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Directions: Complete the following exercises using the code discussed during computer lab. Save your work in an R script as well as a Word document containing the necessary output and comments. Be sure to use notes in the script to justify any computations. If you have any questions, do not hesitate to ask.
1 Simple Linear Regression
- Load the data set pressure from the datasets package in R. Perform a Simple Linear Regression on the two variables. Provide the regression equation, coefficients table, and anova table. Summarize your findings. What is the relationship between the t statistic for temperature and the F statistic in the ANOVA table?
- Refer to the previous exercise. Check the assumptions on the regression model and report your results. Be sure to include the scatterplot with regression equation, normal QQ plot, and residual plot. Explain what you see.
- Refer to exercise 1. Experiment with different transformations of the data to improve the model. What is the best transformation?
2 Multiple Linear Regression
- Load the swiss data set from the ‘datasets’ package in R. Find the correlation matrix and print the pairwise scatterplots. What variables seem to be related?
- Run a Multiple Regression on Fertility using all of the other variables as predictors. Print the model and coefficients table. Explain the meaning of the significant coefficients.
- Check the assumptions using the diagnostic tests mentioned in this section. Discuss your findings.
- Run a stepwise selection method to reduce the dimension of the model using the backward direction. Print the new model and new coefficients table. Check the assumptions and discuss any changes.
- Use Mallow’s Cp to determine the best model. Does your choice match the model in the previous exercise?
3 Principal Component Analysis
- Load the longley data set from the R datasets package. This data set was used to predict a countries GNP based on several variables. Find the correlation matrix of the explanatory variables.
- Refer to the previous exercise. Perform a principal component analysis on the explanatory variables using the correlation matrix. Use a scree plot to determine the optimal number of components and report them. Try to explain the meaning behind each component.
- Refer to the previous exercise. What proportion of variation does each component explain? What is the total cumulative variance explained by the optimal number of components?
Day 2 Lab Activities - Solutions解答
Simple Linear Regression
1. > pressure.lm <- lm(pressure ~ temperature, data = pressure)
代码语言：javascript复制> summary(pressure.lm)
Call:
lm(formula = pressure ~ temperature, data = pressure)
Residuals:
Min 1Q Median 3Q Max
-158.08 -117.06 -32.84 72.30 409.43
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -147.8989 66.5529 -2.222 0.040124 *
temperature 1.5124 0.3158 4.788 0.000171 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 150.8 on 17 degrees of freedom
Multiple R-squared: 0.5742, Adjusted R-squared: 0.5492
F-statistic: 22.93 on 1 and 17 DF, p-value: 0.000171
> anova(pressure.lm)
Analysis of Variance Table
Response: pressure
Df Sum Sq Mean Sq F value Pr(>F)
temperature 1 521530 521530 22.93 0.000171 ***
Residuals 17 386665 22745
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1The temperature coefficient is positive so if there is a significant relationship between temperature and pressure, it is a direct relationship. Since the p-value is less than 0.05, temperature is indeed significant in the model. The relationship between the t statistic and the F statistic is t^2 = F.
2. Linearity: The scatterplot shows a clear violation of the linearity assumption. The data appears to be exponentially increasing. The standardized residual plot reinforces this observation.
Equal Variance: The lack of a linear relationship makes it difficult to determine the equality of variance in observations.
Normality: The Normal Quantile plot shows a lack of linearity at the tails of the data set. A Shapiro-Wilk test verifies that the residuals do not follow a normal distribution.
Shapiro-Wilk normality test
data: rstandard(pressure.lm)
W = 0.8832, **p-value = 0.02438**
3. Using a Box Cox transformation, the optimal transformation is either
or
where λ = 0.01
Multiple Linear Regression
1.
代码语言：javascript复制 Fertility Agriculture Examination Education Catholic Infant.Mortality
Fertility 1.000 0.353 -0.646 -0.664 0.464 0.417
Agriculture 0.353 1.000 -0.687 -0.640 0.401 -0.061
Examination -0.646 -0.687 1.000 0.698 -0.573 -0.114
Education -0.664 -0.640 0.698 1.000 -0.154 -0.099
Catholic 0.464 0.401 -0.573 -0.154 1.000 0.175
![]()Infant.Mortality 0.417 -0.061 -0.114 -0.099 0.175 1.000
*Related Variables:*
Fertility, Agriculture
Fertility, Examination
Fertility, Infant Mortality
Agriculture, Examination
Agriculture, Education
Examination, Education
2.
