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1 Exercise 10.1 (a) Plot wages versus LOS. Describe the relationship. There is one woman with relatively high wages for her length of service. Circle this point and do not use it in the rest of this exercise. (b) Find the least-squares line. Summarize the significance test for the slope. What do you conclude? The least-squares line is ( ) The standard error for the estimate of the slope is The corresponding t- and p-values are 2.85 and , respectively. With a p-value so small, we conclude that there is significant evidence against the mull hypothesis that the slope is zero. (c) State carefully what the slope tells you about the relationship between wages and length of service. Page 1 of 32

2 The slope tells us that for every unit increase in LOS, the average wage will increase by units. (d) Give a 95% confidence interval for the slope. A 95% confidence interval for the slope is [ , ]. > confint( m1) 2.5 % 97.5 % (Intercept) LOS Exercise 10.2 Refer to the previous exercise. Analyze the data with the outlier included. How does this change the estimates of the parameters,, and? What effect does the outlier have on the results of the significance test for the slope? The estimates of the parameters from the previous exercise (without the outlier) were: With the outlier included, the estimates become: The outlier has the effect of lowering the significance of the test for the slope increasing the p-value from to Exercise 10.5 In Example 10.8 we examined the yield in bushels per acre of corn for the years 1966, 1976, 1986, and Data for all years between 1957 and 1996 appear in Table 10.2 (a) Plot the yield versus year. Describe the relationship. Are there any outliers or unusual years? Page 2 of 32

3 Yield Exercise Year Comment: The relationship looks roughly linear. There don t appear to be any outliers. As for unusual years, there are a few years in the 70s and 80s that appear to have a lower yield than what might be expected. (b) Perform the regression analysis and summarize the results. How rapidly has yield increased over time? The least-squares line is ( ) Page 3 of 32

4 The significance tests for intercept and slope are both highly significant the p-value for the test of the intercept being equal to zero was p = 2.77e-15 and for the test of the slope being equal to zero was p = 1.26e-15. The average yield has increases by approximately 1.84 bushels per acre per year. > m5= lm( Yield ~ Year) > summary( m5) Call: lm(formula = Yield ~ Year) Residuals: Min 1Q Median Q Max Coefficients: Estimate Std. Error t value Pr(> t ) (Intercept) e-15 *** Year e-15 *** --Signif. codes: 0 *** ** 0.01 * Residual standard error: on 38 degrees of freedom Multiple R-squared: , Adjusted R-squared: F-statistic: on 1 and 38 DF, p-value: 1.257e-15 Exercise 10.6 (a) Find the equation of the least-squares line. The equation is: ( ) > m6= lm( Y ~ X) > summary( m6) Call: lm(formula = Y ~ X) Residuals: Min 1Q Median Q Max Coefficients: Estimate Std. Error t value Pr(> t ) (Intercept) ** X e-07 *** --Signif. codes: 0 *** ** 0.01 * Page 4 of 32

5 Residual standard error: on 7 degrees of freedom Multiple R-squared: , Adjusted R-squared: F-statistic: 312 on 1 and 7 DF, p-value: 4.596e-07 (b) Test the null hypothesis that the slope is zero and describe your conclusion. The p-value for this test is p = 4.6e-07, so we reject at the 0.05 level. The conclusion is that it is extremely unlikely that the slope is zero. (c) Give a 95% confidence interval for the slope. A 95% confidence interval for the slope is [ ] > confint( m6) 2.5 % 97.5 % (Intercept) X (d) The parameter corresponds to natural gas consumption for cooking, hot water, and other uses when there is no demand for heating. Give a 95% confidence interval for this parameter. A 95% confidence interval for the intercept is [ ] Exercise 10.8 (a) Plot the data. Does the trend in lean over time appear to be linear? Comment: Yes, the trend appears to be linear. Plot: Lean Exercise Year (b) What is the equation of the least-squares line? What percentage of the variation in lean is explained by this line? Page 5 of 32

