Chapter 4: Standardized Scores

Chapter 4: Standardized Scores and the Normal Distribution

1. The Properties of z Scores

To compare two individuals who are in different distributions, it can be a big help to change their raw scores to standardized scores, and then compare the standardized scores. The simplest and most common standardized scores are the ones known as z scores.

1) Above or Below the Mean: Once raw scores have been converted to z scores, it is amazingly easy to tell if a data point lies above or below the mean of its distribution: if it’s a positive value, the data point is above the mean, and if it’s a negative value, the data point is below the mean. Simple as that.

2) Distance from the Mean: The magnitude of a z score tells you immediately how many standard deviations it is from the mean. If a z score is +2 or –2, you know right away that the score is pretty far (i.e., 2 SDs) from the mean, and in a bell-shaped curve, it would be pretty unusual.

3) Comparing Variables on Different Scales: Standardized scores make it possible to compare two raw scores that are measured on very different scales. For example, you could compare the number of hours someone spent stalking people on Facebook in a month to the number of face- to-face dates she went on during the same month. The two z scores would tell you where she fell in the Facebook stalking distribution and where she fell in the dating distribution, respectively (e.g., she might be low on stalking relative to her peers, but relatively high in dating). By converting to z scores for the two different distributions, you’ve managed to compare apples (e.g., number of hours) with oranges (e.g., number of dates).

After converting raw scores to z scores, the mean of your numbers will be zero, and the standard deviation will be 1. The new mean and SD are consequences of using the following formula:

Let’s try an example: Imagine that you need to acquire everyone’s weight in a fraternity house in order to match them as participants in a Greek Olympics Wrestling Tournament. Since disclosing one’s weight can be a touchy subject for some people (the bodybuilder may be proud to announce his muscle mass to the world, but the lanky cross-country runner may be a bit more shy on that front), we can at least try to mask the obvious numbers by using z scores. To keep things simple, we’ll imagine that only eight guys from your frat will be involved in the tournament.

Data: Weights of eight fraternity brothers in pounds: 165, 235, 170, 185, 210, 190, 180, 145.

Step 1: First, find the mean of the weights given: (165+235+170+185+210+190+180+145)/8 = 1480/8 = 185. Next, find the standard deviation of the 8 numbers: (biased) SD = 26.0.

Step 2: Then, plug each raw score into the formula: z = (X – µ)/σ, where µ = 185 and σ = 26. Weight of Each Formula z score Fraternity Brother 165 (165–185)/26 –.77 235 (235–185)/26 1.92 170 (170–185)/26 –.58 185 (185–185)/26 0 210 (210–185)/26 .96 190 (190–185)/26 .19 180 (180–185)/26 –.19 145 (145–185)/26 –1.54

Step 3: If you knew the z scores for weight of the possible opponents of these frat brothers, you could then match opponents based on z scores, without having to know their actual weights. For example, the guy who weighs 235 pounds is nearly 2 standard deviations above the mean and therefore needs to be matched with someone in the same ballpark to avoid an unseemly massacre. Note that the table tells you immediately whether someone is below the average of the eight frat brothers, because he will have a negative z score (–.77, –.58, –.19, –1.54), or above average (+1.92, +.96, +.19), or right at the average (z = 0). Also, note that the mean of the z scores is 0, and the (biased) SD is 1.0 (within rounding error— check for yourself!).

The one problem with the matching system just described is that you won’t know if the opposing team is lighter or heavier on average (or more or less variable), because you are removing the original mean (and standardizing the SD) when you convert to z scores (we’re assuming that the opposing team is presenting its weights in terms of z scores, as well). But you will be good at matching wrestlers who are at the same relative positions in their respective distributions.

2. T-Scores, SAT Scores, and IQ Scores

So now that you’re a pro with z scores, you may be wondering why everyone doesn’t use them all the time. The answer is: Would you want to tell someone that you scored a –.35 on an exam? It may be a bit awkward to post that on the refrigerator! To combat this, other scoring scales have been developed to make people feel more positively about themselves by giving everyone a positively valued score. (To be real, the main reason for these other scales is to avoid having to deal with decimal points and minus signs.) Some common examples include: T-scores, SAT scores and IQ scores.

To create these new scale scores, it makes sense to begin by finding the z score for the raw score you want to convert. Then, you can simply plug the z score into one of the formulas that follow to obtain a more convenient and aesthetically pleasing score.

T-Score = 10z +50  mean = 50; SD = 10.

