Lab 14

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Lab 14

• Curvilinear analysis and detailed example of
  categorical and continuous variables analysis

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Curvilinear Regression

• Linear regression assumes that a straight line
  properly represents the relations between each IV
  and the DV.
• This is not always the case. For example, it has
  been found that the relationship between job
  satisfaction and job tenure (length of time in a
  job) is a curvilinear relationship. Employees
  with low and high tenure have high satisfaction
  and employees with moderate tenure have the
  lowest satisfaction.

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Example of Curvilinear Job Satisfaction and tenure
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How to test this with SAS

• What we do in polynomial regression is to conduct
  a sequence of tests. We start with regressing DV
  on IV.
• Then add IVIV to model to see if that accounts
  for a significant amount of additional variance.
• If it does, we add IVIVIV to see if it adds
  variance. We stop when adding a successive power
  term fails to add variance accounted for.

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Example

• A sports physiologist is interested in the
  effects of diet on strength of athletes. He
  measures strength and the amount of protein
  consumed and he wants to know what the
  relationship is between these two variables.
• Form quadratic and cubic terms.
• Run the regressions to test for trends and
  identify the best model.
• Graph the relations between X and Y for evidence
  of nonlinearity.

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Example program

• data d1
• input protein strength
• create power terms
• protein2proteinprotein
• protein3protein2protein
• cards
• regressions with linear, quadratic, and cubic
  models
• linear
• proc reg
• model strength protein
• plot strengthprotein r.p.
• quadratic
• proc reg
• model strength protein protein2
• plot r.p.
• cubic
• proc reg
• model strength protein protein2 protein3
• plot r.p.

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Output Model 1

• Model MODEL1
• Dependent
  Variable strength
• Analysis
  of Variance
• 
  Sum of Mean
• Source DF
  Squares Square F Value Pr gt F
• Model 1
  16191 16191 646.01 lt.0001
• Error 248
  6215.86885 25.06399
• Corrected Total 249 22407
• Root MSE
  5.00639 R-Square 0.7226
• Dependent Mean
  202.56800 Adj R-Sq 0.7215
• Coeff Var
  2.47146
• Parameter
  Estimates
• 
  Parameter Standard
• Variable DF Estimate
  Error t Value Pr gt t
• Intercept 1 145.33012
  2.27414 63.91 lt.0001
• protein 1 0.81480
  0.03206 25.42 lt.0001

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Model 2 Output

• Model MODEL1
• Dependent
  Variable strength
• Analysis
  of Variance
• 
  Sum of Mean
• Source DF
  Squares Square F Value Pr gt F
• Model 2
  19145 9572.45217 724.73 lt.0001
• Error 247
  3262.43966 13.20826
• Corrected Total 249 22407
• Root MSE
  3.63432 R-Square 0.8544
• Dependent Mean
  202.56800 Adj R-Sq 0.8532
• Coeff Var
  1.79412
• Parameter
  Estimates
• 
  Parameter Standard
• Variable DF Estimate
  Error t Value Pr gt t
• Intercept 1 22.06447
  8.40699 2.62 0.0092
• protein 1 4.42387
  0.24247 18.24 lt.0001
• protein2 1 -0.02589
  0.00173 -14.95 lt.0001

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Model 3 Output

• 
  Sum of Mean
• Source DF
  Squares Square F Value Pr gt F
• Model 3
  19145 6381.64432 481.20 lt.0001
• Error 246
  3262.41104 13.26183
• Corrected Total 249 22407
• Root MSE
  3.64168 R-Square 0.8544
• Dependent Mean
  202.56800 Adj R-Sq 0.8526
• Coeff Var
  1.79776
• Parameter
  Estimates
• 
  Parameter Standard
• Variable DF Estimate
  Error t Value Pr gt t
• Intercept 1 20.15763
  41.90111 0.48 0.6309
• protein 1 4.51006
  1.87112 2.41 0.0167
• protein2 1 -0.02716
  0.02742 -0.99 0.3230
• protein3 1 0.00000613
  0.00013194 0.05 0.9630

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Conclusions

• The b-weight is significant for the quadratic
  model and not for the cubic model, therefore it
  appears that the quadratic equation is the best
  fit for this data (Y22.064.42X1-.026X12) and
  it accounts for 85 of the variance.
• Looking back at the graph (strengthprotein), it
  appears that the benefit of protein is large at
  first and then levels off, where athletes receive
  little to no benefit at around the 70 mark.

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Detailed Example

• Events variable is a person's score on a life
  event scale, indicating the number and severity
  of recent life events. 
• Status variable is a measure of whether a person
  co-habits with a partner (a 0 indicates that they
  do not, and a 1 indicates that they do). 
• Stress variable is the score on self-report
  measure of experienced stress

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Hypotheses

• 1 The more life events, the greater the stress.
• 2 Those who live with their partner will have
  lower stress than participants who dont live
  with a partner.
• 3 The relationship between events and stress is
  predicted to be moderated by status.
  Participants who cohabitate with a partner are
  predicted to be less stressed by life events than
  those who do not live with a partner.

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Evaluate Normality

• Check normality in variables.
• Proc univariate normal plot
• Check normality by Status.
• Proc univariate normal plot
• By status

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Results of normality

• Box plots Stress variable looks normal but
  Events is positively skewed with few people
  having high scores. No evident outliers.
• Shapiro-Wilk supports visual conclusions, Stress
  was not significant (W 0.981, ns) and Events
  was significant (W 0.935, p lt .05) , indicating
  non normality. With a small percentage of
  participants reporting large number of life
  events.
• Good distribution of status, 30 in a relationship
  and 30 not in a relationship.

