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Multiple Regression (Categorical variables -in regression (categorical…

- - - - order effects #
    - - direction depends on the order of coding
- - - - ie height and hair length= no multicollinearity
      - height anf weight - multicollinearity lots of shared variance
      - ASSESING
        
        look at pearsons correlation- want correlations btween -.9 and +.9- values outside of this are big and bad!!
        
        to meet assumption : look at the coefficients table
        
        Coefficients
        
        Tolerance should be more than 0.2
        
        Varience inflation factor (VIF) should be less than 10
        
        no mulitcollineariyy- interpret overall model as usual with the collinearity stats
  - - - use the bar chart and scatter graph for distribution but this is robust and not systematically biased
  - - - similar variance of residuals (errors) across the variable continuum ( high and low outcomes) equall across all scores in accuracy of prediction
        
        we don't want HETEROSCEDASTICITY
        
        variance of residuals (errors) differ across the variable continuum ( high and low) more accurate with some scores than others
        
        want points to be plotted randomly across the scatter graph - in a splodge in the middle - not points pointing in a cone to either side
- - - - r=.83, p <.001
      - more child plays games the more aggression they display
    - - model summary table :
        
        how good your model with predictors is at predicting the outcome
        
        looking at the R2 and adjusted R2
        
        need adjusted R squared
        
        the amount of variance in the outcome variable (aggression)explained by the predictor
        
        adjusted R 2= .667 x 100 = 66.7% variance explained by model
        
        66.7% of the variance in aggressive behaviours can be explained by number of hours child spends playing computer games
      - next look at the ANOVA table
        
        tells you how good the regression model is at predicting the outcome using the predictors - measure if the model is good ?
        
        so ......
        
        ( F ( Df experimental, df error)= F Stat, p significance
        
        model significantly better than just using the means - (F(1,18) = 39.1, p<.001) regression line is not flat
        
        look at the mean score?
      - next look at the coefficient to use as predictor
        
        look at the Beta and the t and sg
        
        intercept is the B Constant
        
        number of aggressive behaviours per day if 0 hours spent = 1.767
        
        slope = b predictor
        
        number of additional aggressive behaviours for each hour spent playing computer games .687
        
        Y=Bo+(B1*X) #
        
        value want to predict = Intercept + (Slope* the value you know : number of hours spent playing computer games per week)
        
        (B=.687, t =6.3, p<.001)
        
        foir each hour a child spends playing computer games they display 0.687 additional aggressive beahviors per day
- - - - B=.551, t=4.9, P<.001
        
        INCREASE
      - physical exercise : B=-.559, t=-2.5, p=.025
        
        DECREASE
      - extraversion
        
        B=-.128, t=-1.0, p=.353
      - THEN INTERPRET
        
        INCREASE OR DECREASE IN OUTCOME
        
        only if significant
  - - - .732= 73.2%
      - model significant predictor of aggression (F(3,16)=18.3, p<.001)
        
        ANOVA TABLE
- - - - when we add predictor varibales into the model would they be better at predicting which group a particular person belongs to than a random model
      - look at the omnibus tests of model coefficients
        
        then look at model summary table
        
        R2 estimations of variance explained by the overall model Cox & Snell: 9.0%, nagelkerke: 11.9%
        
        X2 (1)=4.7, p =.030
        
        need: chi square (Degrees of freedom)= chi square stat, p value
        
        overall model using chi squared
    - - for a binary outcome there is a 50/50 chance of belonging to group 1 if membership is random, can chance explain membership
      - randomly predicting group membership - look at classification table
        
        Predicted (randomly) number of people in each group UN= 0 SN=50
        
        observed : actual number of peoplein each group Unsuccessful N= 24 , successful N=26
        
        52% correct
      - variables in the equation table
        
        step 0 constant: not sig B=.08, Wald (1) = .08, p=.777
        
        random model is not able to sig predict whether participants are successful at giving up smoking or not- adding a predictor variable improve this model?
    - - classification table
      - looking at those observed and those predicted -
        
        sucessful
        
        8 peoples group membership was wrongly predicted
        
        18 peoples membership was correctly predicted
        
        unsucessful
        
        10 peoples membership was correctly predicted
        
        14 peoples membership was wrongly predicted
        
        percentage unsuccessful (n=24) correctly predicts 41.7% of unsuccessful cases
        
        predicting successful cases (n=26) correctly predicts 69.2% of cases
        
        MOdel more accurate at predicting success - overall model correctly predicts 56% of cases
        
        variables In the equation
        
        number of cigarettes smoked per day is a sig predictor of success giving up smoking
        
        B=-.091, Wald (1)=4.2, p=.042
        
        Use the B value to interpret the direction of the predictive relationship
      - B values :
        
        change in probability of group membership for each one unknit increase of the predictor variable
        
        positive B value: as predictor variable increases , chance of belonging to group 1 increases
        
        negative B value: as predictor variable increases , chance of belonging to group 1 decrases ( so belonging to group 0)
        
        with each extra cig smoked he probability of success of quitting reduces
      - add variales to scatterplot - highly sig model - stepped
        
        he more stepped the scatterplot better at predicting the modle is
- - - - aggressin and computer games- r=.827, p<.001
      - aggression and physical exercise : r= -.650, p =.001