代码语言：javascript复制Call:
lm(formula = Fertility ~ Agriculture Examination Education
Catholic Infant.Mortality, data = swiss)
Residuals:
Min 1Q Median 3Q Max
-15.2743 -5.2617 0.5032 4.1198 15.3213
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 66.91518 10.70604 6.250 1.91e-07 ***
Agriculture -0.17211 0.07030 -2.448 0.01873 *
Examination -0.25801 0.25388 -1.016 0.31546
Education -0.87094 0.18303 -4.758 2.43e-05 ***
Catholic 0.10412 0.03526 2.953 0.00519 **
Infant.Mortality 1.07705 0.38172 2.822 0.00734 **
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 7.165 on 41 degrees of freedom
Multiple R-squared: 0.7067, Adjusted R-squared: 0.671
F-statistic: 19.76 on 5 and 41 DF, p-value: 5.594e-10All of the predictors are significant except examination.
3. Residual Plot: There is a random pattern in the residual plot which causes no concern with the model fit.
Normal Q-Q Plot: The data follows the diagonal line quite nicely, indicating that the residuals probably satisfy the normality assumption.
Scale - Location: The data is randomly scattered which indicates that the homoscedasticity assumption is probably met.
4.
代码语言：javascript复制> swiss.step.b <- step(swiss.lm, direction = 'backward')
Start: AIC=190.69
Fertility ~ Agriculture Examination Education Catholic
Infant.Mortality
Df Sum of Sq RSS AIC
- Examination 1 53.03 2158.1 189.86
<none> 2105.0 190.69
- Agriculture 1 307.72 2412.8 195.10
- Infant.Mortality 1 408.75 2513.8 197.03
- Catholic 1 447.71 2552.8 197.75
- Education 1 1162.56 3267.6 209.36
Step: AIC=189.86
Fertility ~ Agriculture Education Catholic Infant.Mortality
Df Sum of Sq RSS AIC
<none> 2158.1 189.86
- Agriculture 1 264.18 2422.2 193.29
- Infant.Mortality 1 409.81 2567.9 196.03
- Catholic 1 956.57 3114.6 205.10
- Education 1 2249.97 4408.0 221.43
Call:
lm(formula = Fertility ~ Agriculture Education Catholic
Infant.Mortality, data = swiss)
Residuals:
Min 1Q Median 3Q Max
-14.6765 -6.0522 0.7514 3.1664 16.1422
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 62.10131 9.60489 6.466 8.49e-08 ***
Agriculture -0.15462 0.06819 -2.267 0.02857 *
Education -0.98026 0.14814 -6.617 5.14e-08 ***
Catholic 0.12467 0.02889 4.315 9.50e-05 ***
Infant.Mortality 1.07844 0.38187 2.824 0.00722 **
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 7.168 on 42 degrees of freedom
Multiple R-squared: 0.6993, Adjusted R-squared: 0.6707
F-statistic: 24.42 on 4 and 42 DF, p-value: 1.717e-10
The new model does not include the examination variable. Now all of the predictors are significant.
Residual Plot: There is a random pattern in the residual plot which causes no concern with the model fit.
Normal Q-Q Plot: The data no longer seems to follow a precise normal distribution. This assumption may now be violated.
Scale - Location: The data is randomly scattered which indicates that the homoscedasticity assumption is probably met.
5. The two models that best fit Mallow's Cp are the model with all 5 variables or the model with the 4 variables Agriculture, Education, Catholic, and Infant.Mortality. We prefer a simpler model in statistics, so the best model choice is the model with four explanatory variables. This is the exact same model that backward selection had identified.
Principal Component Analysis
代码语言：javascript复制 Unemployed Armed.Forces Population Year Employed
Unemployed 1.000 -0.177 0.687 0.668 0.502
Armed.Forces -0.177 1.000 0.364 0.417 0.457
Population 0.687 0.364 1.000 0.994 0.960
Year 0.668 0.417 0.994 1.000 0.971
Employed 0.502 0.457 0.960 0.971 1.000
![]()
2. Comp.1 Comp.2
Unemployed 0.3633 0.5988
Armed.Forces 0.2269 -0.7911
Population 0.5261 0.0435
Year 0.5291 -0.0024
Employed 0.5097 -0.1171The first component is a standardized measure of GNP and the second component is difficult to interpret.
3.
代码语言：javascript复制Component 1: **71.23%** Variance explained
Component2: **23.67%** Variance explained
Cumulative Variance: **94.89%**
