actually, is quite simple. The History test is, by definition, easier.
More examinees got a higher percentage of the items correct on the History
test than on the Science test.
On the History test the average student probably got a score closer to 85% correct,
while on the Science test, the average student was more likely to have had a
score lower than 45% correct.
Item 3
Tom, Dick, and Harry reported their scores on the same science test to their
mothers. Each reported his score as 70.
In Harry’s case, his score was actually 70% of the items correct.
Tom’s score really referred to the 70th percentile.
Dick was actually reporting the number of items he got correct.
The test, which contained 75 items, had a mean of 60 and a standard deviation of
10. Who, if any, had the best performance on the test?
A.
B.
Tom
Dick
C.
D.
Harry
Can’t tell
The correct answer is B. Dick, who scored 70 of 75 items correct did best on the
test. His score is clearly better than Harry’s. Harry, who got 70% of the items
correct only got 52 or 53 items correct.
The only thing we have to check is whether Dick scored above or below Tom’s
70th percentile. To check this we first have to calculate his standard score. A
simple calculation reveals that Dick’s standard score is 1. He scored one
standard deviations above the mean. Since one standard deviation above the
mean equates to the 84th percentile, Dick clearly outscored Tom.
Equation for computing a standard score (z);
z = (Raw Score – Mean Score) / Standard Deviation
Item 4
Rosemary and Angela are new 10th-grade students to Lincoln High School. They
have just moved in to the area from different states, Rosemary from
Virginia and Angela from Texas. Both states have statewide standardized
testing programs. The Lincoln High counselor is trying to place the students
in appropriate classes. Right now he is trying to place the two students in
math and English.
He notes that Rosemary’s standard scores on her previous state’s math and
English tests were .25 and 1.2, respectively. Angela’s scores on her
previous state’s math and English tests were .75 and .95, respectively.
Which of the following statements is true?
A.
B.
C.
D.
Angela should be placed in higher level math and English Classes.
The students should be placed in the same level math and English classes.
Both should be placed in higher than average level classes in math and English.
Not enough information is given to determine placements.
The answer is D: Not enough information is given.
The two girl’s standardized test scores were obtained from different norm groups
and are, hence, not comparable. In order to be able to compare two
individuals’ norm-referenced scores (standard scores, percentiles, grade
equivalent scores, etc.) the scores have to be compared to the same norm
group. Rosemary’s scores are from a test normed in Virginia; Angela’s from a
test normed in Texas. It is possible that an above-average performance on, say,
a math test in Texas would be equivalent to a below-average performance on a
similar test in Virginia.
Item 5
Recall that Rosemary’s standard scores on her previous state’s math and English
tests were .25 and 1.2, respectively, and that Angela’s scores on her
previous state’s math and English tests were .75 and .95, respectively.
Which of the following interpretations can most easily be supported?
A.
B.
C.
Rosemary and Angela both perform better in English than in Math.
Only Rosemary is clearly better at English than at Math.
While Rosemary is clearly better than Angela in English, Angela is better in
math.
The correct answer is B: Only Rosemary is clearly better at English than at math.
The difference between Rosemary’s standard scores in math and English (.25 and
1.2) correspond, roughly, to a difference in percentiles of 51 %tile for math
and the 90th %tile for English. The difference in Angela’s standard scores for
math and English, on the other hand, correspond, again roughly, to a
difference between the the 79th %tile and the 83ed %tile.
Choice C is incorrect for reasons given for the previous item. The girls’ test scores
come from test normed on different populations and, hence, are not
comparable.