BIOSTATS 640 Spring 2018 Unit 2. Regression and Correlation (Part 1 of 2) STATA Users

Transcription

1 Unit Regression and Correlation 1 of - Practice Problems Solutions Stata Users 1. In this exercise, you will gain some practice doing a simple linear regression using a Stata data set called week0.dta. This data set has n=31 observations of boiling points (Y=boiling) and temperature (X=temp). Tip In this exercise you will need to create a new variable newy = 100*log 10 (boiling). generate newy=100*log10(boiling) Carry out an exploratory analysis to determine whether the relationship between temperature and boiling point is better represented using (i) (ii) Y = β 0+βX or 1 ' ' 100 log (Y) = β +βx In developing your answer, try our hand at producing (a) Estimates of the regression line parameters (b) Analysis of variance tables (c) R (d) Scatter plot with overlay of fitted line. Complete your answer with a one paragraph text that is an interpretation of your work. Take your time with this and have fun. sol_regression(1 of ) STATA.docx Page 1 of 18

2 Tip! A really good practice is to maintain a log of your session. Here is how. Step 1: From the main menu at upper left, click: FILE > LOG > BEGIN sol_regression(1 of ) STATA.docx Page of 18

3 Step : At the dialog box SAVE AS: Type in a file name of your choosing. Mine is carol_log Note that I have NOT typed an extension. Stata does this for you. Do NOT click on SAVE just yet. Step 3: At the dialog box WHERE: Click on the drop down menu and select a path of your choosing. Again, do NOT click on SAVE just yet Step 4: At the dialog box FILE FORMAT: click on drop down menu at right. Step 5: From the drop down menu, choose STATA LOG. Now you can click on SAVE sol_regression(1 of ) STATA.docx Page 3 of 18

4 My Stata Session Note I imported my log into an MS Word document and, there, edited all my misspellings and errors Key BLACK commands. Comments begin with asterisk and are in bold. Blue output name: <unnamed> log: /Users/carolbigelow/Desktop/hw_week0.log log type: text **** Turn off screen pause of output. set more off **** Read data that you have already downloaded into stata workspace **** At top left, FILE > OPEN **** Preliminary) Use command codebook with option compact for a quick look at data. codebook, compact Variable Obs Unique Mean Min Max Label temp boiling **** Preliminary) Use command generate to create newy = 100* log10(boiling). generate newy=100*log10(boiling) **** Simple Linear Regression * a-c) Use command regress yvariable xvariable for: fitted line, anova, and R-squared. regress boiling temp Source SS df MS Number of obs = F( 1, 9) = Model Prob > F = Residual R-squared = Adj R-squared = Total Root MSE = boiling Coef. Std. Err. t P> t [95% Conf. Interval] temp _cons sol_regression(1 of ) STATA.docx Page 4 of 18

5 * d) Use command graph twoway for scatter plot with overlay of fitted line.** No frills graph. graph twoway (scatter boiling temp) (lfit boiling temp) * Graph with aesthetics.. graph twoway (scatter boiling temp, symbol(d)) (lfit boiling temp), title("simple Linear Regression") subtitle("y=boiling on X=temp") caption("boiling.png", size(vsmall)) Tip I used the option caption( )so that the name of my saved file (which I am about to save) will appear on the graph itself. Handy! sol_regression(1 of ) STATA.docx Page 5 of 18

6 Tip! Save your graph as a.png image so that you can paste it elsewhere. Here is how. Step 1: Create your graph (see previous page ) Step : With the graph window still active, click on the SAVE icon. Step 3: From drop down menu for File format: Choose PORTABLE NETWORK GRAPHICS (*.png) sol_regression(1 of ) STATA.docx Page 6 of 18

7 **** Simple Linear Regression of Y=100*log10(boiling) on X=temp * a-c) Use command regress yvariable xvariable for: fitted line, anova, and R-squared. regress newy temp Source SS df MS Number of obs = F( 1, 9) = 3.69 Model Prob > F = Residual R-squared = Adj R-squared = Total Root MSE =.9618 newy Coef. Std. Err. t P> t [95% Conf. Interval] temp _cons * d) Scatter plot with overlay of fitted line using the command graph twoway * No frills graph. graph twoway (scatter newy temp) (lfit newy temp) sol_regression(1 of ) STATA.docx Page 7 of 18

8 * Graph with aesthetics. graph twoway (scatter newy temp, symbol(d)) (lfit newy temp), title("simple Linear Regression") subtitle("y=100*log10(boiling) on X=temp") caption("newy.png", size(vsmall)) sol_regression(1 of ) STATA.docx Page 8 of 18

