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Practice. You collect data from 53 females and find the correlation between candy and depression is -.40. Determine if this value is significantly different than zero.
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Practice • You collect data from 53 females and find the correlation between candy and depression is -.40. Determine if this value is significantly different than zero. • You collect data from 53 males and find the correlation between candy and depression is -.50. Determine if this value is significantly different than zero.
Practice • You collect data from 53 females and find the correlation between candy and depression is -.40. • t obs = 3.12 • t crit = 2.00 • You collect data from 53 males and find the correlation between candy and depression is -.50. • t obs = 4.12 • t crit = 2.00
Practice • You collect data from 53 females and find the correlation between candy and depression is -.40. • You collect data from 53 males and find the correlation between candy and depression is -.50. • Is the effect of candy significantly different for males and females?
Hypothesis • H1: the two correlations are different • H0: the two correlations are not different
Testing Differences Between Correlations • Must be independent for this to work
When the population value of r is not zero the distribution of r values gets skewed Easy to fix! Use Fisher’s r transformation Page 746
Testing Differences Between Correlations • Must be independent for this to work
Testing Differences Between Correlations Note: what would the z value be if there was no difference between these two values (i.e., Ho was true)
Testing Differences • Z = -.625 • What is the probability of obtaining a Z score of this size or greater, if the difference between these two r values was zero? • p = .267 • If p is < .025 reject Ho and accept H1 • If p is = or > .025 fail to reject Ho • The two correlations are not significantly different than each other!
Remember this:Statistics Needed • Need to find the best place to draw the regression line on a scatter plot • Need to quantify the cluster of scores around this regression line (i.e., the correlation coefficient)
Regression allows us to predict! . . . . .
Straight Line Y = mX + b Where: Y and X are variables representing scores m = slope of the line (constant) b = intercept of the line with the Y axis (constant)
That’s nice but. . . . • How do you figure out the best values to use for m and b ? • First lets move into the language of regression
Straight Line Y = mX + b Where: Y and X are variables representing scores m = slope of the line (constant) b = intercept of the line with the Y axis (constant)
Regression Equation Y = a + bX Where: Y = value predicted from a particular X value a = point at which the regression line intersects the Y axis b = slope of the regression line X = X value for which you wish to predict a Y value
Practice • Y = -7 + 2X • What is the slope and the Y-intercept? • Determine the value of Y for each X: • X = 1, X = 3, X = 5, X = 10
Practice • Y = -7 + 2X • What is the slope and the Y-intercept? • Determine the value of Y for each X: • X = 1, X = 3, X = 5, X = 10 • Y = -5, Y = -1, Y = 3, Y = 13
Finding a and b • Uses the least squares method • Minimizes Error Error = Y - Y (Y - Y)2 is minimized
. . . . .
Error = Y - Y (Y - Y)2 is minimized . Error = 1 . Error = .5 . . Error = -1 . Error = 0 Error = -.5
Finding a and b • Ingredients • COVxy • Sx2 • Mean of Y and X
Regression Ingredients Mean Y =4.6 Mean X = 3 Covxy = 3.75 S2X = 2.50
Regression Ingredients Mean Y =4.6 Mean X = 3 Covxy = 3.75 S2x = 2.50
Regression Ingredients Mean Y =4.6 Mean X = 3 Covxy = 3.75 S2x = 2.50
Regression Equation Y = a + bx Equation for predicting smiling from talking Y = .10+ 1.50(x)
Regression Equation Y = .10+ 1.50(x) How many times would a person likely smile if they talked 15 times?
Regression Equation Y = .10+ 1.50(x) How many times would a person likely smile if they talked 15 times? 22.6 = .10+ 1.50(15)
Y = 0.1 + (1.5)X . . . . .
Y = 0.1 + (1.5)XX = 1; Y = 1.6 . . . . . .
Y = 0.1 + (1.5)XX = 5; Y = 7.60 . . . . . . .
Y = 0.1 + (1.5)X . . . . . . .
Mean Y = 14.50; Sy = 4.43Mean X = 6.00; Sx= 2.16 Quantify the relationship with a correlation and draw a regression line that predicts aggression.
∑XY = 326 ∑Y = 58 ∑X = 24 N = 4
∑XY = 326 ∑Y = 58 ∑X = 24 N = 4
COV = -7.33 • Sy = 4.43Sx= 2.16
COV = -7.33 • Sy = 4.43Sx= 2.16
Regression Ingredients Mean Y =14.5 Mean X = 6 Covxy = -7.33 S2X = 4.67
Regression Ingredients Mean Y =14.5 Mean X = 6 Covxy = -7.33 S2X = 4.67
Regression Equation Y = a + bX Y = 23.92 + (-1.57)X
Y = 23.92 + (-1.57)X . 22 20 . 18 16 . 14 . 12 10
Y = 23.92 + (-1.57)X . . 22 20 . 18 16 . 14 . 12 10