Example: If z = –0.35, then T = 10 (–0.35) + 50 = 46.5 (a big improvement over a negative score!)

So, unless someone scores 5 SDs below the mean (an extremely rare event), his/her score will be a positive number. A common use of the T-score is for various psychological tests that are measured originally on arbitrary scales that have no intrinsic meaning, such as a self-esteem rating, which may have been measured originally as the sum of a bunch of 5-point Likert scales. If the original score of someone being tested is transformed to a T score of 40, it becomes obvious that the person is one SD below the mean (his/her z score would be –1, which is much more awkward to deal with).

SAT Score = 100z +500  mean = 500; SD = 100

Ever wonder how you ended up with a score of 670 on your verbal SAT, when there were only 45 questions on the test? Well, here’s your answer! Again, the raw score is first converted into a z score (in the case of 670, the corresponding z score is +1.7), and then the z score is plugged into the SAT formula. You should note that SAT scores can be thought of as T scores that have been multiplied by a factor of 10.

IQ Score (Stanford-Binet) = 16z + 100  mean =100; SD = 16

Now it should no longer be a mystery to you why the average IQ score is 100, a number that was obviously chosen for its simplicity! IQ scores could just have easily been on the T score scale, but for reasons that are known only to Stanford and Binet (), the common IQ formula took the form shown in that equation. As noted previously, if you know someone with an IQ score 2 SDs above 100 (i.e., 132), you know you’re dealing with someone who is unusually intelligent (someone above the 95th percentile as we will soon show). [Please note that the WAIS IQ scale is based on the formula: IQ = 15z + 100.]

Now try a few examples using the data from the exercises in the previous chapter:

1. Convert these body temperatures into z-scores: 97.6, 98.7, 96.9, 99.0, 93.2, 97.1, 98.5, 97.8, 94.5, 90.8, 99.7, 96.6, 97.8, 94.3, 91.7, 98.2, 95.3, 97.9, 99.6, 89.5, 93.0, 96.4, 94.8, 95.7, 97.4.

a) How many of these scores are above the mean? b) What is the spread between the highest and lowest z-score? What does this tell us?

2. Convert the ratings for each dorm (separately) into z-scores: Happy Hall: 5.5, 4.5, 6.0, 7.0, 3.0, 1.0, 5.5, 9.0, 2.0, 3.5; Terrific Tower: 6, 7, 8.5, 7.5, 6.5, 9, 9, 4, 6.5, 8

a) Which dorm has a larger gap between the highest and lowest z-score? b) Does this correspond with the raw data as well?

3. You just started your first teaching gig, and to ensure the students couldn’t decipher what the list depicted, you are given a list of their IQs as z-scores, based on the entire school population. Convert each one to an actual IQ score, using the Stanford-Binet formula: 2.1, –.8, .3, .25, 1.5, –1.6, 1.8, 2, 0, .2, 2.8, –1.2, –.6, 1.3, 1.6.

a) What is the average IQ score? Is this above or below the mean? b) Without first converting each z-score to an IQ score, how could you have figured out the average IQ score by only using the z–score data?

3. The Normal Distribution

The normal distribution (aka the normal curve) is an elusive entity that exists only in theoretical terms, since the tails of the curve continue endlessly. (It is considered to be theoretical, because in real life, there are almost always actual endpoints on each side of the curve.) Nonetheless, it is an important concept to understand, because it shows up (albeit not an exact replica) quite often. A good example (illustrated here) of an approximate normal distribution would be the curve for IQ scores (based on the WAIS scale); as discussed in the last chapter, the top of the distribution (the center point) would be the 100 mark, and all of the rest of the scores would fall elsewhere on the bell-shaped curve. Note in this example it is obviously only an approximation of the normal distribution because the tails are finite, as no one could score below zero, and any particular IQ test has to have a maximum score—no matter what kind of genius takes it!

Areas of Distributions

The area under the curve is considered equal to 100%, with standard cutoff points at each standard deviation marker as you stray from the mean.

As an example, let’s try to find out how Casey’s IQ fares in comparison to the rest of the population. (His brother is always teasing him that he’s a meathead whose only talent is catching a football, but Casey refuses to believe that. Sure, he’s not the best student and rarely cracks a textbook, but he knows he could go neck and neck with the AP students if he put his mind to it. Or so he hopes . . .) So his IQ score is 130, which puts him at two standard deviations above the mean on the WAIS IQ scale (remember that M = 100, SD = 15). When we glance at the normal distribution in the illustration, we can see that two standard deviations above the mean would equal 97.72% (13.59% + 34.13 % + 50 %)—which means that Casey scores as high, if not higher than, 97.72% of the population. Looks like it’s time for Casey to toss the football aside and pick up that statistics book!

