PSY201: Chapter 5: The Normal Curve and Standard ...
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
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+ 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
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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
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a z score is a transformed score that designated how many standard deviation units the corresponding raw score is above or below the mean
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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
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+ 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
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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!!!
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+ 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
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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
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+ 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
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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!!!