6 The equation of the least-squares line is ( ) From the R-squared value, approximately 98.7% of the variation is explained by this line. > m8= lm( Lean ~ Year) > summary( m8) Call: lm(formula = Lean ~ Year) Residuals: Min 1Q Median Q Max Coefficients: Estimate Std. Error t value Pr(> t ) (Intercept) * Year e-12 *** --Signif. codes: 0 *** ** 0.01 * Residual standard error: on 11 degrees of freedom Multiple R-squared: 0.988, Adjusted R-squared: F-statistic: on 1 and 11 DF, p-value: 6.503e-12 (c) Give a 95% confidence interval for the average rate of change (tenths of a millimeter per year) of the lean. A 95% confidence interval for the average rate of change (tenths of a millimeter per year) of the lean is [ ] Exercise 10.9 (a) In 1918 the lean was (The coded value is 71.) Using the least-squares equation for the years 1975 to 1987, calculate a predicted value for the lean in The predicted value for the lean in 1918 is tenths of a millimeter. > predict( m8, newdata=data.frame( Year=c( 18)), interval="prediction", level=0.95) fit lwr upr (b) Although the least-squares line gives an excellent fit to the data for 1975 to 1987, this pattern did not extend back to Write a short statement explaining why this conclusion follows from the information available. Use numerical and graphical summaries to support your explanation. The reason why the conclusion (that the linear pattern does not extend back to 1918) follows from the information available is because the predicted value for lean in 1918 ( Page 6 of 32

7 meters/coded value 106.6) does not match the value of the actual lean in 1918 ( meters/ coded value 71). As for the part of the question asking for numerical and graphical summaries, I m not sure which numerical/graphical summaries the author has in mind; also, I don t see how any numerical/graphical summary would explain why the pattern does not extend back to Exercise (a) The engineers working on the Leaning Tower of Pisa are most interested in how much the tower will lean if no corrective action is taken. Use the least-squares equation to predict the tower s lean in the year The predicted value for the lean in 1997 is tenths of a millimeter. > predict( m8, newdata=data.frame( Year=c( 97)), interval="prediction", level=0.95) fit lwr upr (b) To give a margin of error for the lean in 1997, would you use a confidence interval for a mean response or a prediction interval? Explain your choice. We would use a prediction interval because this was a prediction. Exercise Exercise 10.6 gives information about the regression of natural gas consumptionon degree-days for a particular household. (a) What is the t statistic for testing? The t statistic was (b) For the alternative, what critical value would you use for a test at the significance level? Do you reject at this level? We would use the critical value. Yes, we would reject at this level. (c) How would you report the P-value for this test? We would report p= e-07. > pt( q=17.663, df=7, lower.tail=f) [1] e-07 Exercise (a) Find and ( We have: ) from the data. and from the equation: Page 7 of 32

8 ( ) we have: ( ) ( ) Thus ) ( ( ) > mean( Spheres) [1] > m15= lm( Vein ~ Spheres) > summary( m15) Call: lm(formula = Vein ~ Spheres) Residuals: Min 1Q Median Q Max Coefficients: Estimate Std. Error t value Pr(> t ) (Intercept) Spheres e-06 *** --Signif. codes: 0 *** ** 0.01 * Residual standard error: on 8 degrees of freedom Multiple R-squared: , Adjusted R-squared: F-statistic: on 1 and 8 DF, p-value: 4.733e-06 (b) We expect x and y to be positively associated. State hypotheses in terms of the slope of the population regression line that expresses this expectation, and carry out a significance test. What conclusions do you draw? The hypotheses are Page 8 of 32

9 The p-value of the significance test, carried out on a t-value of from a distributions is 2.365e-06 (get this either by dividing the p-value from the two-sided test from the output from part (a) above (i.e. p= 4.73e-06) by 2 or by using the pt() function in R). The conclusion is that the slope is positive. Either > 4.73e-06/2 [1] 2.365e-06 or > pt( q=10.810, df=8, lower.tail=f) [1] e-06 (c) Find a 99% confidence interval for the slope. A 99% confidence interval is [ ] > confint( m15, level=.99) 0.5 % 99.5 % (Intercept) Spheres (d) Suppose that we observe a value of Spheres equal to 15.0 for one dog. Give a 90% interval for predicting the variable Vein for that dog. A 90% prediction interval for Vein corresponding to a value of 15.0 for Spheres is [ ] > predict( m15, newdata=data.frame( Spheres=c( 15.0)), interval="prediction", level=0.90) fit lwr upr Exercise (a) Plot the data. Are there any outliers or unusual points? There are not outliers or unusual points. Plot: Page 9 of 32

10 I 2.0 Exercise V (b) Find the least-squares fit to the data, and estimate 1/R for this wire. Then give a 95% confidence interval for 1/R. The least-squares fit is ( ) the estimate for 1/R is and a 95% confidence interval for 1/R is [ > m17= lm( I ~ V) > summary( m17) Call: lm(formula = I ~ V) Page 10 of 32 ]

