Day4: Where am I
An entire distribution can often be reduced to a mean and standard deviation. A z-score uses that information to indicate the location of an individual score. Essentially, z-scores indicate how many standard deviations you are away from the mean. If z = 0, you’re at the mean. If z is positive, you’re above the mean; if negative, you’re below the mean. In practical terms, z scores can range from -3 to +3.
Composed of two parts, the z-score has both magnitude and sign. The magnitude can be interpreted as the number of standard deviations the raw score is away from the mean. The sign indicates whether the score is above the mean (+) or below the mean (-). To calculate the z-score, subtract the mean from the raw score and divide that answer by the standard deviation of the distribution. In formal terms, the formula is
Using this formula, we can find z for any raw score, assuming we know the mean and standard deviation of the distribution. What is the z-score for a raw score of 110, a mean of 100 and a standard deviation of 10? First, we find the difference between the score and the mean, which in this case would be 110-100 = 10. The result is divided by the standard deviation (10 divided by 10 = 1). With a z score of 1, we know that the raw score of 110 is one standard deviation above the mean for this distribution being studied.
Applications
First, z-scores can be used for describing the location of an individual score. If your score is at the mean, your z score equals zero. If z = 1, you are one standard deviation above the mean. If z = -2, you are two standard deviations below the mean. If z = 1.27, you score is a bit more than one and 1/4th standard deviations above the mean.
What is the z-score for a raw score of 104, a mean of 110 and a standard deviation of 12? 104-110 equals -6; -6 divided by 12 equals -.5. The raw score of 104 is one-half a standard deviation below the mean.
Second, raw scores can be evaluated in relation to some set z-score standard; a cutoff score. For example, all of the scores above a cutoff z-score of 1.65 could be accepted. In this case, z-scores provide a convenient way of describing a frequency distribution regardless of what variable is being measured.
Each z score’s location in a distribution is associated with an area under the curve. A z of 0 is at the 50th percentile and indicates that 50% of the scores are below that point. A z score of 2 is associated with the 98th percentile. If we wanted to select the top 2% of the individuals taking a musical ability test, we would want those who had a z score of 2 or higher. Z scores allow us to compare an individual to a standard regardless of whether the test had a mean of 60 or 124.
Most statistics textbooks have a table that shows the percentage of scores at any given point of a normal distribution. You can begin with a z score and find an area or begin with an area and find the corresponding z score. Areas are listed as decimals: .5000 instead of 50%. In order to save space, tables only list positive values are shown. The tables also assume you know that 50% of the scores fall below the mean and 50% above the mean. The table usually has 3 columns: the z score, the area between the mean and z, and the area beyond z.
The area between the mean and z is the percentage of scores located between z and the mean. A z of 0 has an area between the mean and z of 0 and the area beyond (the area toward the end of the distribution) as .5000. Although there are no negatives, notice that a z score of -0 would also have an area beyond (toward the low end of the distribution) of .5000.
A z score of .1, for example, has an area between the mean and z of .0398. That is, 3.98% of the scores fall within this area. And the third column shows that the area beyond (toward the positive end of the distribution) is .4602. If the z has -.1, the area from the mean down to that point would account for 3.98% of the scores and the area beyond (toward the negative end of the distribution) would be .4602.
Areas under the curve can be combined. For example, to calculate the percentile of a z of .1, the area between the mean and z (.0398) is added to the area below z (which you know to be .5000). So the total percentage of scores below a z of .1 is 53.98 (that is, .0398 plus .5000). A z score of .1 is at the 53.98th percentile.
Third, an entire variable can be converted to z-scores. This process of converting raw scores to z-scores is called standardizing and the resulting distribution of z-scores is a normalized or standardized distribution. A standardized test, then, is one whose scores have been converted from raw scores to z-scores. The resultant distribution always has a mean of 0 and a standard deviation of 1.
Standardizing a distribution gets rid of the rough edges of reality. If you’ve created a nifty new test of artistic sensitivity, the mean might be 123.73 and the standard deviation might be 23.2391. Interpreting these results and communicating them to others would be easier if the distribution was smooth and conformed exactly to the shape of a normal distribution. Converting each score on your artistic sensitivity test to a z score, converts the raw distribution’s bumps and nicks into a smooth normal distribution with a mean of 0 and a standard deviation of 1. Z scores make life prettier.
Fourth, once converted to a standardized distribution, the variable can be linearly transformed to have any mean and standard deviation desired. By reversing the process, z-scores are converted back to raw score by multiplying each by the desired standard deviation and add the desired mean. Most intelligence tests have a mean of 100 and a standard deviation of 15 or 16. But these numbers didn’t magically appear. The original data looked as wobbly as your test of artistic sensitivity. The original distribution was converted to z scores and then the entire distribution was shifted.
To change a normal distribution (a distribution of z scores) to a new distribution, simply multiply by the standard deviation you want and add the mean you want. It’s easy to take a normalized distribution and convert it to a distribution with a mean of 100 and a standard deviation of 20. Begin with the z scores and multiply by 20. A z of 0 (at the mean) is still 0, a z of 1 is 20 and a z of -1 is -20. Now add 100 to each, and the mean becomes 100 and the z of 1 is now 120. The z of -1 becomes 80, because 100 plus -20 equals 80. The resulting distribution will have a mean of 100 and a standard deviation of 20.
Fifth, two distributions with different means and standard deviations can be converted to z-scores and compared. Comparing distributions is possible after each distribution is converted into z’s. The conversion process allows previously incomparable variables to be compared. If a child comes to your school but she old school used a different math ability test, you can estimate her score on your school’s test by converting both to z scores.
If her score was 65 on a test with a mean of 50 and a standard deviation of 10, her z score was 1.5 on the old test (65-50 divided by 10 equals 1.5). If your school’s test has a mean of 80 and a standard deviation of 20, you can estimate her score on your test as being 1.5 standard deviations above the mean; a score of 110 on your test.
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Correlation
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