From Z Scores to Probability in Decision Making
We have calculated a series of statistics moving from the mean to the standard deviation to the z score. We then used the Z table to understand the relationship between the z score and standard deviation and the area underneath the normal curve or normal distribution. In the last assignment, you then used the Z table to identify the proportion of scores in a group that fall below a particular score in a unit of measure that corresponds to the proportion of total scores, i.e. area under the curve. But how does this apply to decision-making that relates statistics to probability?
There are several ideas to think through. First, understand that the area under the curve can represent one set of scores, say the math scores for a class of students. The area under the curve can also represent an infinitely large set of scores that exist in probability terms only. That is, if the mid-point or mathematical middle of a set is at the middle of the curve, then—because numbers continue into infinity both above and below that mid-point—there is an endless quantity of scores above and below the mean of a group that could be observed in theory. However, as these scores move away from the mean, moving further away from the mean or typical score in a group, it becomes increasingly unlikely that one of those extreme scores would be observed in reality. We can say this same thing a number of ways to assist understanding:
Since the Z score of a raw score is interpreted in standard deviation units and can be both positive or negative, then the higher the z score (in absolute terms, that is the further away from the mean of 0.00) the lower the probability of actually observing that score in the group.
Looking at the Z Table again: Find the z score of 0.00 (remember, it is the very first score on the table). Column 4 (for positive z scores; column 3 when looking at a negative z score: remember for negatives we flip the column headings) tells me that for a z of 0.00, .5000 or 50% of the area under the curve is below this score. I can (and should) read this column as the probability of obtaining this score due to random chance in a group is = or < .50 (equal to or less than 50%). Mathematically, we know this to be so. If I “flip” a two-sided coin, then the probability or “odds” of one side or the other coming up is 50/50 or 50%.
So, Column 4 (for positive z scores) tells me the probability, in an infinitely large set of scores, of getting a particular z score due to random chance. To state this differently for understanding, for any given score, column 4 tells me the probability that some student will obtain that score. (Lisa, for example: your z score for the raw score of 100 was 1.3. If I find this 1.3 on the Z Table, then I observe in column 4 that approximately 10% of the scores are above this score, or 10% of the area under the curve is above this score and 90% is below this score. Interpreted in probability terms, the probability of observing this score due to random chance in a group is <.10 or <10%. We would say that the p value of this score is .10, and write this as p<.10.
This skill, i.e. assigning probability of a score occurring to particular scores has been an historically important part of the practice of education. In gifted and talented education, and also in special education or intervention, historically an IQ or achievement score that was greater than 2 standard deviations above or below the mean of the group has been the primary basis for testing and/or assignment of students for intervention services. If you look at the Z Table for a z score of 2.00 (remember, the z score is read as standard deviations), you observe in column 4 that only 2% of scores are above this, and 98% of scores are below this. In probability terms, we read this as a p value of .02, or p<.02. This tells me the same thing: in probability terms, it would be expected that less than 2 percent of students would obtain IQ or achievement scores at or above this point. Historically, that indicator has been a demarcation of the line that tells the teacher when a student is “so unlike the group” that he or she really needs additional assistance.
Assignment: In yet another column of your Excel Spreadsheet, make a column heading of “p”. For each of your z scores, tell me the p value of the score. Remember that you want column 4 for positive z scores and column 3 for negative z scores. Do you have any scores which are above or below the 2.00? If this were a real set of scores, that would be a score of a student who was in particular need of intervention for the topic tested.
This is your final assignment. From this point, students would move into the intermediate statistics course (EDFN 509), although we are phasing that course out as it is no longer required for licensure in advanced coursework.