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# Week 4 Notes - Psychology 211 Psychology 211

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This 22 page Class Notes was uploaded by Kennedy Patterson on Monday February 2, 2015. The Class Notes belongs to Psychology 211 at University of Alabama - Tuscaloosa taught by Andre Souza in Spring2015. Since its upload, it has received 222 views. For similar materials see Psychology 211-003 Elem Statistical Methods in Psychlogy at University of Alabama - Tuscaloosa.

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Date Created: 02/02/15

Week 4 1261513015 1 a Statistics is about variation b What is variation c The greater the variability in your data the greater your uncertainty i Uncertainty about what Uncertainty about the parameter that you want to estimate d i Wrote down the amount of money spend on drinks for 11 consecutive weekends Weekend Amount ZXL 1375129156119712 13 n 11 7 5 12 9 15 6 11 9 7 I tI LKOOOVGU39IwaI t HQ 12 N a Variability provides a quantitative measure of the differences between scores in a distribution b It describes the degree to which the scores are spread out or clustered together c Variability measures how well an individual score represents an entire distribution d What is the single number that represents the amount of dollars I spend on drinks i 964 e Depending on variability the mean is a good representation of the entire distribution 3 a Range is the distance covered by the scores in a distribution from the smallest score to the largest score b The range is the distance between the minimum and the maximum values c One important characteristic of the range is that only two data points contribute to it Week 4 1261513015 4 a b a b What ifI want all data points to contribute to the variability measure i We find the mean and how far from the mean each data point it The distance between each data point and the mean X I The longer the residual lines the more variability in the data Why is high variability a bad thing i Variability brings uncertainty c The higher the variability the higher the uncertainty d What is the best way to come up with one number to represent all the residuals i We can average all residuals e Whathappens if you sum all the residuals ZX39J The sum of all residuals will be weekend Amount Res39dua39 zero or close to zero due to 1 13 33936 rounding error 2 7 39264 This happens because the positive 3 5 39464 residuals cancel out the negative 4 12 236 ones 5 9 39064 How can you find the mean for the 6 15 536 residuals if the sum of the residuals 7 6 39364 will always be zero 8 11 136 The solution is you need to get rid 9 9 O64 of the negative signs 10 7 264 11 12 236 0 You can get rid of the negative signs by using absolute values 2 IXI39I I 0 Alternatively you can square all the residuals 2 IXi39f UZ 5 Week 4 1261513015 1 answer is zero 2 distance between each data point and mean 2Xi39f O 3 XI39IJ 4 ZX39f2 get most important measure of variability in statistics a SS sum of squares sum of deviation 6 How many glasses per week 1O61 8 Number of Residual Squared ZX 8 Glasses Residual f 2 1 1 1 14161 22 o 2 4 2xr2 22 6 4 16 ss 2x2 g k x2 1 1 n 1O618 1 1 ZX 8 ss 38 00 1O36138 4 ii6 2x2 38 ss 22 a Represents the total variability b What units i Glasses of wine SQUARED ii SS is always the squared of whatever original unit c What happens to SS if another data point is added X2 T 2 Answer is the same becausethe data 1 point added was the mean I O 1O612 10 0 If you added a new data point of 5quot 1O3614 42 36 SS 42 10 2 22 answer would be 29 Increase 1 LS If you added a new data point of O 4 answer would be 252 increase d The more data points added to the set n the bigger the SS e More numbers bigger total variation more uncertainty f To avoid have to make the SS not dependent on sample size Week 4 1261513015 9 Divide the SS by sample size I I mean squared deviation n When you add data galif increases Also increases n Average of Squared Deviation ZX I n ZX 2 total variation mint 2 average variation n 8 the number of independent pieces of information remaining after estimating one or more parameters a Sample A with 5 numbers n5 and mean I4 i What is the sum of all 5 numbers a 