As a quick cheat sheet, relevant to all normal distribution curves, you should memorize the following values to make your life a little easier when you’re trying to better understand these percentage values.

Area to the left of the mean + 1 SD = 84.13% Area to the left of the mean + 2 SDs = 97.72% Area to the left of the mean + 3 SDs = 99.87% Area to the left of the mean + 4 SDs = 99.99999%

Looking at the value for 4 SDs should help illustrate why it’s unique to find some value that is 4 SDs (or beyond) above the mean. As an example, at 6'6'', Michael Jordan’s height is only roughly 3 SDs above the mean. On the other hand, Yao Ming is 7'6'' and clocks in as the second tallest person in the world; admittedly, he is 6 SDs above the mean, but yeah, it takes being second tallest IN THE WORLD to get to that point! One other way to view these values is from the standpoint of figuring out how much of the population you capture within each standard deviation, starting at the center point (which is the mean) and working toward the tails.

The quick cheat sheet for those values is as follows:

1 SD in both directions from the mean = 68.26% 2 SDs in both directions from the mean = 95.44% 3 SDs in both directions from the mean = 99.74% 4 SDs in both directions from the mean = 99.999999%

Again, being 4 SDs out from the mean captures almost the entire population; except for the occasional outlier pretty much everyone is within 4 SDs from the mean. Whereas the first set of cheat sheet figures translates to percentile ranks, these values will help you to understand where the middle XX% fall with respect to standard deviations away from the mean.

As a note, these percentages could all be expressed as values from 0–1.00 (i.e., proportions), which will be more relevant when we discuss probabilities in later chapters. For example, someone with an IQ score of 115 is at the 84.13%tile, so we can also say that they beat .8413 of the people in their distribution.

Now you try a few examples:

4. What (approximate) percentile corresponds to an IQ score of:

a) 109? ______

b) 135? ______

c) 90? ______

d) 75? ______

Parameters of the Normal Distribution

Although the normal distribution (ND) has the same basic shape for each one created, the central point (its mean) and the width or spread of the curve (the standard deviation) are the parameters that give each ND its uniqueness.

As an example, look at the difference between men’s and women’s heights, both shown in the same graph here. As you can see, the women’s curve is narrower and taller, while the men’s curve is shorter and wider. More concretely, women are on average shorter than men (when comparing means), and there is more variability in men’s heights than in the heights of women. They are both normal distributions, but with differing means and SDs. Can you think of a variable in nature that would fall into a normal distribution?

Table of the Standard Normal Distribution

To stave off having to do integral calculus (remember that evilness from your high school days?) to determine an area underneath the curve every time there is a different mean and standard deviation, the standard normal distribution was created, with an accompanying table of values. Keep in mind, the standard normal distribution is based on the mean equaling zero and the standard deviation being 1, which should sound somewhat familiar to you. Remember those useful z-scores? Well, they’re back! But this time, they will have abundantly more meaning to you, since you’re now a burgeoning statistician.

So once you’ve transformed your values into z-scores, you can look up the area under the curve in Table A of your text. Keep in mind, Table A provides only the percent of area between the mean and the z- score, which means it is going to cap off at 50% (the top half of the curve), which is OK, because the curve is symmetrical.

Let’s do an example . . .

If I told you that the z-score for the average running speed for the QB of the UT Austin football team (when compared to all other QBs in college football) is +2.58, what percentage of QBs would be faster? First, look at the row for 2.5, and then skim across it to find the column for .08, and you’ll come to the value 49.51. With this information, you can assume that roughly only .49% (50.00 – 49.51) of QBs run faster than the UT QB. I smell a victory in UT’s future this year!

Now, you try to find these values on your own, and explain what each one means.

5. The z-score for the fraternity pledge who ate 14 guppies as part of his hazing process is +1.89. About what percentage of pledges did he “beat” in his attempt to please his fellow brothers?

6. A coffee shop at USC sells an absurd amount of coffee in the mornings and afternoons. However, around 11 P.M., their sales plummet dramatically. In comparison to every other hour of the day, the 11 P.M. time slot has a z-score for sales of –2.97. How dismal are the sales for this hour, in terms of its percentile rank? 7. The photography club at NYU has a budget of $14,500 per year for equipment purchases, which ranks it at 94.50% among U.S. private universities. What is the corresponding z-score?

