Midterm
Enkhtsatsral 111035098
2023-11-10
#Chapter 2 #C9
library(wooldridge)
data <- wooldridge::kielmcInterpret the coefficient on prpblck
cat("The coefficient on prpblck is the estimated change in psoda for a one-unit change in prpblck.\n")## The coefficient on prpblck is the estimated change in psoda for a one-unit change in prpblck.cat("In this context, it represents the change in the price of soda for a 1% increase in the proportion\n")## In this context, it represents the change in the price of soda for a 1% increase in the proportioncat("of the population that is black. Whether it is economically large depends on the magnitude and significance\n")## of the population that is black. Whether it is economically large depends on the magnitude and significancecat("of the coefficient, which can be determined from the summary output.\n\n")## of the coefficient, which can be determined from the summary output.(iii) Compare the estimate from part (ii) with the simple regression estimate from psoda on prpblck.
simple_model <- lm(psoda ~ prpblck, data = discrim)
summary(simple_model)##
## Call:
## lm(formula = psoda ~ prpblck, data = discrim)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.30884 -0.05963 0.01135 0.03206 0.44840
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 1.03740 0.00519 199.87 < 2e-16 ***
## prpblck 0.06493 0.02396 2.71 0.00702 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.0881 on 399 degrees of freedom
## (9 observations deleted due to missingness)
## Multiple R-squared: 0.01808, Adjusted R-squared: 0.01561
## F-statistic: 7.345 on 1 and 399 DF, p-value: 0.007015(iv) Estimate the model log(psoda) = B0 + B1prpblck + B2log(income) + u
log_model <- lm(log(psoda) ~ prpblck + log(income), data = discrim)
summary(log_model)##
## Call:
## lm(formula = log(psoda) ~ prpblck + log(income), data = discrim)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.33563 -0.04695 0.00658 0.04334 0.35413
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -0.79377 0.17943 -4.424 1.25e-05 ***
## prpblck 0.12158 0.02575 4.722 3.24e-06 ***
## log(income) 0.07651 0.01660 4.610 5.43e-06 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.0821 on 398 degrees of freedom
## (9 observations deleted due to missingness)
## Multiple R-squared: 0.06809, Adjusted R-squared: 0.06341
## F-statistic: 14.54 on 2 and 398 DF, p-value: 8.039e-07Calculate the estimated percentage change in psoda for a 20% increase in prpblck
percentage_change <- exp(coef(log_model)[2] * 0.20) * 100 - 100
cat("Estimated percentage change in psoda for a 20% increase in prpblck:", percentage_change, "\n\n")## Estimated percentage change in psoda for a 20% increase in prpblck: 2.46141(v) Add the variable prppov to the regression in part (iv)
model_with_prppov <- lm(log(psoda) ~ prpblck + log(income) + prppov, data = discrim)
summary(model_with_prppov)##
## Call:
## lm(formula = log(psoda) ~ prpblck + log(income) + prppov, data = discrim)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.32218 -0.04648 0.00651 0.04272 0.35622
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -1.46333 0.29371 -4.982 9.4e-07 ***
## prpblck 0.07281 0.03068 2.373 0.0181 *
## log(income) 0.13696 0.02676 5.119 4.8e-07 ***
## prppov 0.38036 0.13279 2.864 0.0044 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.08137 on 397 degrees of freedom
## (9 observations deleted due to missingness)
## Multiple R-squared: 0.08696, Adjusted R-squared: 0.08006
## F-statistic: 12.6 on 3 and 397 DF, p-value: 6.917e-08(vi) Find the correlation between log(income) and prppov
correlation_log_income_prppov <- cor(discrim$lincome, discrim$prppov)
cat("Correlation between lincome and prppov:", correlation_log_income_prppov, "\n\n")## Correlation between lincome and prppov: NA(vii) Evaluate the following statement
cat("The statement 'Because lincome and prppov are so highly correlated, they have no business being in the same regression.'\n")## The statement 'Because lincome and prppov are so highly correlated, they have no business being in the same regression.'cat("The high negative correlation between lincome and prppov suggests multicollinearity between these two variables.