11 Residuals: Coefficients: Estimate Std. Error t value (Intercept) V Signif. codes: 0 *** ** Pr(> t ) *** * Residual standard error: on 3 degrees of freedom Multiple R-squared: , Adjusted R-squared: F-statistic: on 1 and 3 DF, p-value: > confint( m17) 2.5 % 97.5 % (Intercept) V (c) If estimates 1/R, then 1/ estimates R. Estimate the resistance R. Similarly, if L and U represent the lower and upper confidence limits for 1/R, then the corresponding limits for R are given by 1/U and 1/L, as long as L and U are positive. Use this face and your answer to (b) to find a 95% confidence interval for R. A 95% confidence interval for R is [ ] [ ] (d) Ohm s law states that in the model is 0. Calculate the test statistic for this hypothesis and give an approximate P-value. The test statistic for this hypothesis (see Code and Output for part (b)) is: and the corresponding p-value from a two-sided test is approximately: Exercise (a) Plot the data. Are there any outliers or unusual points? There don t appear to be any outliers. There are a few different VO2 values corresponding to the same HR value, which might be unusual. Plot: Page 11 of 32

12 V Exercise HR (b) Compute the least-squares regression line for predicting oxygen uptake from hear rate for this individual. The least-squares regression line is ( ) > m19= lm( V02 ~ HR) > m19 Call: lm(formula = V02 ~ HR) Coefficients: (Intercept) HR (c) Test the null hypothesis that the slope of the regression line is 0. Explain in words the meaning of your conclusions from this test. Page 12 of 32

13 The p-value for a two-sided test of the slope being 0 is 1.00e-11. The meaning is that there is a statistically significant linear relationship between the two variables. > summary( m19) Call: lm(formula = V02 ~ HR) Residuals: Min 1Q Median Q Coefficients: Estimate Std. Error t value (Intercept) HR Signif. codes: 0 *** ** 0.01 Max Pr(> t ) 4.59e-09 *** 1.00e-11 *** * Residual standard error: on 17 degrees of freedom Multiple R-squared: , Adjusted R-squared: F-statistic: on 1 and 17 DF, p-value: 1.000e-11 (d) Calculate a 95% confidence interval for the oxygen uptake of this individual on a future occasion when his heart rate is 95. Repeat the calculation for heart rate % confidence intervals for the oxygen uptake of this individual when his heart rate is 95 and 110 are [ ] and [ ] respectively. > predict( m19, newdata=data.frame( HR=c( 95, 110)), interval="confidence", level=0.95) fit lwr upr (e) From what you have learned in (a), (b), (c), and (d) of this exercise, do you think that the researchers should use predicted VO2 in place of measured VO2 for this individual under similar experimental conditions? Explain your answer. Yes, researchers may use predicted VO2 in place of measured VO2 for this individual. I don t know how to explain it using parts (a), (b), (c), and (d), but I would say, using the Adjusted R-squared value of that a straight-line model does a good job of explaining the relationship between VO2 and heart rate. Exercise Calculate the t statistic for testing Specify an appropriate alternative hypothesis for this Page 13 of 32

14 problem and give an approximate p-value for the test. Then explain your conclusion in words a physician can understand. The t statistic is, an appropriate alternative hypothesis is and an approximate p-value is The conclusion is that there is a statistically significant straight line relationship between the traditional procedure and the new procedure; and every unit increase in the new method results in 0.83 units increase in the old method? > 2* pt( 0.83/0.065, df=81-2, lower.tail=f) [1] e-21 Exercise (a) It is reasonable to suppose that greater airflow will cause more evaporation. State hypotheses to test this belief and calculate the test statistic. Find an approximate P-value for the significance test and report your conclusion. The hypotheses are The test statistic is and an approximate P-value is The conclusion is that greater airflow will cause more evaporation. (b) Construct a 95% confidence interval for the additional evaporation experience when airflow increases by 1 unit. A 95% confidence interval or the slope is ( )( ) [ ] Exercise Return to the data on current versus voltage given in the Ohm s law experiment in Exercise (a) Compute all values for the ANOVA table. > anova( m17) Analysis of Variance Table Response: I Df Sum Sq Mean Sq F value Page 14 of 32 Pr(>F)

