PSY201: Chapter 5: The Normal Curve and Standard Scores Introduction: Normal curve + a very important distribution in behavior sciences + three principal reasons why... - 1. many of the variables measured in behavioral science research have distributions that quite closely approximate the normal curve (ie: height, weight, intelligence and achievement are few examples) - 2. many of the inference tests used in analyzing experiments have sampling distributions that become normally distributed with increasing sample size. (ie: sign test & Mann-Whitney U test) - 3. many inference tests require sampling distributions that are normally distributed. The z test, Student's t test, and the F test are examples of inference tests that depend on this point much of importance of normal curve occurs in conjunction with inferential statistics. The Normal Curve: normal curve is a theoretical distribution of population scores. + a theoretical curve and is only approximated by real data + bell-shaped curve that is described by equation: curve has two inflection points, one on each side of the mean + inflection points are located where the curvature changes direction + ie: inflection points are located where curve changes from being convex downward to being convex upward - if the bell-shaped curve is a normal curve, inflection points are at 1 standard deviation from the mean ( and ) - as the curve approaches the horizontal axis, it is slowly changing its Y value. - the curve never quite reaches the axis - it approaches the horizontal axis and gets closer and closer to it, but it never quite touches it. - curve is asymptotic to the horizontal axis infection points under the curve, horizontal... in the diagram on page 97 Area Contained Under the Normal Curve: in distributions that are normally shaped, there is a special relationship between the mean and the standard deviation with regard to the area contained under the curve when a set of scores is normally distributed, 34.13% of the area under the curve is contained between the mean (u) and a score that is equal to u + 1o ; 13.59% of the area is contained between a score equal to + 1 and a score of u+ 2 o; 2.15%of the area is contained between scores of u+ 2o and u + 3o ; and 0.13% of the area exists beyond u+ 3o. This accounts for 50% of the area + since curve is symmetrical, same percentages hold for scores below the mean + since frequency is plotted on vertical axis, these percentages represent the percentage of scores contained within the area
ie: + have a population of 10,000 IQ scores + distribution normally shaped with u = 100 and o = 16 + since scores are normally distributed, 34.13% of scores are contained between scores of 100 and 116 ( u+ 1o = 100 + 16 = 116), 13.59% between 116 and 132 ( u+ 2o = 100+32 = 132), 2.15% between 132 and 148, and 0.13% above 148 + similarly, 34.13% of scores fall between 84 and 100, 13.59% between 68 and 84, 2.15% between 52 and 68, and 0.13% below 52. to calculate the number of scores in each area, multiply the relevant percentage by the total number of scores there are 34.13% x 10,000 = 3413 scores between 100 and 116, 13.59% x 10.000 = 1359 scores between 116 and 132, and 215 scores between 132 and 148; 13 scores are greater than 148. + for other half of distribution, there are 3413 scores between 84 and 100, 1359 scores between 68 and 84, and 215 scores between 52 and 68; there are 13 scores below 52. + these frequencies would be true only if distribution is exactly normally distributed + in actual practice, the frequencies would vary slightly depending on how close the distribution is to this theoretical model Standard Scores (z Scores): IQ of 132... + a score is meaningless unless you have a reference group to compare against + without one, can't tell whether the score is high, average, or low score is one of the 10,000 scores of distributions gives IQ of 132 some meaning + ie: can determine the percentage of scores in distribution that are lower than 132 determining the percentile rank of score of 132 (percentile rank of a score is defined as the percentage of scores that is below the score in question) 132 is 2 standard deviations above the mean + in normal curve, there are 34.13 + 13.59 = 47.72% of the scores between the mean and a score that is 2 standard deviations above the mean + to fine percentile rank of 132, need to add this percentage the 50.00% that lie below the mean 97.72% (47.72 + 50.00) of the scores fall below your IQ score of 132. + should be happy to be intelligent to solve problem, had to determine how many standard deviations the raw score of 132 was above or below the mean + transformed the raw score into a standard score, also called a z score a z score is a transformed score that designated how many standard deviation units the corresponding raw score is above or below the mean
