Sections 18-20 Standard Scores, Transformed, and the Normal Curve Flashcards
Standard Scores (z-scores) (Part 1)
The STANDARD SCORE (Also called the z-score) describes WHERE an individual stands in a group.
RAW SCORES (the number of points earned) mean little on their own because it doesn’t really tell you anything about the score or the distribution.
Ex: A Raw Score of 89 doesn’t mean much unless you have some other information, like what the maximum possible score was or how the others who took the test scored. It only has meaning if the score has some CONTEXT.
Ex: If you get your blood pressure taken and the doctor tells you it is 110/70, you may not know if that is good or bad unless you were told what the ‘normal’ range is.
- If you happen to already know the normal range for blood pressure, then you are engaging in a NORM-REFERENCED INTERPRETATION (meaning you have a useful REFERENCE POINT for INTERPRETATION)
STANDARD SCORES (also called z-scores) indicates how many standard deviation units a person’s score is from the mean and whether his or her score is above or below the mean.
- So z-score is represented as a positive or negative number of STANDARD DEVIATIONS.
- A POSITIVE number means the score is ABOVE the MEAN.
- A NEGATIVE number means the score is BELOW the MEAN.
Ex: Suppose a person is told that he has a z-score of 1.00. This indicates that he is…
- …exactly ONE STANDARD DEVIATION unit from the MEAN.
- …is 1 Standard Deviation ABOVE the mean (because the number is positive).
- We also know that a z-score of 1.00 puts him above about 84% of the participants in a NORMAL DISTRIBUTION.
- How do we know this? (Refer to figure 1. below)
- First, 50% of the participants lie BELOW the MEAN in a NORMAL DISTRIBUTION.
- Second, recall that the % of a normal distribution that lies within 1 SD from the mean (1 SD below the mean plus 1 SD above the mean) is about 68%. Since we are only looking at the 1 SD ABOVE the mean, we would divide the 68% by 2 to get 34%.
- In other words, about 34% of the cases lie between the mean and 1 standard deviation unit above the mean.
- Finally, add the 50% (which represents all scores below the mean) and 34% (which represents the scores between the mean and 1 SD above the mean) to get 50% + 34% = 84%
- How do we know this? (Refer to figure 1. below)
Standard Scores (z-scores) (Part 2)
To CALCULATE a z-score, use the following formula (see toward the bottom for a prettier representation of this same formula).
z = (X - M) / S
- z = z-score
- X = actual score (RAW SCORE)
- M = MEAN of ALL SCORES
- S = STANDARD DEVIATION of ALL SCORES
Ex: Suppose that a child weighs 55.00 pounds and that, for her age, height, and gender, the mean is 60.00 pounds and the standard deviation is 5.00. Entering this information in the formula…
Note: that pounds are raw scores.
- z = z-score = ?
- X = actual score (RAW SCORE) = 55.00
- M = MEAN of ALL SCORES = 60.00
- S = STANDARD DEVIATION of ALL SCORES = 5.00
Thus, using z = (X - M) / S
- z = (55.00 - 60.00) / 5.00 = (-5.00) / 5.00 = -1.00
…we find that she is ONE standard deviation unit BELOW the mean, which is much more meaningful than knowing just her weight expressed in pounds.
We also know that she is TALLER than ABOUT 16% of her peers and SHORTER than ABOUT 84% of her peers.
- How do we know this?
- First, 50% of the participants lie ABOVE the MEAN in a NORMAL DISTRIBUTION – So ALL of these participants (50% of the total) are TALLER than her.
-
BETWEEN 1 Standard Deviation BELOW THE MEAN and the MEAN lie another 34% of the total – these are all taller than her as well.
- REMEMBER that she was at -1.00, so this portion of the participants – this 34% of the total that is BOTH taller than her AND shorter than the mean – must be ADDED to the 50% that are TALLER than the mean to see where she stands:
- So, 50% + 34% = 84% of participants were TALLER than her.
- Because 84% of the participants were taller than her, that means that 100% - 84% = 16% were SHORTER than her
NOTE: 34% 1 SD BELOW the MEAN PLUS 34% 1SD ABOVE the MEAN = the 68% that we are familiar with – that 68% of the scores are within 1 SD (above or below) the MEAN.
Standard Scores (z-scores) (Part 3)
Here are some IMPORTANT Facts about STANDARD SCORES (z-scores)
- Suppose that an individual has a z-score of 0.00. This means the person has a score with the EXACT value of the MEAN (i.e., zero standard deviation units above or below the mean).
- If someone has a z-score of -2.00. This indicates that the person is 2 standard deviation units below the mean.