15 V Residuals *** (b) State the null hypothesis tested by the ANOVA F statistic, and explain in plain language what this hypothesis says. The null hypothesis test by the ANOVA F statistic is (Moore and McCabe 2006, p 655): In plain language, this hypothesis says that y is not linearly related to x. (c) What is the distribution of this F statistic when is true? Find an approximate P-value for the test of The distribution of this F statistic when is true is an ( ) distribution. An approximate P-value is. Exercise (a) The correlation between monthly income and birth weight was r=0.39. Calculate the t statistic for testing the null hypothesis that the correlation is 0 in the entire population of infants. The t statistic for testing the null hypothesis that the correlation is 0 is given by (Moore and McCabe, p 664) (b) The researchers expected that higher birth weights would be associated with higher incomes. Express this expectation as an alternative hypothesis for the population correlation. The alternative hypothesis expressing this expectation is: (c) Determine a P-value for versus the alternative that you specified in (b). What conclusion does your test suggest? The P-value is This suggests that monthly income and birth weight are related specifically, it suggests that there is a positive correlation between the two. Code and output: > pt( , df=38, lower.tail=f) [1] Exercise (a) The correlation between parental control and self-esteem was r = Calculate the t statistic for testing the null hypothesis that the population correlation is 0. The t statistic is: Page 15 of 32

16 ( ) (b) Find an approximate P-value for testing versus the two-sided alternative and report your conclusion. An approximate p-value for testing the null hypothesis against the two-sided alternative is p = e-07. The conclusion is that the correlation coefficient is significant and that since the value of the correlation coefficient is only -0.19, we can conclude that there is no strong linear relationship? Exercise (a) Plot the data and describe the pattern. Is it reasonable to summarize this kind of relationship with a correlation? The pattern looks somewhat linear. It seems reasonable to summarize this kind of relationship with a correlation because the correlation between the variables is Humerus Exercise Femur Femur Humerus (b) Find the correlation and perform the significance test. Summarize the results and report your conclusion. The correlation is, the t statistic is, and the corresponding p-value for testing the null hypothesis against the two-sided alternative is Page 16 of 32

17 > cor.test( Humerus, Femur) Pearson's product-moment correlation data: Humerus and Femur t = , df = 3, p-value = alternative hypothesis: true correlation is not equal to 0 95 percent confidence interval: sample estimates: cor Exercise (a) Plot the data and describe the relationship between the two scores. There appears to be a somewhat linear relationship between the two scores Round Exercise Round1 (b) Find the correlation between the two scores and test the null hypothesis that the population correlation is 0. Summarize your results. The correlation is, the t statistic is, and the corresponding p- value for testing the null hypothesis against the two-sided alternative is summary, there is evidence of a linear relationship. Page 17 of 32 In

18 > cor.test( Round1, Round2) Pearson's product-moment correlation data: Round1 and Round2 t = , df = 10, p-value = alternative hypothesis: true correlation is not equal to 0 95 percent confidence interval: sample estimates: cor (c) The plot shows one outlier. Recompute the correlation and redo the significance test without this observation. Write a short summary explaining the effect of the outlier on the correlation and significance test in (b). The correlation becomes, the t statistic, and the corresponding p-value for testing the null hypothesis against the two-sided alternative In summary, the outlier in part (b) had the effect of reducing the correlation and increasing the p-value. > > > > detach( data) data=data[-8,] attach( data) cor.test( Round1, Round2) Pearson's product-moment correlation data: Round1 and Round2 t = , df = 9, p-value = alternative hypothesis: true correlation is not equal to 0 95 percent confidence interval: sample estimates: cor Exercise (a) Find the equation of the least-squares line for predicting GHP from FVC. The equation is ( ) where the slope and intercept were found using the equations (Moore and McCabe, p 157) Page 18 of 32

19 and (b) Give the results of the significance test for the null hypothesis that the slope is 0. (Hint: What is the relation between this test and the test for a zero correlation?) Testing for the null hypothesis that the slope is 0 is the same as testing for the null hypothesis that the correlation is zero? Recall that the t statistic for testing zero correlation is and hence the p-value for a test of zero correlation against the two-sided alternative is Exercise (a) Plot the data with SAT on the x axis and ACT on the y axis. Describe the overall pattern and any unusual observations. The overall relationship looks linear. There s a potential outlier (observation 42). Plot: Page 19 of 32

