1 / 11

Stat 6601 project Linear Statistical Models Analysis of Covariance Example

Stat 6601 project Linear Statistical Models Analysis of Covariance Example. By Gadir Marian Myrna Moreno. Data.

minh
Download Presentation

Stat 6601 project Linear Statistical Models Analysis of Covariance Example

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Stat 6601 project Linear Statistical ModelsAnalysis of Covariance Example By Gadir Marian Myrna Moreno

  2. Data • ‘Whiteside’ data, Mr. Derek recorded weekly gas consumption and average external temperature at his house during two ‘heating seasons’ one before and after cavity-wall insulation was installed. • Variables: - Insul (levels: before or after insulation) - Temp (the average outside temperature in degrees Celsius) -Gas (The weekly gas consumption in 1000 cubic feet units)

  3. Goal • Assess the effect of the insulation on gas consumption.

  4. Plotting the data

  5. Method • Linear Model for Analysis of Covariance Y=  +X +  Where:  is a random effect due to treatment.  is a fixed effect due to covariate.  is a random error.

  6. Method(continued) • Using R: -A primary model is fitted using a “model fitting function” lm (formula, data, weights, subset, na.action) - A resulting “fitted model object” can be analysed, interrogated or modified.

  7. Codes • require(latice) • xyplot(Gas ~ Temp | Insul, whiteside, panel = • function(x, y, ...) { • panel.xyplot(x, y, ...) • panel.lmline(x, y, ...) • }, xlab = "Average external temperature (deg. C)", • ylab = "Gas consumption (1000 cubic feet)", aspect = "xy", • strip = function(...) strip.default(..., style = 1)) • gasB <- lm(Gas ~ Temp, whiteside, subset = Insul=="Before") • gasA <- update(gasB, subset = Insul=="After") • summary(gasB) • summary(gasA) • gasBA <- lm(Gas ~ Insul/Temp - 1, whiteside) • summary(gasBA) • gasQ <- lm(Gas ~ Insul/(Temp + I(Temp^2)) - 1, whiteside) summary(gasQ)$coef • gasPR <- lm(Gas ~ Insul + Temp, whiteside) • anova(gasPR, gasBA) • options(contrasts = c("contr.treatment", "contr.poly")) • gasBA1 <- lm(Gas ~ Insul*Temp, whiteside) • summary(gasBA1)$coef

  8. Results • The output from fitting regression model: Residuals: Min 1Q Median 3Q Max -0.97802 -0.18011 0.03757 0.20930 0.63803 Coefficients: Estimate Std. Error t value Pr(>|t|) InsulBefore 6.85383 0.13596 50.41 <2e-16 *** InsulAfter 4.72385 0.11810 40.00 <2e-16 *** InsulBefore:Temp -0.39324 0.02249 -17.49 <2e-16 *** InsulAfter:Temp -0.27793 0.02292 -12.12 <2e-16 *** Residual standard error: 0.323 on 52 degrees of freedom

  9. Results(continued) The output by fitting quadratic regression model: Estimate Std. Error t value Pr(>|t|) InsulBefore 6.759215179 0.150786777 44.826312 4.854615e-42 InsulAfter 4.496373920 0.160667904 27.985514 3.302572e-32 InsulBefore:Temp -0.31765873 0.062965170 -5.044991 6.362323e-06 InsulAfter:Temp -0.137901603 0.073058019 -1.887563 6.489554e-02 InsulBefore:I(Temp^2) -0.008472572 0.006624737 -1.278930 2.068259e-01 InsulAfter:I(Temp^2) -0.014979455 0.007447107 -2.011446 4.968398e-02

  10. Results(continued) The output from the ANOVA Estimate Std. Error t value Pr(>|t|) (Intercept) 6.8538277 0.13596397 50.409146 7.997414e-46 InsulAfter -2.1299780 0.18009172 -11.827185 2.315921e-16 Temp -0.3932388 0.02248703 -17.487358 1.976009e-23 InsulAfter:Temp 0.1153039 0.03211212 3.590665 7.306852e-04

  11. Summary • Whiteside data • Fitting Linear Regression Model • Fitting Quadratic Regression Model

More Related