process which by the raw score is altered score transformation + z transformation results in a distribution having a mean of 0 and a standard deviation of 1 + reason z scores are called standard deviation is they are expressed relative to a distribution mean of 0 and a standard deviation of 1 in conjunction with a normal curve, z scores allow to determine the number or percentages of scores that fall above or below any score in the distribution + z scores allow comparison between scores in different distributions, even when the units of the distributions are different ie: suppose that weights of all the rats in the housed in a university vivarium are normally distributed with u = 300 and o = 20 grams + what is the percentile rank of a rat weighing 340 grams? + first need to convert the raw score of 340 grams to its corresponding z score: since scores are normally distributed, 34.13 + 13.59 = 47.72% of the scores are between the score and the mean + adding the remaining 50.00% that lie below the mean, we arrive at a percentile rank of 47.72 + 50.00 = 97.72% for the weight of 340 grams. + thus the IQ score of 132 and the rat's weight of 340 grams have something in common... they both occupy the same relative position in their respective distributions the rat is heavy as you are smart z scores used to compare scores that are not otherwise directly comparable + ordinarily, not be able to compare intelligence and weight... they are measured on different scales and have different units + but by converting the scores to their z-transformed scores, we eliminate the original units and replace them with a universal unit, the standard deviation score of 132 IQ units becomes a score of 2 standard deviation units above the mean and the rat's weight of 340 grams also becomes a score of 2 standard deviation units above the mean it's possible to compare anything with anything as long as the measuring scales allow computation of the mean and standard deviation + ability to compare scores that are measured on different scales is of fundamental importance to the topic of correlation computing z scores using sample data
Characteristics of z Scores: three characteristics 1. z scores have the same shape as the set of raw scores + transforming raw scores into their corresponding z scores doesn't change the shape of the distribution nor do the scores change their relative positions all that's changed are the score values + (5.5): IQ scores and their corresponding z scores - used in conjunction with the normal distribution, all z distributions are not normally shaped can be calculated for distributions of any shape - resulting z scores will take on the shape of the raw scores 2. the mean of the z-scores always equals to zero (uz = 0) + scores located at the mean of the raw scores will also be the mean of the zero scores + z-value for raw scores will also be at the mean of the z-scores + z-value for raw scores at the mean equals zero + ie: z-transformation for a score at the mean of the IQ distribution is given by (5.5) ~~~ the mean of the z-distribution equals zero - 3. standard deviation of z scores always equals 1 + follows because a raw score that is 1 standard deviation above the mean has a z score of +1 Finding the Area Given the Raw Score: instead of an IQ of 132, desire to find the percentile rank of an IQ of 142 + assume same population parameters + first, draw a curve showing the population and locate the relevant area by entering the score 142 on horizontal axis
+ then shade the area desired + then calculate z: - first column of table (column A) contains the z score - column B lists the proportion of the total area between a given z score and the mean - column C lists the proportion of the total area that exists beyond the z-score use Table A to find percentile rank of 142 + locate the z-score of 2.62 in column A + determine from column B the proportion of the total area between the z-score and the mean + for a z score of 2.62, this area equals 0.4956 must add 0.5000 to this value to take into account the scores lying below the mean the proportion of scores that lie below an IQ of 142 is 0.4956 + 0.5000 = 0.9956 + to convert this proportion to a percentage, must multiply by 100 percentile rank of 142 is 99.56 + table A used to find the area for any z-score provided the scores are normally distributed - when using it, it it's sufficient to round z-values to two decimal place accuracy
Finding the Raw Score given the Area know the area and want to determine the corresponding score + ie: find the raw score that divides the distribution of aptitude scores such that 70% of the scores are below it this problem is the reverse of the previous one + given the area and need to determine the score + don't know what the raw score value is, can determine its corresponding z score from Table A + once know the z-score, can solve for the raw score using the z-equation if 70% of scores lie below the raw score, then 30% must lie above it + can find z-score by searching in Table A, column C, until we locate the area closest to 0.3000 (30%) and then noting that the z-score corresponding to this area is -.52 + to find the raw score, all we need to do is substitute the relevant values in the z-equation and solve for X:
PG 554 FOR TABLE A!!!