20 (b) Find the least-squares regression line and draw it on your plot. Give the results of the significance test for the slope. The least-squares line is ( ) The significance test for the slope yields a p-value of Thus we strongly reject the null hypothesis that the slope is zero. Code and Outputs: > m39=lm( ACT ~ SAT)#-> a= , b= > summary(m39) Call: lm(formula = ACT ~ SAT) Residuals: Min 1Q Median Q Max Page 20 of 32

21 Coefficients: Estimate Std. Error t value Pr(> t ) (Intercept) SAT e-15 *** --Signif. codes: 0 *** ** 0.01 * Residual standard error: on 58 degrees of freedom Multiple R-squared: 0.667, Adjusted R-squared: F-statistic: on 1 and 58 DF, p-value: 1.796e-15 (c) What is the correlation between the two tests? The correlation is. Exercise (a) What is the mean of these predicted values? Compare it with the mean of the ACT scores. The mean of predicted values is The mean of the ACT scores is also Code and Outputs: > ACT.predicted=predict( m39, newdata=data.frame( HR=SAT)) > mean( ACT.predicted) [1] (b) Compare the standard deviation of the predicted values with the standard deviation of the actual ACT scores. If least-squares regression is used to predict ACT scores for a large number of students such as these, the average predicted value will be accurate but the variability of the predicted scores will be too small. The standard deviation of predicted values is , while the standard deviation of actual ACT scores is > s.act.predicted=sd( ACT.predicted)#== > s.act=sd( ACT)#== (c) Find the SAT score for a student who is one standard deviation above the mean ( ( ) ). Find the predicted ACT score and standardize this score. (Use the means and standard deviations from this set of data for these calculations.) Student #6 scored a 1440 on the SAT, which is standard deviations above the mean. The predicted ACT score for this student is , which when standardized also becomes in other words it is standard deviations above the mean predicted ACT score. > predict( m39, newdata=data.frame( SAT=c(1440)))#== > ( mean( ACT.predicted))/s.ACT.predicted#== (d) Repeat part (c) for a student whose SAT score is one standard deviation below the mean (z = 1). Page 21 of 32

22 ACT ACT Student #7 scored a 490 on the SAT, which is standard deviations below the mean. The predicted ACT score for this student is , which is also standard deviations below the mean predicted ACT score. (e) What do you conclude from parts (c) and (d)? Perform additional calculation for different z s if needed. We conclude that when using this least-squares line to predict values, the prediction will be the same number of standard deviations above/below the mean of predicted values as the explanatory variable is above/below the mean of explanatory variables. Exercise (a) Using the data in Table 10.4, find the values of and. Using the formula We get the values and, which seems wrong. However, if we use the formula we get the values and (b) Plot the data with the least-squares line and the new prediction line. Exercise 10.41: Using a.1= s.x/s.y formula least-squares line new prediction line SAT Exercise 10.41: Using a.1= s.y/s.x formula least-squares line new prediction line SAT (c) Use the new line to find predicted ACT scores. Find the mean and the standard deviation of these scores. How do they compare with the mean and standard deviation of the ACT scores? Page 22 of 32

23 Using the formula instead of the formula to determine the slope of the least-squares equation yields new predicted ACT scores with a mean of and a standard deviation of Compared with the mean and standard deviation of the ACT scores, they are the same. Cod and Outpue: > ACT.predicted.new <- (function(satscore){ *SATscore})( SAT) > mean( ACT.predicted.new)#== [1] > mean( ACT)#== [1] > sd( ACT.predicted.new)#== [1] > sd( ACT)#== [1] Page 23 of 32

24 Exercise (a) Based on the output for model (3.7) a business analyst concluded: [ ] Provide a detailed critique of this conclusion. The discernable pattern indicates that an improper model has been fit. Also, the outlier (observation 13) in the plot of studentized residuals versus Distance warrants some concern. (b) Does the ordinary straight line regression model (3.7) seem to fit the data well? If not, carefully describe how the model can be improved. Page 24 of 32

25 Standardized Residuals The ordinary straight line regression does sort of fit the data well, but the model can be improved by removing the outlier. Even after the model is refit with this outlier removed, there is still a discernable pattern in the plot of studentized residuals versus Distance: Exercise 3.4.1: Standardized Residuals versus Distance Distance The pattern indicated that a quadratic term should be added. Addition of a quadratic term yields: Page 25 of 32