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Normality with by Status

• Participants not in a relationship had higher
  means on events in life than those in a
  relationship. Similar variability in the both
  status groups across the event variable.
• Participants not in a relationship had higher
  means on stress variable than those in a
  relationship, providing visual support for
  hypothesis 1. There were two outliers in the
  relationship group and the variance appears
  smaller in the relationship group.

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Descriptive stats

• Means, SD, and correlations.
• Proc means
• Proc corr

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Proc mean and corr results

• Both independent variables, Status and Events,
  had significant relationships with stress.
• Status had a significant negative relationship
  with stress (r(58) -.49, p lt.05 0doesnt
  cohabit and 1does cohabit).
• Events had a significant positive relationship
  with stress (r(58) .41, p lt .05).
• Independent variables not significantly
  correlated with one another (status and events
  r(58) -.12, ns), which indicates that
  collinearity is not a problem with these data.

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Linearity, Outliers, and Homoscedasticity

• Look at plots for heteroscedasiticity and
  nonlinearity.
• proc gplot
• plot stressevent
• proc gplot
• plot stressevent
• by status

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Graphs

• No evidence of heteroscedasticity or non linear
  trends.
• There does appear to be a stronger relationship
  between stress and events for those participants
  who do not live with a partner.

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Statistical test for Curvilinear data

• Create power terms
• Event2eventevent
• Standardize variables
• Proc standard m0
• Run regression on linear and quadratic models
• proc reg
• model stress event
• proc reg
• model stress event event2

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Results of curvilinear analysis

• Linear model is significant and accounts for 17
  of the variance in stress (F(1, 58) 11.85, p lt
  .05).
• Quadratic model is also significant (F(2,57)
  6.80, p lt .05) and accounts for 19 of the
  variance, but the beta-weight for the quadratic
  term is not significant (b(57) 1.27, ns).
• Therefore, the linear model appears to be the
  best fit for this data.

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Data fit, outliers and homoscedasticity

• Run regression and check for outliers.
• Proc reg
• Model stress event status/ stb R influence
• Plot p.r. stressp.

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Results outliers

• Predicted by residuals plot showed no apparent
  heteroscedasticity. The values appeared to be
  randomly scattered around the zero residual line.
• Predicted by actual demonstrates a positive
  relationship. No apparent outliers

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Results outliers (cont.)

• 3 outliers were identified with a studentized
  residual greater than 2, 10, 29, and 54.
• Leverage gt 2(k1)/N .10.
• Cooks D gt.2
• DF Betas gt .26

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Outlier conclusions

• There doesnt appear to be any large problems
  with outliers. 29 did have some influence so we
  will try running the regression analysis without
  it at the end and see if there are differences in
  the significance.

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Collinearity

• Analyze regression with collinearity diagnostics
  included.
• Proc reg
• Model stress event status/ vif tol collin

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Collinearity results

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Analyze Regression Results

• Create interaction term
• inter statusevent
• Run regression analysis with and without
  interaction.
• Proc Reg
• Model stress status event inter/stb
• Go to flow chart on the next slide.

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• Y a b1X1(groupvar) b2X2(continvar)
  b3X1X2(inter)

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Regression results

• Overall model without the interaction was
  significant (F(2,57) 16.51, p lt .05) and
  accounted for 37 of the variance.
• Both life events (ß .36, t(58) 3.35, plt.05)
  and status (ß -.45, t(58) -4.21, plt.05) were
  significant predictors of stress.
• The overall model with the interaction was also
  significant (F(3,56) 12.98, p lt .05) and
  accounted for 41 of the variance.
• The interaction was significant (ß -.40, t(58)
  -2.00, plt.05), but status was no longer
  significant (ß -.13, t(58) -.67, ns).
• Therefore, The slopes of the two groups differ
  Compute separate regressions for each group

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Produce regression on the same graph, correlation
by status, proc means

• Proc Means
• By status
• Run correlation by group
• Proc corr
• Var stress event
• By status
• Overlay regressions for two groups
• symbol1 colorblue interpolr1 valuenone
• symbol2 colorblack interpolr2 valuenone
• Proc Sort by status
• Proc gplot
• plot stressevent status

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Conclusions

• For participants who did not live with a partner,
  the correlation between stress and life events
  was not significant (r(28) .10, ns).
• For participants who did live with a partner, the
  correlation between stress and life events was
  significant (r(28) .62, p lt .05).
• The graph of the two regression lines illustrate
  the interaction effect, with almost no slope for
  those not living with a partner and a moderate
  slope for those living with a partner.
• Those participants living with a partner did show
  lower levels of stress (M 18.3, SD 5.47) than
  participants who do not live with a partner (M
  24.3, SD 6.14), but this difference was not
  significant when the interaction was added to the
  model.

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Oops, one last thing, we forgot to run the model
again deleting participant 29

• Delete participant 29 and rerun the analysis.
• If _n_ 29 then delete

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Conclusions after deleting

• After deleting that one case, the interaction
  term is no longer significant (ß -.32, t(57)
  -1.62, ns). You would want to look at that one
  value and see if it was an error.
• If you feel that the data point is a true score
  you should probably report results before and
  after.
• A big limitation of this example is the low
  sample size.
• If the sample size was larger, the interaction
  would probably be significant. There seemed to
  be a large effect. Even after the outlier was
  deleted, the correlations for the two groups were
  .62 and .19.
• Might try testing for difference in significance
  between the two correlations, even though this
  test generally has less power.