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Psych 210 Week 4 Notes: Indicators of Relative Standing

by: Victoria Snow

Psych 210 Week 4 Notes: Indicators of Relative Standing 210

Marketplace > University of North Carolina at Chapel Hill > PSYC > 210 > Psych 210 Week 4 Notes Indicators of Relative Standing
Victoria Snow

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About this Document

This week we talked about the indicators of relative standing (percentiles), the normal/standard distribution, and Z scores. I have detailed notes with examples and pictures of what we discussed in...
Statistical Principles of Psychology
Dr. Harrison
Class Notes
Psychology, Statistics, z, score, percentile
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This 6 page Class Notes was uploaded by Victoria Snow on Sunday September 18, 2016. The Class Notes belongs to 210 at University of North Carolina at Chapel Hill taught by Dr. Harrison in Fall 2016. Since its upload, it has received 19 views. For similar materials see Statistical Principles of Psychology in PSYC at University of North Carolina at Chapel Hill.


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Date Created: 09/18/16
PSYC 210 Week 4 Indicators of Relative Standing 9/12/16 Relativestanding-whereyou fall incomparison to other things Ex: Statue height. Lady Liberty compared to other statues. Percentilesand percentilemarks-relativestanding ofscores insample. Ex: Ifyou’re in the 80 percentile, you arehigher than 80% of people in that value/category. Datatransformations-changes inthemean and variance as a function of adding and multiplying constants Z-scores andthe Standard Normal Distribution Relativestanding using Z-scores and standard normal distributions -Percentile Ranks describe what a particular data point means relative to other data points. -The Pth percentile point is the valuebelow which the proportion of the cases fall -P.75  point below which 75% of the scores fall P( N ) n b X = L+ [ i ] n w X= Percentile point L=lower real limit P= percentile rank of required point N= total samplesize n(w)= number of cases withinreal limits n(b)= number of cases falling belowL i=sizeof score measurement unit To usethis formula:  Order (ascending)cases byscorevalue  Determine N  Determine location of desired point o CalculateP(N)-round to desired precision o Locate P(N)th caseindistribution  Determine other required values o N(w)= number of cases atdesiredpoint o N(b)= number of cases belowpoint o L=lower real limitof point o i =level of precision of measurement (1for whole numbers, .01 for two decimal places,etc.) o Example: Find 40 percentile from the list: 2,3,4,4,5,6,8,8,9,9 P( N )n b X = L[+ i ] nw P(N)= .4(10)= 4 4 pt from the bottom  2,3,4,4,5,6,8,8,9,9 N(b)= 2 (becausethere areonly 2 values below 4: 2 and 3) N(w)= 2 (becausethere aretwo “4”s indata set) i=1bc whole numbers X= 3.5 +((0.4(10)-2)/2)! X= 3.5 +(4-2)/2 X= 3.5 +2/2 X=4.5 40 percentile = 4.5  40% of the scores are below 4.5 Example: Find 50 percentile from GPAs P(N)= .5(31)= 15.5 Go to data,seethat 3.4is the desired point (between 15 and 16 lowest spots) So, L= 3.395, n(w)= 1, n(b)=15, and i=0.01 X= 3.395 + ((15.5-15)/1)(.01)= 3.4 What percentileisagiven score? nw( X L + (i)() n ) b P = (N ( ) )i All same variable X= your score P= percentile you are in Indicators of Relative Variability 9/21/16 Standardizing scores takes advantage of the “standard normal distribution” In order to do so,we need to “standardize” the scores Adding each number by a constant changes the mean by the constant  No effect on the varianceand standard deviation When multiplying each scoreby a constant, the mean changes by the constant  Increases the standard deviation and variance What a scoremeans in a distribution depends inpart on the mean and the variance Z Transformation -2 samples may contain the same highestscore,but in 1 case,thehighest scoreis very far from the mean and in the other it could be very close. S1: 5,6,7,8,20 S2: 16,17,18,19,20 Mean and variance for S1 and S2 are very differenct Mean S1= 9.2S(S1)= 6.14 Mean S2= 18 S(S2)= 1.58 -Although 20 is the highestin both distributions, 20doesn’t necessarily“mean” the samething in eachdistribution -Transforming things to Z scores allows us to compare across different samples with different characteristics-itequates the two distributions -Equates the means by making them 0 and equates std dev. by making them 1 X X Z = i Z for sample s x X i  X Z = Z for population  X Making theZ score: -First, the numerator of the Z transformation equation creates new distribution with mean of 0. - Referred to as “centering” a variable. - Makes the mean equal to 0 -Second, dividing by standard deviation creates a new distribution with a stddev of 1.0 Example: Web surfing data: Mean= 9.061 s=6.319 If another class hadsomeone with 15 hours per week, where does that put him/her? Z=(15 –9.061)/6.319 = 0.94 0.940 Z-scoreunits almost1 almost 1std dev above the mean If another class hadsomeone with 2hours per week, where does that put him/her? Z=(2-9.061)/6.319 =-1.12  almost 1 std dev less thanthe mean -Bystandardizing, we seethem within and across distributions with common and mean variance -Ifwe assumethe population distribution (is normal), we can useZ scores to calculate percentile ranks and points. Normal/Z Distribution: In a normal distribution:  Mean=0  S=1  Total area under curve=1.0  Curve is symmetrical  Usecurve to estimate the proportion ofscores that should fall at/below any given z score  TableA (p. 470) provides proportions for Z scores Example 1: 1) Calculate13hours z score: a. Z=(13-9.06)/6.32= 0.623 2) Look up 0.623 in Z table  0.2324= 23.2% above mean 3) Add 0.5 (to account for those below the mean) + 0.2324  0.7324 = 73.2% or the 73 percentile Example 2: 1) Find Z scorethat would leave40% of curve to the left. This is -.25. a. -.25 =(x –9.06)/6.32 b. x= 7.48  7.48 hours is our estimate of the 40 percentile point for web surfing Conclusion: - Using distributions like Z help us to make inferencesabout the populations from samples - Z is justthe firstthat we will use. We will come back it and more later.


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