data 605 assignment 12
Tyler Baker
4/14/2022
Week 12 Assignment
Multi-Factor Regression
Tyler Baker
Assignment Description
Can be found here: https://github.com/tylerbaker01/data-605-week-12-assignment/blob/main/Assignment%20Description
Data
who_data <- read.csv("https://raw.githubusercontent.com/tylerbaker01/data-605-week-12-assignment/main/who.csv")Question 1
Provide a scatterplot of LifeExp~TotExp, and run simple linear regression. Do not transform the variables. Provide and interpret the F statistics, R^2, standard error,and p-values only. Discuss whether the assumptions of simple linear regression met.
q_one.lm <- lm(LifeExp ~ TotExp, data=who_data)
with(who_data, plot(LifeExp, TotExp))
abline(q_one.lm)
From looking at the plot, the model does not seem to fit the data but let’s look at the summary statistics still.
summary(q_one.lm)##
## Call:
## lm(formula = LifeExp ~ TotExp, data = who_data)
##
## Residuals:
## Min 1Q Median 3Q Max
## -24.764 -4.778 3.154 7.116 13.292
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 6.475e+01 7.535e-01 85.933 < 2e-16 ***
## TotExp 6.297e-05 7.795e-06 8.079 7.71e-14 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 9.371 on 188 degrees of freedom
## Multiple R-squared: 0.2577, Adjusted R-squared: 0.2537
## F-statistic: 65.26 on 1 and 188 DF, p-value: 7.714e-14The F-Statistic
The F-statistic provides a way to globally test if any of the independent variables is related to the dependent variable. If the p-value associated with the F-Statistic is >= 0.05 then we can say with 95% confidence that the dependent variable is not dependent on the given independent variables. In our case, we have a p-value much smaller tham 0.05. Thus, we can conclude that TotExp does have an relationship with LifeExp. ### R^2 The R^2 is very low, both are about 25%. This means that about 25% of the predictions acount for observational data. Although we have a low R^2, this does not neccessarily mean that we have a bad model. We should look at the residuals plot.
plot(fitted(q_one.lm), resid(q_one.lm))
From looking at this we see that the residuals are not equally spread out. This is a good indicator that our model is not strong. ### Standard Errors The standard error should be about 5-10 times smaller than the cooresponding coefficient. In this case ours is about 86 times smaller. Again, this is not a good indicator. ### Conclusion While we can say that TotExp does have a role in the outcome of LifeExp, we can say that it is not a linear relationship. From looking at the scatterplot, we should note that there was an almost exponential relationship. ## Question 2 2. Raise life expectancy to the 4.6 power (i.e., LifeExp^4.6). Raise total expenditures to the 0.06 power (nearly a log transform, TotExp^.06). Plot LifeExp^4.6 as a function of TotExp^.06, and r re-run the simple regression model using the transformed variables. Provide and interpret the F statistics, R^2, standard error, and p-values. Which model is “better?” ### Setting Up the Data
who_data_two <- who_data
who_data_two$LifeExp <- who_data_two$LifeExp ^4.6
who_data_two$TotExp <- who_data_two$TotExp^.06Rerunning the Regression
q_two.lm <- lm(LifeExp ~ TotExp, data=who_data_two)
with(who_data_two, plot(LifeExp, TotExp))
abline(q_two.lm)
From the plot alone, this appears that it will lead to a better model. Let’s find out.
summary(q_two.lm)##
## Call:
## lm(formula = LifeExp ~ TotExp, data = who_data_two)
##
## Residuals:
## Min 1Q Median 3Q Max
## -308616089 -53978977 13697187 59139231 211951764
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -736527910 46817945 -15.73 <2e-16 ***
## TotExp 620060216 27518940 22.53 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 90490000 on 188 degrees of freedom
## Multiple R-squared: 0.7298, Adjusted R-squared: 0.7283
## F-statistic: 507.7 on 1 and 188 DF, p-value: < 2.2e-16F-Statistic
Well, the cooresponding p-value shows that there is a strong relationship between the two variables. ### R^2 The R^2 is much better in this model. This model is account for almost 73% of observational data. However, we must again check the plot.