94ZX5420 n 5 ii 51OZ 20 1 Za na 54 20 iii What are the 5 numbers 1 nl 2 Degrees of Freedom df 3 Cannot choose last number freely sanu es choose m Cannot Sanu es Choose H 039 H n3 g XIf 2 degrees of freedom sample size number of le n1 parameters estimated from the data Cannot Week 4 1261513015 SS Sum of Squares SS 9 measure of total variability SS increases with sample size SS for popuation 9 population mean u 9 have to estimate because you don t know u n1 lose degree of freedom nl by estimating 9 gives us variance most common measure of average variability a b c d Eggff 9 not estimating because it is population 9 don t lose n degree of freedom Variance Population ZX2 S2 2X2 n n1 e Will be used to measure reliability of an estimate i Used to calculate confidence intervals ii Hypothesis testing iii In squared units 9 S2 sample 1 Population o 2 Difference between a Zx 2 9 sum of squares total variability b Leary 9 total average n1 3 How many texts were sent at 3am 4 a What is the average variation Number of texts Squared Scores 2X2 4 16 S2 ZX2 n 6 36 nl 5 25 52 386338 685 texts2 11 121 7 7 49 685 262 9 81 9 262 avg of residual g 499 distance between number and mean Week 4 1261513015 Standard Deviation 12XLZ S 2X2 n S ZX Xiff n1 391 2xi 2 32 zzff s gig2 n1 n1 in sq Units not in sq units Same thin 5 a 6 Can never get a negative number for standard deviation 7 The more variability around the mean the larger the S 8 Standard Deviation 9 the average distance between the data point and mean 9 Can get 0 as the standard deviation when all the numbers have the same value a number of friends m S gt 0 you hang outwith a week 3 5 N o o o o s o OOOOQO 4 1 2 3 4 Weeks 1 l the property of a statistic whose longrange average is not equal to the parameter it estimates the longrange average of a statistic over repeated samples 12 a graphical representation of the dispersion of a sample 13 a graphical representation of the dispersion of a sam le h the location of the quartile in an ordered series Median Location 1 2 15 line from top and bottom of the box to the farthest point that is no more than 15 times the interquartile range from the box 10 11 14 5 Week 4 1261513015 16 replaces the extreme observations with the highest and lowest value remaining after trimming 17 mince of a winsorized sample 18 the standard deviation of a winsorized sample Graph 1 Graph 2 00 00 l l 0 m LG LG lt1quot m N m H N 1 2 3 4 x 1 1 2 3 4 The range for Graph 1 is smaller than the range of Graph 2 The mean for Graph 1 is no different than the mean for Graph 2 The variance for Graph 1 is smaller than the variance for Graph 2 What is the variance and standard deviation for this sample I 58 S2 2X2 2X2 n Subject Scores X I XI2 1 B 65 7 49 24308 7 c 41 17 289 6 D 47 11 121 E 66 8 64 52 24308 23548 F 72 14 196 6 G 62 4 16 s2 126667 S 113 Week 4 1261513015 What is the variance and standard deviation for this population I 58 02 ZXJ 2 Subject Scores Xu Xu 02 m A 47 8 64 5 B 73 18 324 02 943 C 56 1 1 a 9 7 D 46 9 81 E 66 2 4 The mean of the following values is 89625 88967284l95l1ool92l90 The Deviations from the mean have been calculated as follows 1625 6376 17625 5625 i 5375 l1o375l 2375 i 0375 If this is a sample data If this is a population data the variance amp standard the variance amp standard deviation is deviation is s2 2xi2 o2 2xu2 n 1 n 52 527875 02 527875 7 8 2 z 8735 02 6598 39 a 812 The mean of the following values is 84875 16o93848ol95l99l78l90 The Deviations from the mean have been calculated as follows 24875 8125 o875 4875 l 10125 l 14125 i 6875 i 5125 If this is a sample data If this is a population data the variance amp standard the variance amp standard deviation is deviation is Week 4 1261513015 2 S2 ZxT2 02 M n 1 39 8 7 52 15498 0392 13561 S 1245 039 1165 Suppose the smallest value of 99 in the data was misrecorded as 999 If you were to recalculate the variance and standard deviation with the 999 instead of the 99 your new values for the variance and standard deviation would be Larger Use this BoxandWhisker plot to answer the questions 4120 2095 W ii WW 3 1200 800 400 000 400 800 1200 1600 The boxandwhisker plot shows Z observations identified as outliers Low values outliers correspond to extreme losses