- Because 3 standard deviation units on both sides of the mean encompass 99.7% of the cases in a normal curve, z-scores are seldom higher than 3.00 or lower than -3.00. Figure 2 illustrates the effective range of z-scores:
Referring to a NORMAL DISTRIBUTION, we know that:
- The area lying within 1 SD of the mean contains 68% of scores (34% are 1 SD BELOW the mean, and 34% are 1 SD ABOVE the mean)
- The area lying within 2 SD of the mean contains APPROXIMATELY 95% of scores (47.50% are 2 SD BELOW the mean, and 47.50% are 2 SD ABOVE the mean)
- The area lying within 3 SD of the mean contains APPROXIMATELY 99.7% of scores (49.85% are 3 SD BELOW the mean, and 49.85% are 3 SD ABOVE the mean)
Knowing this, you can calculate the % of the total scores in each section of the NORMAL CURVE:
- Between 0 and +1 SD are about 34% of the scores
- Between 0 and -1 SD are about 34% of the scores
- Between -1 SD and +1 SD are about 68% of the scores
- Between 0 and +2 SD are about 47.5% of the scores
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Between 0 and -2 SD are about 47.5% of the scores
- Between -2 SD and +2 SD are about 95% of the scores
- Between 0 and +3 SD are about 49.85% of the scores
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Between 0 and -3 SD are about 49.85% of the scores
- Between -3 SD and +3 SD are about 99.7% of the scores
Transformed Standard Scores
Recall that STANDARD SCORES (z-scores) describe how many standard deviations a participant is from the mean,
- that standard scores range from -3.00 to 3.00.
- And a z-score of 0.00 indicates that a participant is at the mean-that is, exactly average.
A PROBLEM with STANDARD SCORES is that they’re not very intuitive to NON-statisticians. Consider an individual who knows she is about average and has worked hard on a test. If she is told that her score is 0.00, she is likely to be confused.
TRANSFORMED STANDARD SCORES (TSS) are the solution to this problem. TSS is another SCALE that does NOT have zeroes as the average and does NOT have negative values.
The TRANSFORMATION is very simple:
- T = (z)(10) + 50
So the TRANSFORMED score (the T-score) is simply the z-score multiplied by 10 plus 50. (refer to figure 1. below). This does a couple of things:
- The z-score (which for all practical purposes ranges between -3 and +3) would now RANGE BETWEEN +20 (50 - 30 = 20) and +80 (50 + 30 = 80) (after being multiplied by 10).
- “+50” would SHIFT THE AVERAGE from 0.00 in the z-score to 50 in the T-score.
- Together, these create a NORMAL DISTRIBUTION centered on 50 with each Standard Deviation represented by 10.
Ex: Say you have a z-score of -1.00. What is the T-score?
- using: T = (z)(10) + 50
- T = (-l.00)(10) + 50 = -10.00 + 50 = 40.00 = 40
- So the T-score = 40
IMPORTANT: There are DIFFERENT TYPES of TRANSFORMATIONS
- The one above is very common
- One used to TRANSFORM IQ z-scores is T = (z)(15) + 100 which sets the average IQ at 100 (See figure 2. below)
- NOTICE how the AVERAGE SHIFTED in this NORMAL CURVE (from 50 in figure 1. to 100 in figure 2.) and how the value of each STANDARD DEVIATION changed (from 10 in figure 1. to 15 in figure 2.)
- figure 1 and figure 2. have nothing to do with one another. I was simply pointing out how the TRANSFORMATION formula determines the values and their location on the curve.
- In General, TRANSFORMATIONS are created with this format:
- TSS = (z)(new standard deviation) + (new mean)
- If a particular TRANSFORMATION formula is NOT presented, then use the original one we used above.
Standard Scores and the Normal Curve
For any given STANDARD SCORE (z-score), we can determine the PERCENTAGE of CASES ABOVE and BELOW it by using the table of the normal curve in Table 1 near the BACK of the BOOK.
- In the left-most column, you will find a list of z-values that increase by 0.1.
- In the top row, you will see another list of z-values that increase by 0.01.
- (Refer to figure 1. below) To locate a z-value of 1.00, we look at Table 1. in the row labeled “1.00,” then choose the column with the appropriate hundredths place.
- Our z-value has a 0 in the hundredths place, so we look under the heading for “.00” to find a value of .8413 (if our z-value were 1.01, we would look under the “.01” label to find a value of .8438).
- Our value of .8413 is the probability of scores AT OR BELOW a z-value of 1.00. The probability of a score at or higher than a z-value of 1.00 is 1 minus .8413, or .1587.
- Figure 1 illustrates these numbers as percentages, which sum to 100%.
- It is important to note that Table 1 has two separate pages: one for NEGATIVE z-values and one for POSITIVE z-values.
EX: (Refer to figure 2. below for visual representation) As we know, going out 1.96 standard deviation units in both directions from the mean captures 95% of the cases.
- This can be confirmed by looking up z-scores of -1.96 and 1.96 in Table 1 near the end of this book.
- For a z-value of -1.96, you will find a probability of .0250 (or 2.5%) to the left of z (meaning 2.5% of the scores lie to the left of that z-score on the normal curve)
- For a z-value of 1.96, you will find a probability of .9750 (or 97.5%) to the left of z (meaning 97.5% of the scores lie to the left of that z-score on the normal curve).
- To identify the percentage of scores above 1.96, we subtract 97.5% from 100% to get 2.5%.
- We know now that 5% of the scores are beyond 1.96 standard units from the mean and that 95% (100 - 5.00 = 95.00) of the cases lie within this interval – confirmed.