\n")## The high negative correlation between lincome and prppov suggests multicollinearity between these two variables.cat("Multicollinearity can lead to unstable coefficient estimates, making it challenging to interpret the individual effects\n")## Multicollinearity can lead to unstable coefficient estimates, making it challenging to interpret the individual effectscat("of the variables. However, the decision to include or exclude variables should be based on the specific research question,\n")## of the variables. However, the decision to include or exclude variables should be based on the specific research question,cat("theoretical considerations, and the goals of the analysis.\n")## theoretical considerations, and the goals of the analysis.cat("In some cases, including both variables in the regression model might still be justified if they capture different aspects\n")## In some cases, including both variables in the regression model might still be justified if they capture different aspectscat("of the relationship with the dependent variable and contribute to a more comprehensive understanding of the phenomenon under study.\n")## of the relationship with the dependent variable and contribute to a more comprehensive understanding of the phenomenon under study.#Ch4 #3 # Given coefficients and standard errors
coeff_log_sales <- 0.321
se_log_sales <- 0.216
coeff_profmarg <- 0.50
se_profmarg <- 0.46a) Estimated percentage point change in Rdintens for a 10% increase in sales
percentage_change <- coeff_log_sales * 10
cat("a) Estimated percentage point change in Rdintens for a 10% increase in sales:", percentage_change, "\n")## a) Estimated percentage point change in Rdintens for a 10% increase in sales: 3.21b) Test for log(sales) coefficient
t_stat_log_sales <- coeff_log_sales / se_log_sales
p_value_log_sales <- 2 * (1 - pt(abs(t_stat_log_sales), df = 29)) # two-tailed test
cat("b) p-value for the test on log(sales) coefficient:", p_value_log_sales, "\n")## b) p-value for the test on log(sales) coefficient: 0.1480413cat(" (At 5% level):", ifelse(p_value_log_sales < 0.05, "Reject H0", "Fail to reject H0"), "\n")## (At 5% level): Fail to reject H0cat(" (At 10% level):", ifelse(p_value_log_sales < 0.10, "Reject H0", "Fail to reject H0"), "\n")## (At 10% level): Fail to reject H0c) Interpretation of the coefficient on profmarg
cat("c) Coefficient on profmarg:", coeff_profmarg, "\n")## c) Coefficient on profmarg: 0.5d) Test for profmarg coefficient
t_stat_profmarg <- coeff_profmarg / se_profmarg
p_value_profmarg <- 2 * (1 - pt(abs(t_stat_profmarg), df = 29)) # two-tailed test
cat("d) p-value for the test on profmarg coefficient:", p_value_profmarg, "\n")## d) p-value for the test on profmarg coefficient: 0.2860082cat(" (At 5% level):", ifelse(p_value_profmarg < 0.05, "Reject H0", "Fail to reject H0"), "\n")## (At 5% level): Fail to reject H0cat(" (At 10% level):", ifelse(p_value_profmarg < 0.10, "Reject H0", "Fail to reject H0"), "\n")## (At 10% level): Fail to reject H0#ch4 #c8 # Load the wooldridge library
library(wooldridge)Load the dataset
data("k401ksubs")(i) How many single-person households are there in the data set?
single_person_households <- subset(k401ksubs, fsize == 1)
num_single_person_households <- nrow(single_person_households)
cat("Number of single-person households:", num_single_person_households, "\n\n")## Number of single-person households: 2017(ii) Use OLS to estimate the model: nettfa = B0 + B1inc + B2age + u
model <- lm(nettfa ~ inc + age, data = single_person_households)
summary(model)##
## Call:
## lm(formula = nettfa ~ inc + age, data = single_person_households)
##
## Residuals:
## Min 1Q Median 3Q Max
## -179.95 -14.16 -3.42 6.03 1113.94
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -43.03981 4.08039 -10.548 <2e-16 ***
## inc 0.79932 0.05973 13.382 <2e-16 ***
## age 0.84266 0.09202 9.158 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 44.68 on 2014 degrees of freedom
## Multiple R-squared: 0.1193, Adjusted R-squared: 0.1185
## F-statistic: 136.5 on 2 and 2014 DF, p-value: < 2.2e-16Interpret the slope coefficients
cat("Interpretation of slope coefficients:\n")## Interpretation of slope coefficients:cat("B1 (inc): The estimated change in nettfa for a one-unit change in inc (annual family income).\n")## B1 (inc): The estimated change in nettfa for a one-unit change in inc (annual family income).cat("B2 (age): The estimated change in nettfa for a one-unit change in age.\n")## B2 (age): The estimated change in nettfa for a one-unit change in age.cat("There might be surprises depending on the context and expectations of the relationship between variables.\n\n")## There might be surprises depending on the context and expectations of the relationship between variables.(iii) Does the intercept from the regression in part (ii) have an interesting meaning? Explain.