while highvalues outliers correspond to extreme gains You can define a maximum lower whisker of the first quartile QM15 x IQR and a maximum upper whisker of the third quartile Q315 x IQR for all nonoutlying values where IQR is the interquartile range Values below the maximum lower whisker or above the maximum upper whisker are considered outliers In the above boxandwhisker plot the outliers are either below the maximum lower whisker of 595 or above the upper limit of 693 Q1 1120 Q1153215 595 Q3 2095 Q3153215 693 Q3 Q1 3215 Week 4 1261513015 There are n10 women in a group their depression levels from a screening test are listen below in numerical order 5 17 l 28 l 31 44 l 45 l 47 l 49 51 63 l The Mean depression level is 3800 3800 10 17 l 28 l 31 44 l 45 l 47 l 49 51 The 10 Trimmed mean depression level is 3900 1728314445474951 312 3900 8 17 17 l 28 l 31 44 l 45 l 47 l 49 51 51 l The 10 Winsorized mean depression level is 3800 17172831444547495151 380 3800 10 Calculate the variance of the trimmed mean and its winsorized counterpart The variance of the 10 trimmed mean is 14829 The variance of its winsorized counterpart is 18400 Trimmed Variance Winsorized Variance ZX I2 1046 14929 ZXI2 1656 18400 n 1 7 n 1 9 Calculate the Standard Deviation of the trimmed mean and the winsorized sample The standard deviation of the 10 trimmed mean is 1218 The standard deviation of its winsorized counterpart is 1356 Trimmed Mean Winsorized Counterpart 1 14929 1218 1 184 1356 Week 4 1261513015 Answer the following questions based on the following graph There are 6 scores on the seesaw when balanced rests level when it pivots at a point equal to the sample mean What is the mean and what is the 6th score that needs to be placed on the seesaw The sample mean is f 5 Complete the table for the 5 scores that are displayed on the graph Score Distance from the Mean X 2 3 points below the mean X 3 2 points below the mean X 3 2 points below the mean X 4 1 point below the mean X 9 4 points above the mean There is exactly one possible value for an unknown X when the mean and all the other values of X are known M g 5 23349x 5 21X 21X 65 n 6 6 X 3021 X 9 Complete the following statements for the 5 scores that are displayed on the graph The total distance below the mean is g 3221 8 The total distance above the mean is A su of all 6 deviations from the mean is Q T e 9 5 9 5 45 1 441223 O 3 5 3 5 2 5 Week 4 1261513015 1 a Statistics is about variation b What is variation c The greater the variability in your data the greater your uncertainty i Uncertainty about what Uncertainty about the parameter that you want to estimate d i Wrote down the amount of money spend on drinks for 11 consecutive weekends Weekend Amount ZXL 1375129156119712 13 n 11 7 5 12 9 15 6 11 9 7 I tI LKOOOVGU39IwaI t HQ 12 N a Variability provides a quantitative measure of the differences between scores in a distribution b It describes the degree to which the scores are spread out or clustered together c Variability measures how well an individual score represents an entire distribution d What is the single number that represents the amount of dollars I spend on drinks i 964 e Depending on variability the mean is a good representation of the entire distribution 3 a Range is the distance covered by the scores in a distribution from the smallest score to the largest score b The range is the distance between the minimum and the maximum values c One important characteristic of the range is that only two data points contribute to it Week 4 1261513015 4 a b a b What ifI want all data points to contribute to the variability measure i We find the mean and how far from the mean each data point it The distance between each data point and the mean X I The longer the residual lines the more variability in the data Why is high variability a bad thing i Variability brings uncertainty c The higher the variability the higher the uncertainty d What is the best way to come up with one number to represent all the residuals i We can average all residuals e Whathappens if you sum all the residuals ZX39J The sum of all residuals will be weekend Amount Res39dua39 zero or close to zero due to 1 13 33936 rounding error 2 7 39264 This happens because the positive 3 