plot(fitted(q_two.lm), resid(q_two.lm))
This appears to be in much better shape then the previous model. Our residuals seem to be pretty equally spread out across 0. ### Standard Errors I do not know how to interpret a negative t-value. ### Conclusion The second version is “better” model. However, we have changed the data in a way that makes interpretation very difficult. What does a life expectancy raised to the power of 4.6 mean exactly? Life expectancy in United States is about 77 years. So how do we interpret 77^4.6 years old? ## Question 3 Using the results from 2, forecast life expectancy when TotExp^.06 =1.5. Then forecast life expectancy when TotExp^.06=2.5.
coef(q_two.lm)## (Intercept) TotExp
## -736527909 620060216So we will calculate using y=mx+b, where m and b are the coefficients.
a <- -736527909 + 620060216*1.5
print(a)## [1] 193562415b <- -736527909 + 620060216*2.5
print(b)## [1] 813622631Again, I do not know how we can interpret this information. ## Question 4 4. Build the following multiple regression model and interpret the F Statistics, R^2, standard error, and p-values. How good is the model? LifeExp = b0+b1 x PropMd + b2 x TotExp +b3 x PropMD x TotExp ### Multi-Factor Regression
who_data$value <- who_data$PropMD *who_data$TotExp
q_four.lm <- lm(LifeExp ~ value + TotExp + PropMD, data=who_data)Now let’s look at the summary statistics.
summary(q_four.lm)##
## Call:
## lm(formula = LifeExp ~ value + TotExp + PropMD, data = who_data)
##
## Residuals:
## Min 1Q Median 3Q Max
## -27.320 -4.132 2.098 6.540 13.074
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 6.277e+01 7.956e-01 78.899 < 2e-16 ***
## value -6.026e-03 1.472e-03 -4.093 6.35e-05 ***
## TotExp 7.233e-05 8.982e-06 8.053 9.39e-14 ***
## PropMD 1.497e+03 2.788e+02 5.371 2.32e-07 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 8.765 on 186 degrees of freedom
## Multiple R-squared: 0.3574, Adjusted R-squared: 0.3471
## F-statistic: 34.49 on 3 and 186 DF, p-value: < 2.2e-16F-Statistic
The p-value corresponding to the F-statistic is again very small. Which is saying that we’re over 99% confident that the independent variables have an impact on the dependent variables. ### R^2 The R^2 is low. Which means that our model only accounts for about 36% of the observations. A low R^2 doesn’t necessarily mean that it is a bad model.
plot(fitted(q_four.lm), resid(q_four.lm))
Viewing this plot it is hard to be sure that the residuals are spread out equally across 0, but they very well could be. ### Standard Errors I’m not sure how to interpret standard errors on the coefficients. The significance code shows that they are all as significant as possible. ### Conclusion I do not think our model is “good” because the residuals don’t seem to follow a nearly normal distribution. I know that our choice of independent variables is good but I imagine the points follow a different distribution than what are model is providing. ## Question 5 Forecast LifeExp when PropMD=.03 and TotExp = 14. Does this forecast seem realistic? Why or why not?
coef(q_four.lm)## (Intercept) value TotExp PropMD
## 6.277270e+01 -6.025686e-03 7.233324e-05 1.497494e+03We will note y= 6.277270e+01 + -6.025686e-03(value) + 7.233324e-05(TotExp) + 1.497494e+03(PropMD)
y<- 62.77270 + (-0.06025686*(.42)) + .00007233324*(14) + 1497.494*(.03)
print(y)## [1] 107.6732This is saying that the average life expectancy would be 107. This doesn’t seem correct. However, if we can really trust our model then it is saying something pretty remarkable. Simply increasing the percent of doctors has a large impact on life expectancy.