26 Standardized Residuals Exercise 3.4.1: Standardized Residuals versus Distance The problem of non-random residuals appears to have been fixed, however the improvement in adjusted R-squared when going from the Fare ~ Distance model to the Fare ~ Distance + DistanceSquared Distance model is small the change is only from 99.63% to 99.87%. It might be possible to improve the model further by looking for more outliers and then refitting the model with them removed. Page 26 of 32

27 Exercise Is the following statement true or false? If you believe that the statement is false, provide a brief explanation. Response: Suppose that a straight line regression model has been fit to bivariate data set of the form ( ) ( ) ( ) Furthermore, suppose that the distribution of X appears to be normal while the Y variable is highly skewed. A plot of standardized residuals from the least squares regression line produce a quadratic pattern with increasing variance when plotted against ( ). In this case, one should consider adding a quadratic term in X to the regression model and thus consider a model of the form. I agree. Regarding the plot of standardized residuals, Sheather writes that if a plot of residuals against X that produces a discernible pattern, then the shape of the pattern provides information on the function of x that is missing from the model (p 49). He goes on to write that if the residuals from the straight-line fit of Y and X have a quadratic pattern, then we can conclude that there is need for a quadratic term to be added to the original straight-line regression model (p 50). Regarding the issue of increasing variance, Sheather writes that there are two methods to deal with it transformations and weighted least squares. Exercise Part A (a) Develop a simple linear regression model based on least squares that predicts advertising revenue per page from circulation (i.e. feel free to transform either the predictor or the response variable or both variables). Ensure that you provide justification for you choice of model. A simple linear regression model is ( ) ( ) ( ) Justification: The plot of AdRevenue ~ Circulation has x values that are spread too far a log transformation on the Xs will help bring them closer together. The plot of AdRevenue ~ log( Circulation) has y values that are too care apart a log transformation on the Ys will, again, help bring them closer together. The final model ( ) ( ) looks good visiually. A plot of all three models is below: Page 27 of 32

28 AdRevenue AdRevenue 800 Exercise 3.4.3: AdRevenue ~ log( Circulation) 800 Exercise 3.4.3: AdRevenue ~ Circulation Circulation log(circulation) log(adrevenue) 6.5 Exercise 3.4.3: log( AdRevenue) ~ log( Circulation) log(circulation) (b) Find a 95% prediction interval for the advertising revenue per page for magazines with the following circulations: 1) 0.5 million A 95% prediction interval for advertising revenue is [ ]. 2) 20 million ]. A 95 prediction interval for advertising revenue is [ > logcirculation= log( Circulation) > logadrevenue= log( AdRevenue) > m343= lm( logadrevenue ~ logcirculation) Page 28 of 32 3

29 > logadrevenue.predicted= predict( m343, newdata=data.frame( logcirculation= log( c( 0.5, 20))), interval="prediction", level=.95) > AdRevenue.predicted= exp( logadrevenue.predicted) > AdRevenue.predicted fit lwr upr (c) Describe any weaknesses in your model. Interpretation of the least squares coefficients becomes difficult with the log transformation applied to both explanatory and response variables. Part B (a) Develop a polynomial regression model based on least-squares that directly predicts the effect on advertising revenue per page of an increase in circulation of 1 million people (i.e. do not transform either the predictor nor the response variable). Ensure that you provide detailed justification for your choice of model. [Hint: Consider polynomial model of order up to 3.] A polynomial regression model based on least-squares that directly predicts the effect on advertising revenue per page of an increase in circulation of 1 million people is: ( ( ) ( ) ) Detailed Justification: This is how we arrived at the above model: 1) we first fit three models: i. ii. iii. 2) we identified the leverage points 3) For each model, we identified which of the leverage points were bad (using the rule that identifies points outside the interval -2 to 2 as bad) the bad leverage points were 2, 20, and 49 for the first model; 2, 4, 20, 49 for the second; and 2, 8, 20, 49 for the third. 4) we removed the bad leverage points for each model and then refit each model to the new data set (the data set with the bad leverage points removed) 5) Finally, we compared the R squared vales for all three models. Model 3 resulted in the highest adjusted R- squared value, so that is why it was chosen. Code and Outputs: #IDENTIFY LEVERAGE POINTS: leverage.vals= lm.influence(m343b.2)$hat Page 29 of 32