cat("The intercept (B0) represents the estimated net financial wealth (nettfa) when both inc and age are zero.\n")## The intercept (B0) represents the estimated net financial wealth (nettfa) when both inc and age are zero.cat("In this context, it may not have a meaningful interpretation, as having zero income and age is not practically meaningful.\n\n")## In this context, it may not have a meaningful interpretation, as having zero income and age is not practically meaningful.(iv) Find the p-value for the test H0: B₂ = 1 against H₁: B₂ < 1. Do you reject H0 at the 1% significance level?
test_result <- summary(model)$coefficient[3, "Pr(>|t|)"]
cat("p-value for the test H0: B₂ = 1 against H₁: B₂ < 1:", test_result, "\n")## p-value for the test H0: B₂ = 1 against H₁: B₂ < 1: 1.265959e-19cat("At the 1% significance level, we would reject H0 if the p-value is less than 0.01.\n")## At the 1% significance level, we would reject H0 if the p-value is less than 0.01.if (test_result < 0.01) {
cat("We reject H0; there is evidence that B₂ is less than 1.\n\n")
} else {
cat("We do not reject H0; there is not enough evidence to conclude that B₂ is less than 1.\n\n")
}## We reject H0; there is evidence that B₂ is less than 1.(v) If you do a simple regression of nettfa on inc, is the estimated coefficient on inc much different from the estimate in part (ii)? Why or why not?
simple_model <- lm(nettfa ~ inc, data = single_person_households)
summary(simple_model)##
## Call:
## lm(formula = nettfa ~ inc, data = single_person_households)
##
## Residuals:
## Min 1Q Median 3Q Max
## -185.12 -12.85 -4.85 1.78 1112.66
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -10.5709 2.0607 -5.13 3.18e-07 ***
## inc 0.8207 0.0609 13.48 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 45.59 on 2015 degrees of freedom
## Multiple R-squared: 0.08267, Adjusted R-squared: 0.08222
## F-statistic: 181.6 on 1 and 2015 DF, p-value: < 2.2e-16cat("Comparison of the estimated coefficient on inc:\n")## Comparison of the estimated coefficient on inc:cat("The estimated coefficient on inc in the simple regression is compared to the estimate in part (ii).\n")## The estimated coefficient on inc in the simple regression is compared to the estimate in part (ii).cat("Differences may arise due to the inclusion of age in the multiple regression model, which may affect\n")## Differences may arise due to the inclusion of age in the multiple regression model, which may affectcat("the relationship between nettfa and inc. The context and goals of the analysis will determine\n")## the relationship between nettfa and inc. The context and goals of the analysis will determinecat("whether the inclusion of age improves the model.\n")## whether the inclusion of age improves the model.#ch5 #Load the wooldridge library
library(wooldridge)Load the dataset
data('econmath')
# Load necessary libraries
library(ggplot2)
library(stats)Assuming “ECONMATH” is your data frame and “score” is the variable of interest
data <- econmath$scoreCreate a histogram
hist_data <- hist(data, breaks = 30, plot = FALSE)Fit a normal distribution
mu <- mean(data)
sigma <- sd(data)
x <- seq(min(data), max(data), length = 100)
y <- dnorm(x, mean = mu, sd = sigma)Plot the histogram and the fitted normal distribution
hist_plot <- ggplot() +
geom_histogram(aes(x = data, y = ..density..), bins = 30, fill = "gold1", color = "black") +
geom_line(aes(x = x, y = y), color = "red", size = 1) +
labs(title = "Histogram and Fitted Normal Distribution",
x = "Score",
y = "Density") +
theme_minimal()## Warning: Using `size` aesthetic for lines was deprecated in ggplot2 3.4.0.
