[This article was first published on

Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.

**denis haine » R**, and kindly contributed to R-bloggers]. (You can report issue about the content on this page here)Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.

In the previous post, I forgot to show an example of Box-Cox transformation when there’s a lack of normality. The Box-Cox procedure computes values of which best “normalises” the errors.

value | Transformed value of Y |
---|---|

2 | |

1 | |

0.5 | |

0 | |

-0.5 | |

-1 | |

-2 |

For example:

lm.wpc2 <- lm(wpc ~ vag_disch + milk120 + milk120.sq, data = daisy3) library(MASS) boxcox(lm.wpc2, plotit = TRUE)

The plot indicates a log transformation.

** Matrix Representation **

We can use a matrix representation of the regression equation. The regression equation can be written as where , , and

.

With the following regression:

lm.wpc3 <- lm(wpc ~ milk120 + hs100_ct, data = daisy3) (lm.wpc3.sum <- summary(lm.wpc3)) Call: lm(formula = wpc ~ milk120 + hs100_ct, data = daisy3) Residuals: Interval from wait period to conception Min 1Q Median 3Q Max -71.83 -36.57 -15.90 23.86 211.18 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 75.884063 6.115894 12.408 <2e-16 *** milk120 -0.001948 0.001858 -1.049 0.294 hs100_ct 18.530340 2.106920 8.795 <2e-16 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 50.53 on 1522 degrees of freedom Multiple R-squared: 0.0495, Adjusted R-squared: 0.04825 F-statistic: 39.63 on 2 and 1522 DF, p-value: < 2.2e-16 anova(lm.wpc3) Analysis of Variance Table Response: wpc Df Sum Sq Mean Sq F value Pr(>F) milk120 1 4865 4865 1.9053 0.1677 hs100_ct 1 197512 197512 77.3518 <2e-16 *** Residuals 1522 3886309 2553 --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

We start by making the X matrix and response :

X <- cbind(1 , daisy3[, c(8, 24)]) # or alternatively: model.matrix(lm.wpc3) y <- daisy3$wpc

Then we get the :

XX <- t(X) %*% X # or alternatively: crossprod(X, X) XX 1 milk120 hs100_ct 1 1525.00000 4907834.4 -29.87473 milk120 4907834.39954 16535517546.0 -120696.45468 hs100_ct -29.87473 -120696.5 576.60900

We calculate the inverse of :

XX.inv <- solve(t(X) %*% X) # or alternatively: qr.solve(XX) # or also: lm.wpc3.sum$cov.unscaled XX.inv 1 milk120 hs100_ct 1 1.464864e-02 -4.348902e-06 -1.513557e-04 milk120 -4.348902e-06 1.351675e-09 5.761289e-08 hs100_ct -1.513557e-04 5.761289e-08 1.738495e-03

Now to get the coefficients estimates:

XX.inv %*% t(X) %*% y # or alternatively: solve(t(X) %*% X, t(X) %*% y) # or also: XX.inv %*% crossprod(X, y) # or: qr.solve(X, y) [,1] 1 75.884062540 milk120 -0.001948437 hs100_ct 18.530340385

The fitted values:

crossprod(t(X), qr.solve(X, y)

SSE and MSE;

SSE <- crossprod(y, y) - crossprod(y, yhat) SSE [,1] [1,] 3886309 MSE <- SSE / (length(y) - 3) MSE [,1] [1,] 2553.423

The residual standard error:

sqrt(sum(lm.wpc3$res^2) / (1525 - 3))

The standard errors of the coefficients and the :

sqrt(diag(XX.inv)) * 50.5314 1 milk120 hs100_ct 6.115893490 0.001857794 2.106920139 1 - sum(lm.wpc3$res^2) / sum((y - mean(y))^2) [1] 0.04949679

To

**leave a comment**for the author, please follow the link and comment on their blog:**denis haine » R**.R-bloggers.com offers

**daily e-mail updates**about R news and tutorials about learning R and many other topics. Click here if you're looking to post or find an R/data-science job.

Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.