30 leverage.vals[ leverage.vals > 4/ length( Circulation)]# these points are levereage points: 2, 4, 6, 8, 20, 46, 49. #CALCULATE STANDARDIZED RESIDUALS FOR ABOVE LEVERAGE PTS AND DETERMINE WHICH ONES ARE OUTLIERS/BAD: rstandard.vals1= rstandard( m343b.1)[ c( 2, 4, 6, 8, 20, 46, 49)] rstandard.vals1[ rstandard.vals1 > 2 rstandard.vals < -2]# these are bad leverage points: 2, 20, 49 rstandard.vals2= rstandard( m343b.2)[ c( 2, 4, 6, 8, 20, 46, 49)] rstandard.vals2[ rstandard.vals2 > 2 rstandard.vals < -2]# these are bad leverage points: 2, 4, 20, 49 rstandard.vals3= rstandard( m343b.3)[ c( 2, 4, 6, 8, 20, 46, 49)] rstandard.vals3[ rstandard.vals3 > 2 rstandard.vals < -2]# these are bad leverage points: 2, 8, 20, 49 #REMOVE BAD LEVERAGE POINTS (i.e., OUTLIERS) AND REFIT EACH MODEL: ad.new1= ad[ -c(2, 20, 49),] detach( ad) attach( ad.new1) m343b.1= lm( AdRevenue ~ Circulation) ad.new2= ad[ -c(2, 4, 20, 49),] detach( ad.new1) attach( ad.new2) m343b.2= lm( AdRevenue ~ Circulation + CirculationSquared) ad.new3= ad[ -c(2, 8, 20, 49),] detach( ad.new2) attach( ad.new3) m343b.3= lm( AdRevenue ~ Circulation + CirculationSquared + CirculationCubed) #COMPARE THE THREE MODELS AGAIN WHEN OUTLIER FOR EACH MODEL ARE REMOVED: > summary( m343b.1) Call: lm(formula = AdRevenue ~ Circulation) Residuals: Min 1Q Median 3Q Max Coefficients: Estimate Std. Error t value Pr(> t ) (Intercept) <2e-16 *** Circulation <2e-16 *** --- Signif. codes: 0 *** ** 0.01 * Residual standard error: on 65 degrees of freedom Multiple R-squared: , Adjusted R-squared: F-statistic: on 1 and 65 DF, p-value: < 2.2e-16 Page 30 of 32

31 > summary( m343b.2) Call: lm(formula = AdRevenue ~ Circulation + CirculationSquared) Residuals: Min 1Q Median Q Max Coefficients: Estimate Std. Error t value Pr(> t ) (Intercept) e-13 *** Circulation < 2e-16 *** CirculationSquared e-06 *** --Signif. codes: 0 *** ** 0.01 * Residual standard error: on 63 degrees of freedom Multiple R-squared: , Adjusted R-squared: F-statistic: on 2 and 63 DF, p-value: < 2.2e-16 > summary( m343b.3) Call: lm(formula = AdRevenue ~ Circulation + CirculationSquared + CirculationCubed) Residuals: Min 1Q Median Q Max Coefficients: Estimate Std. Error t value Pr(> t ) (Intercept) e-06 *** Circulation e-10 *** CirculationSquared ** CirculationCubed ** --Signif. codes: 0 *** ** 0.01 * Residual standard error: on 62 degrees of freedom Multiple R-squared: 0.933, Adjusted R-squared: F-statistic: on 3 and 62 DF, p-value: < 2.2e-16 (b) Find a 95% prediction interval for the advertising revenue per page for magazines with the following circulations: (i) 0.5 million A 95% prediction interval for the first model is [ ]. A 95% prediction interval for the second model is [ ]. A 95% prediction interval for the third model is [ ]. (ii) 20 million Page 31 of 32

32 Part C A 95% prediction interval for the first model is [ ]. A 95% prediction interval for the second model is [ ]. A 95% prediction interval for the third model is [ ]. (c) Describe any weaknesses in your model. The weakness in our model (i.e. model 3) is how much greater the lengths of its predication intervals are compared to the other two models. (a) Compare the model in Part A with that in Part B. Decide which provides a better model. Give reasons to justify your choice. Not sure. Part A? Because the models in Part B give such different prediction intervals? (b) Compare the prediction intervals in Part A with those in Part B. In each case, decide which interval you would recommend. Give reasons to justify each choice. Not sure the intervals are all so different and it isn t clear why any particular one would be better than the others. Page 32 of 32

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