9 Putting it all together (a) Estimates of Regression Line Parameters i. Y= *X ii. 100log 10 (Y)= *X Table 1. Parameters estimations for dependent = y = boiling (Boiling Point) Parameter Estimates Parameter Standard Variable Label DF Estimate Error t Value Pr > t Intercept Intercept <.0001 Temp Temperature <.0001 Table. Parameters estimations for dependent = y = newy (100 log 10 (Boiling Point)) Parameter Estimates Parameter Standard Variable Label DF Estimate Error t Value Pr > t Intercept Intercept Temp Temperature <.0001 (b) Analysis of Variance Tables Y=boiling (Boiling Point) Analysis of Variance Sum of Mean Source DF Squares Square F Value Pr > F Model <.0001 Error Corrected Total Y = newy (100 log 10 (Boiling Point)) Analysis of Variance Sum of Mean Source DF Squares Square F Value Pr > F Model <.0001 Error Corrected Total sol_regression(1 of ) STATA.docx Page 9 of 18

10 (c) R Y=boiling (Boiling Point): R = Y = newy (100 log 10 (Boiling Point)) R = (d) Scatterplot with Overlay of Fitted Line Y=boiling (Boiling Point) Y = newy (100 log 10 (Boiling Point)) Solution (one paragraph of text that is interpretation of analysis): Did you notice that the scatter plot of these data reveal two outlying values? Their inclusion may or may not be appropriate. If all n=31 data points are included in the analysis, then the model that explains more of the variability in boiling point is Y=boiling point modeled linearly in X=temperature. It has a greater R (9% v 89%). Be careful - It would not make sense to compare the residual mean squares of the two models because the scales of measurement involved are different. sol_regression(1 of ) STATA.docx Page 10 of 18

11 . A psychiatrist wants to know whether the level of pathology (Y) in psychotic patients 6 months after treatment could be predicted with reasonable accuracy from knowledge of pretreatment symptom ratings of thinking disturbance (X 1 ) and hostile suspiciousness (X ). (a) The least squares estimation equation involving both independent variables is given by Y = (X 1 ) 7.147(X ) Using this equation, determine the predicted level of pathology (Y) for a patient with pretreatment scores of.80 on thinking disturbance and 7.0 on hostile suspiciousness. How does the predicted value obtained compare with the actual value of 5 observed for this patient? Ŷ = X X with X =.80 and 1 X =7.0 Ŷ = This value is lower than the observed value of 5 (b) Using the analysis of variance tables below, carry out the overall regression F tests for models containing both X 1 and X, X 1 alone, and X alone. Source DF Sum of Squares Regression on X Residual Source DF Sum of Squares Regression on X Residual sol_regression(1 of ) STATA.docx Page 11 of 18

12 Source DF Sum of Squares Regression on X 1, X 784 Residual Model Containing X 1 and X F = SSQ regression on X and X /Regression df 1 SSQ residual / Residual df =,784 / 11,008 / 50 = 6.37 on DF=,50 p-value= à Application of the null hypothesis model has led to an extremely unlikely result (p-value =.00356), prompting statistical rejection of the null hypothesis. The fitted linear model in X 1 and X explains statistically significantly more of the variability in level of pathology (Y) than is explained by Y (the intercept model) alone. Model Containing X 1 ALONE F = SSQ Regression on X /DF Regression 1 SSQ residual/df Residual = 1546 / 1 1,46 / 51 = on DF=1,51 p-value= Here, too, application of the null hypothesis model has led to an extremely unlikely result (p-value =.014), prompting statistical rejection of the null hypothesis. The fitted linear model in X 1 explains statistically significantly more of the variability in level of pathology (Y) than is explained by Y (the intercept model) alone. Model Containing X ALONE F = SSQ Regresion on X /DF Regression SSQ Residual/DF Residual = 160 / 1 13,63 / 51 = on DF=1,51 p-value= Here, application of the null hypothesis model has not led to an extremely unlikely result (p-value =.44). The null hypothesis is therefore not rejected. The fitted linear model in X does not explain statistically significantly more of the variability in level of pathology (Y) than is explained by Y (the intercept model) alone. (c) Based on your results in part (b), how would you rate the importance of the two variables in predicting Y? X explains a significant proportion of the variability in Y when modelled as a linear predictor. 1 X does not. (However, we don't know if a different functional form might have been important.) sol_regression(1 of ) STATA.docx Page 1 of 18

13 (d) What are the R values for the three regressions referred to in part (b)? Total SSQ= (Regression SSQ) + (Residual SSQ) is constant. Therefore total SSQ can be calculated from just one anova table: Total (SSQ)= 1, ,46 = 13,79 ( ) R X1 only = (Re gression SSQ)/(Total SSQ) R (X only) = (160)/(13,79) = ( ) ( ) = (1546)/(13,79) = R X1and X = 784 /(13, 79) = (e) What is the best model involving either one or both of the two independent variables? Eliminate from consideration model with X only. Compare model with X alone versus 1 X and 1 X using partial F test. Partial F = {(SSQ Regression on X,X ) - (SSQ Regression on X )}/VDF 1 1 = SSQ Residual for model w X 1,X /Residual DF = on DF =1,50 P-value =0.016 Addition of X to model containing X 1 is statistically significant (p-value =.0). More appropriate model includes X 1 and X ( ) / 1 (11,008) / 50 sol_regression(1 of ) STATA.docx Page 13 of 18