5 39464 residuals cancel out the negative 4 12 236 ones 5 9 39064 How can you find the mean for the 6 15 536 residuals if the sum of the residuals 7 6 39364 will always be zero 8 11 136 The solution is you need to get rid 9 9 O64 of the negative signs 10 7 264 11 12 236 0 You can get rid of the negative signs by using absolute values 2 IXI39I I 0 Alternatively you can square all the residuals 2 IXi39f UZ 5 Week 4 1261513015 1 answer is zero 2 distance between each data point and mean 2Xi39f O 3 XI39IJ 4 ZX39f2 get most important measure of variability in statistics a SS sum of squares sum of deviation 6 How many glasses per week 1O61 8 Number of Residual Squared ZX 8 Glasses Residual f 2 1 1 1 14161 22 o 2 4 2xr2 22 6 4 16 ss 2x2 g k x2 1 1 n 1O618 1 1 ZX 8 ss 38 00 1O36138 4 ii6 2x2 38 ss 22 a Represents the total variability b What units i Glasses of wine SQUARED ii SS is always the squared of whatever original unit c What happens to SS if another data point is added X2 T 2 Answer is the same becausethe data 1 point added was the mean I O 1O612 10 0 If you added a new data point of 5quot 1O3614 42 36 SS 42 10 2 22 answer would be 29 Increase 1 LS If you added a new data point of O 4 answer would be 252 increase d The more data points added to the set n the bigger the SS e More numbers bigger total variation more uncertainty f To avoid have to make the SS not dependent on sample size Week 4 1261513015 9 Divide the SS by sample size I I mean squared deviation n When you add data galif increases Also increases n Average of Squared Deviation ZX I n ZX 2 total variation mint 2 average variation n 8 the number of independent pieces of information remaining after estimating one or more parameters a Sample A with 5 numbers n5 and mean I4 i What is the sum of all 5 numbers a 94ZX5420 n 5 ii 51OZ 20 1 Za na 54 20 iii What are the 5 numbers 1 nl 2 Degrees of Freedom df 3 Cannot choose last number freely sanu es choose m Cannot Sanu es Choose H 039 H n3 g XIf 2 degrees of freedom sample size number of le n1 parameters estimated from the data Cannot Week 4 1261513015 SS Sum of Squares SS 9 measure of total variability SS increases with sample size SS for popuation 9 population mean u 9 have to estimate because you don t know u n1 lose degree of freedom nl by estimating 9 gives us variance most common measure of average variability a b c d Eggff 9 not estimating because it is population 9 don t lose n degree of freedom Variance Population ZX2 S2 2X2 n n1 e Will be used to measure reliability of an estimate i Used to calculate confidence intervals ii Hypothesis testing iii In squared units 9 S2 sample 1 Population o 2 Difference between a Zx 2 9 sum of squares total variability b Leary 9 total average n1 3 How many texts were sent at 3am 4 a What is the average variation Number of texts Squared Scores 2X2 4 16 S2 ZX2 n 6 36 nl 5 25 52 386338 685 texts2 11 121 7 7 49 685 262 9 81 9 262 avg of residual g 499 distance between number and mean Week 4 1261513015 Standard Deviation 12XLZ S 2X2 n S ZX Xiff n1 391 2xi 2 32 zzff s gig2 n1 n1 in sq Units not in sq units Same thin 5 a 6 Can never get a negative number for standard deviation 7 The more variability around the mean the larger the S 8 Standard Deviation 9 the average distance between the data point and mean 9 Can get 0 as the standard deviation when all the numbers have the same value a number of friends m S gt 0 you hang outwith a week 3 5 N o o o o s o OOOOQO 4 1 2 3 4 Weeks 1 l the property of a statistic whose longrange average is not equal to the parameter it estimates the longrange average of a statistic over repeated samples 12 a graphical representation of the dispersion of a sample 13 a graphical representation of the dispersion of a sam le h the location of the quartile in an ordered series Median Location 1 2 15 line from top and bottom of the box to the farthest point that is no more than 15 times the interquartile range from the box 10 11 14 Week 4 1261513015 16 replaces the extreme observations with the highest and lowest value remaining after trimming 17 the variance of a winsorized sample 18 the standard deviation of a winsorized sample Graph 1 Graph 2 00 00 l l 0 0 L0 L0 lt1 dd m N m x l N 1 2 3 4 x 1 1 2 3 4 The