4. Finding Areas for Normal Distributions

One thing you need to be aware of is that sometimes you need to determine the area under the curve that is between two z-scores, as opposed to between a z-score and the mean (which the table readily supplies). For example, what if you wanted to determine how much of the population has an IQ (WAIS) between 115 and 130?

First, find the value for 130 (47.72—look at 2.0 SDs from the mean, since 130 is exactly 2 SDs above 100), and then find the value for 115 (34.13—look at 1.0 SDs from the mean, since 115 is exactly 1 SD above the mean). Now, to find the area BETWEEN these two values, just subtract one from the other: 47.72 – 34.13 = 13.59. You now know that 13.59% of the population falls between the WAIS IQ scores of 115 and 130; you now also know that between one and two standard deviations on the normal distribution, you end up with an area of 13.59%.

Now you try a few examples . . .

8. What is the area between the following pairs of z-scores?

a) 1.05 and 1.15 ______b) 2.30 and 2.85 ______c) 0.00 and 2.15 ______d) –.34 and –.12 ______e) –2.30 and –2.85 ______f) –1.41 and +1.41 ______g) –.34 and 1.56 ______h) –3.0 and 3.00 ______i) 0.00 and 4.00 ______

2. Using SAT scores, what percentage of the population falls between?

a) 650 and 750 ______b) 450 and 500 ______c) 210 and 790 ______d) 500 and 800 ______e) 200 and 500 ______

Answers to Exercises 1. Mean = 96.08, (biased) SD = 2.7041; z-scores: +.56, +.97, +.30, +1.08, -1.07, +.38, +.89, +.64, -.58, -1.95, +1.34, +.19, +.64, -.66, -1.62, +.78, -.29, +.67, +1.30, -2.43, -1.14, +.12, -.47, -.14, +.49 a) Fifteen scores are positive, and therefore above the mean. b) The spread between the highest (1.34) and lowest (-2.43) is 3.77, which is the number of standard deviations apart they are.

2. Happy Hall mean = 4.7, (biased) SD = 2.2825; z-scores: +.35, -.09, +.57, +1.01, -.74, -1.62, +.35, +1.88, -1.18, -.53

Terrific Tower mean = 7.2, (biased) SD = 1.4697; z-scores: -.82, -.14, +.88, +.20, -.48, +1.22, +1.22, -2.18, -.48, +.54 a) Happy Hall has the larger gap (3.5) between the highest and lowest z-score. b) Yes, it does correspond with the raw data.

3. IQ scores: 133.6, 87.2, 104.8, 104, 124, 74.4, 128.8, 132, 100, 103.2, 144.8, 80.8, 90.4, 120.8, 125.6 a) The average IQ score is 110.29, which is above the population mean of 100. b) You could find the average z-score, and convert that one score into a raw IQ score.

4. What (approximate) percentile corresponds to an IQ of a) 109? ___≈ 70th____ b) 135? ___≈ 99th_____ c) 90? ___≈ 25th_____ d) 75? ___≈ 5th____

5. He beat 1 – .0294, or about 97% of his fellow pledges.

6. Sales for the 11 P.M. timeslot are in the .15 percentile rank.

7. The z-score associated with 94.50% is 1.60.

8. What is the area between the following pairs of z-scores? a) 1.05 and 1.15 __.8749 – .8531 = .0218___ b) 2.30 and 2.85 __.9978 – .9893 = .0085___ c) 0.00 and 2.15 __.9842 – .5000 = .4842___ d) –.34 and –.12 __.4522 – .3669 = .0853___ e) –2.30 and –2.85 __.0107 – .0022 = .0085___ f) –1.41 and +1.41 __.4207 + .4207 = .8414___ g) –.34 and 1.56 __.1331 + .4406 = .5737___ h) –3.0 and 3.00 __.4987 + .4987 = .9974___ i) 0.00 and 4.00 __.5000 – .99997 = .49997__

9. Using SAT scores, what percentage of the population falls between: a) 650 and 750 _1.5 and 2.5: .9938 – .9332 = 6.06%__ b) 450 and 500 _-.5 and 0: .6915 – .5 = 19.15%______c) 210 and 790 _-2.9 and 2.9:_.4981 + .4981 = 99.62% d) 500 and 800 _0 and 3:_.9987 – .5 = 49.97%______e) 200 and 500 _-3 and 0: .9987 – .5 = 49.97%______