14 3. In an experiment to describe the toxic action of a certain chemical on silkworm larvae, the relationship of log 10 (dose) and log 10 (larva weight) to log 10 (survival) was sought. The data, obtained by feeding each larva a precisely measured dose of the chemical in an aqueous solution and then recording the survival time (ie time until death) are given in the table. Also given are relevant computer results and the analysis of variance table. Larva Y = log 10 (survival time) X 1 =log 10 (dose) X =log 10 (weight) Larva Y = log 10 (survival time) X 1 =log 10 (dose) X =log 10 (weight) Y = (X 1 ) Y = (X ) Y = (X 1 ) + ).871 (X ) Source DF Sum of Squares Regression on X Residual Source DF Sum of Squares Regression on X Residual Source DF Sum of Squares Regression on X 1, X Residual sol_regression(1 of ) STATA.docx Page 14 of 18

15 (a) Test for the significance of the overall regression involving both independent variables X 1 and X. X 1 and X F = (SSQ regression on X and X ) / 1 = (0.464) / = on DF =,1 (SSQ Residual) / 1 (0.0471) / 1 P value < Application of the null hypothesis model has led to an extremely unlikely result (p-value =.0001), prompting statistical rejection of the null hypothesis. The fitted linear model in X 1 and X explains statistically significantly more of the variability in log 10 (survival time) (Y) than is explained by Y (the intercept model) alone. (b) Test to see whether using X 1 alone significantly helps in predicting survival time. X 1 alone F = (0.3633) / 1 (0.1480) / 13 = (SSQ Regression on X ) / 1 1 = on DF = 1,13 (SSQ Residual) / 13 P value = Application of the null hypothesis model has led to an extremely unlikely result (p-value =.00008), prompting statistical rejection of the null hypothesis. The fitted linear model in X 1 explains statistically significantly more of the variability in log 10 (survival time) (Y) than is explained by Y (the intercept model) alone. (c) Test to see whether using X alone significantly helps in predicting survival time. X alone F = (SSQ Regression on X ) / 1 (SSQ Residual) / 13 P value = = (0.3367) / 1 = 5.07 on DF = 1,13 (0.1746) / 13 Application of the null hypothesis model has led to an extremely unlikely result (p-value =.0007), prompting statistical rejection of the null hypothesis. The fitted linear model in X explains statistically significantly more of the variability in log 10 (survival time) (Y) than is explained by Y (the intercept model) alone. sol_regression(1 of ) STATA.docx Page 15 of 18

16 (d) Compute R for each of the three models. TotalSSQ = ( ) R X1 and X = / = R (X1 alone) = / = ( ) R X alone = / = (e) Which independent predictor do you consider to be the best single predictor of survival time? Using just the criteria of the overall F test and comparison of containing X is better. 1 R, the single predictor model (f) Which model involving one or both of the independent predictors do you prefer and why? Partial F for comparing model with X1alone versus model with X 1 and X ( ΔRegressionSSQ)/ΔReg DF ( = = )/ -1 ( ResidualSSQ-model w X 1, X )/ResidualDF-model w X 1, X /1 = on DF=1,1 P-value= Choose model with both X 1 and X ( ) sol_regression(1 of ) STATA.docx Page 16 of 18

17 4. Using Stata, try your hand at reproducing the analysis of variance tables you worked with in problem #3. The following assumes you: a) downloaded larvae.dta; and b) opened it Stata commands describe and notes to get information on data set Tip You need to issue the command clear before you load larvae.dta.describe Contains data from /Users/carolbigelow/Desktop/larvae.dta obs: 15 PubHlth 640 Unit Regression - Larvae data vars: 4 10 Feb :36 size: 40 (_dta has notes) storage display value variable name type format label variable label id float %9.0g larva id y float %9.0g log10(survival) x1 float %9.0g log10(dose) x float %9.0g log10(weight) Sorted by:. notes _dta: 1. "Week 3 homework assignment exercises and 3" **** ) Fit model containing x1 alone using command regress yvariable xvariable. regress y x1 Source SS df MS Number of obs = F( 1, 13) = Model Prob > F = Residual R-squared = Adj R-squared = Total Root MSE =.1067 y Coef. Std. Err. t P> t [95% Conf. Interval] x _cons sol_regression(1 of ) STATA.docx Page 17 of 18

18 **** ) Model containing x alone. regress y x Source SS df MS Number of obs = F( 1, 13) = 5.07 Model Prob > F = Residual R-squared = Adj R-squared = 0.63 Total Root MSE = y Coef. Std. Err. t P> t [95% Conf. Interval] x _cons **** ) Model containing both x1 and x. regress y x1 x Source SS df MS Number of obs = F(, 1) = Model Prob > F = Residual R-squared = Adj R-squared = Total Root MSE =.0663 y Coef. Std. Err. t P> t [95% Conf. Interval] x x _cons sol_regression(1 of ) STATA.docx Page 18 of 18