range for Graph 1 is smaller than the range of Graph 2 The mean for Graph 1 is no different than the mean for Graph 2 The variance for Graph 1 is smaller than the variance for Graph 2 What is the variance and standard deviation for this sample I 58 S2 2X2 2 Subject Scores X I XI2 A 53 5 25 B 65 7 49 52 4062 C 41 17 289 24308 7 D 47 11 121 6 E 66 8 64 F 72 14 196 52 24308 23548 G 62 4 16 6 s2 126667 S 113 Week 4 1261513015 What is the variance and standard deviation for this population I 58 02 m 2 Subject Scores Xu Xu2 Xn B 73 18 324 5 D 46 9 81 a 97 E 66 2 4 The mean of the following values is 89625 88967284l95l1ool92l90 The Deviations from the mean have been calculated as follows 1625 6376 17625 5625 i 5375 l1o375l 2375 i 0375 If this is a sample data If this is a population data the variance amp standard the variance amp standard deviation is deviation is s2 2xi2 o2 2xu2 n 1 n 52 527875 02 527875 7 8 2 z 8735 02 6598 39 a 812 The mean of the following values is 84875 16o93848ol95l99l78l90 The Deviations from the mean have been calculated as follows 24875 8125 o875 4875 l 10125 l 14125 i 6875 i 5125 If this is a sample data If this is a population data the variance amp standard the variance amp standard deviation is deviation is Week 4 1261513015 2 S2 ZxT2 02 M n 1 39 8 7 52 15498 0392 13561 S 1245 039 1165 Suppose the smallest value of 99 in the data was misrecorded as 999 If you were to recalculate the variance and standard deviation with the 999 instead of the 99 your new values for the variance and standard deviation would be Larger Use this BoxandWhisker plot to answer the questions 4120 2095 W ii WW 3 1200 800 400 000 400 800 1200 1600 The boxandwhisker plot shows Z observations identified as outliers Low values outliers correspond to extreme losses while highvalues outliers correspond to extreme gains You can define a maximum lower whisker of the first quartile QM15 x IQR and a maximum upper whisker of the third quartile Q315 x IQR for all nonoutlying values where IQR is the interquartile range Values below the maximum lower whisker or above the maximum upper whisker are considered outliers In the above boxandwhisker plot the outliers are either below the maximum lower whisker of 595 or above the upper limit of 693 Q1 1120 Q1153215 595 Q3 2095 Q3153215 693 Q3 Q1 3215 Week 4 1261513015 There are n10 women in a group their depression levels from a screening test are listen below in numerical order 5 17 l 28 l 31 44 l 45 l 47 l 49 51 63 l The Mean depression level is 3800 3800 10 17 l 28 l 31 44 l 45 l 47 l 49 51 The 10 Trimmed mean depression level is 3900 1728314445474951 312 3900 8 17 17 l 28 l 31 44 l 45 l 47 l 49 51 51 l The 10 Winsorized mean depression level is 3800 17172831444547495151 380 3800 10 Calculate the variance of the trimmed mean and its winsorized counterpart The variance of the 10 trimmed mean is 14829 The variance of its winsorized counterpart is 18400 Trimmed Variance Winsorized Variance ZX I2 1046 14929 ZXI2 1656 18400 n 1 7 n 1 9 Calculate the Standard Deviation of the trimmed mean and the winsorized sample The standard deviation of the 10 trimmed mean is 1218 The standard deviation of its winsorized counterpart is 1356 Trimmed Mean Winsorized Counterpart 1 14929 1218 1 184 1356 Week 4 1261513015 Answer the following questions based on the following graph There are 6 scores on the seesaw when balanced rests level when it pivots at a point equal to the sample mean What is the mean and what is the 6th score that needs to be placed on the seesaw The sample mean is f 5 Complete the table for the 5 scores that are displayed on the graph Score Distance from the Mean X 2 3 points below the mean X 3 2 points below the mean X 3 2 points below the mean X 4 1 point below the mean X 9 4 points above the mean There is exactly one possible value for an unknown X when the mean and all the other values of X are known M g 5 23349x 5 21X 21X 65 n 6 6 X 3021 X 9 Complete the following statements for the 5 scores that are displayed on the graph The total distance below the mean is g 3221 8 The total distance above the mean is A su of all 6 deviations from the mean is Q T e 9 5 9 5 45 1 441223 O 3 5 3 5 2 5

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