Question

Not all bivariate relationships are linear. When you plot a scatterplot, sometimes you observe a curved relationship. In those cases, we can apply a transformation to the data that will make the relationship approximately linear. [We generally prefer simpler models. And linear regression models are simpler than curved regression models. Also, transforming data is common in statistical practice.] One of the most common transformation methods is the log transformation.

In this problem, you apply the log transformation to both variables in the data for the previous problem to experiment the model fitness. Create another variable 'lnCost' by taking the natural log of cost. Do the same for the length of stay (los), and create another variable 'lnLos'. Then obtain the least squares regression line for 'lnCost' as a function of 'lnLos', and graph the line on the scatterplot.

Does a linear regression line seem to fit these transformed variables? Better than the one without transformation? Explain.

Create the residual plot of this plot in the same way as the previous problem. How are the residuals different from the previous problem?

Create the normal plot of the residuals. How is this plot different from the one from the previous problem?

In your opinion, is the linear regression model made from the transformed variables a better model? Explain in detail.

cost	los
13728	13
8062	8
4805	13
5099	6
14963	33
4295	2
4046	9
3193	13
15486	16
9413	11
9034	19
8939	20
17596	26
1884	3
1763	5
1233	1
6286	30
2849	4
2818	4
2265	2
1652	9
1846	4
25460	18
4570	16
12213	10
5870	12
24484	52
4735	19
13334	9
35381	85
5681	8
7161	20
10592	41

Answer 1

Worksheet 1 + C1 cost 13728 C5 C6 C7 C8 C9C10C11 C2 los 13 8062 C3 Incost 9.5272 8.9949 8.4774 8.5368 9.6133 8.3652 C4 Inlost 0.00000 2.07944 2.56495 1.79176 3.49651 0.69315 4805 5099 14963 4295 13 6 33 2 Rex ED Pro... 0 X Rex Current Worksheet: Worksheet 1 Editable 11:05 o ) ENG 16/02/2020 F "* + 1 NOI FI 127 prt sc delete home end po up

This image contain original and log transformed observations.

D File Edit Data Calc Stat Graph Editor Tools WindowHelp Assistant A 32 RUYENXQ TOO. UM Session Regression Analysis: los versus cost Analysis of Variance Source Regression cost Error Total DE Adj ss Adj MS F-Value 5826 5826.4 54.61 5826 5826.4 54.61 3307 106.7 32 9134 P-Value 0.000 0.000 31 Model Summary S R -sg 10.3292 63.79% R-sq(adj) 62.62% R-sq (pred) 48.48% Coefficients Term VIF Coef 1.30 0.001713 SE Coef T-Value P-Value 2.72 0.48 0.637 0.000232 7 .39 0.000 cost 1.00 Regression Equation Current Worksheet: Worksheet 1

Minitab - Untitled HOOGT10@ Eile Edit Data Calc Stat Graph Editor Tools Window Help Assistant xa TOOLIM Residual Plots for los Residual Plots for los Normal Probability Plot Versus Fits Percent Residual -15 15 30 Residual 1530 Fitted Value Histogram Versus Order Frequency Residual 12 Residual 20 Observation Order 6 4295 8.3652 0.69315 La Current Worksheet: Worksheet 1
The above two images contain linear reg model of lost  and cost

File Edit Data at at -- A OTDOOM Session Regression Analysis: Inlost versus Incost Analysis of Variance Source Regression Incost Error Total DF Adj SS Adj MS 1 18.73 18.7256 18.73 18.7256 12.10 0.3903 30.82 F-Value P-value 4 7.98 0.000 47.980.000 Model Summary S R-sq 0.624722 60.753 R-sg(adj) 59.488 R-sq (pred) 55.203 Coefficients Term Constant Incost Coef SE Coef T-Value -5.30 1.11 -4.76 0.879 0.127 6.93 VIF P-Value 0.000 0.000 1.00 Regression Equation Inlost = -5.30 + 0.979 Incost Current Worksheet: Worksheet 1

D/+P II TO LIM Session F1 Residual Plots for Inlost Residual Plots for Inlost Normal Probability Plot Versus Fits Percent Residual Residual Fitted Value Histogram Versus Order C10 C11 Frequency Residual Connect line 10 05 -05 0. 0 Residual 1 5 10 15 20 Observation Order 4046 9 8.3055 nncny 2.19722 ECAME Current Worksheet: Worksheet 1 A 19 ENG 10:59 16/02/20208
This above two images containa reg model of transformed variables.

In this both models the variation explained by independent variable(R square value) is almost same. Because the data is already normally distributed but the error in transformed model is very less as compared to original model. So we can say that transformed model is good.

The residual plot of original variable looks like little bit of outward funnel where as the residual plot of transformed variables is shows constant variance which satisfied the assumption of homoscedasticity.

Both the model data is normally distributed. Histogram of transformed observation little bit streched at middle.

Yes, transformed model is good because as compared to original model it gives very less error.

I have one note for this topic- Mostly transformation works good when the difference bet dependent and independent variable are sufficiently large otherwise it gives almost same results.

If it is right please like this

Thank You