## ℹ Please use `linewidth` instead.
## This warning is displayed once every 8 hours.
## Call `lifecycle::last_lifecycle_warnings()` to see where this warning was
## generated.print(hist_plot)## Warning: The dot-dot notation (`..density..`) was deprecated in ggplot2 3.4.0.
## ℹ Please use `after_stat(density)` instead.
## This warning is displayed once every 8 hours.
## Call `lifecycle::last_lifecycle_warnings()` to see where this warning was
## generated.
i. Probability that “score” exceeds 100 using the normal distribution
prob_exceed_100 <- 1 - pnorm(100, mean = mu, sd = sigma)
cat("i. Probability that 'score' exceeds 100 using the normal distribution:", prob_exceed_100, "\n")## i. Probability that 'score' exceeds 100 using the normal distribution: 0.02044288ii. Assess the fit in the left tail visually or using statistical tests
qqnorm(data)
qqline(data)
shapiro.test(data) # Perform a Shapiro-Wilk test for normality##
## Shapiro-Wilk normality test
##
## data: data
## W = 0.96973, p-value = 2.454e-12#ch5 #c1 # Install and load necessary libraries
library(wooldridge)
library(ggplot2)Load the ‘wage1’ dataset
data("wage1")
wage_data <- wage1(i) Estimate the equation wage = b0 + b1educ + b2exper + b3tenure + u.
model_level <- lm(wage ~ educ + exper + tenure, data = wage_data)Save residuals
residuals_level <- residuals(model_level)Display summary statistics for the level-level model
summary(model_level)##
## Call:
## lm(formula = wage ~ educ + exper + tenure, data = wage_data)
##
## Residuals:
## Min 1Q Median 3Q Max
## -7.6068 -1.7747 -0.6279 1.1969 14.6536
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -2.87273 0.72896 -3.941 9.22e-05 ***
## educ 0.59897 0.05128 11.679 < 2e-16 ***
## exper 0.02234 0.01206 1.853 0.0645 .
## tenure 0.16927 0.02164 7.820 2.93e-14 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 3.084 on 522 degrees of freedom
## Multiple R-squared: 0.3064, Adjusted R-squared: 0.3024
## F-statistic: 76.87 on 3 and 522 DF, p-value: < 2.2e-16(ii) Repeat part (i), but with log(wage) as the dependent variable.
model_log_level <- lm(log(wage) ~ educ + exper + tenure, data = wage_data)Save residuals
residuals_log_level <- residuals(model_log_level)Plot histogram of residuals for the log-level model
hist(residuals_log_level, main = "Histogram of Residuals (Log-Level Model)", col = "gold1", border = "black")
Display summary statistics for the log-level model
summary(model_log_level)##
## Call:
## lm(formula = log(wage) ~ educ + exper + tenure, data = wage_data)
##
## Residuals:
## Min 1Q Median 3Q Max
## -2.05802 -0.29645 -0.03265 0.28788 1.42809
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 0.284360 0.104190 2.729 0.00656 **
## educ 0.092029 0.007330 12.555 < 2e-16 ***
## exper 0.004121 0.001723 2.391 0.01714 *
## tenure 0.022067 0.003094 7.133 3.29e-12 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.4409 on 522 degrees of freedom
## Multiple R-squared: 0.316, Adjusted R-squared: 0.3121
## F-statistic: 80.39 on 3 and 522 DF, p-value: < 2.2e-16(iii) Q-Q plots for normality assessment
par(mfrow = c(2, 2)) # Set up a 2x2 grid for Q-Q plots
qqnorm(residuals_level, main = "Q-Q Plot - Level-Level Model")
qqline(residuals_level)
qqnorm(residuals_log_level, main = "Q-Q Plot - Log-Level Model")
qqline(residuals_log_level)
Reset the plotting layout
par(mfrow = c(1, 1))Print summary statistics to the console for easier interpretation
cat("\nSummary Statistics - Level-Level Model:\n")##
## Summary Statistics - Level-Level Model:summary(model_level)##
## Call:
## lm(formula = wage ~ educ + exper + tenure, data = wage_data)
##
## Residuals:
## Min 1Q Median 3Q Max
## -7.6068 -1.7747 -0.6279 1.1969 14.6536
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -2.87273 0.72896 -3.941 9.22e-05 ***
## educ 0.59897 0.05128 11.679 < 2e-16 ***
## exper 0.02234 0.01206 1.853 0.0645 .
## tenure 0.16927 0.02164 7.820 2.93e-14 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 3.084 on 522 degrees of freedom
## Multiple R-squared: 0.3064, Adjusted R-squared: 0.3024
## F-statistic: 76.87 on 3 and 522 DF, p-value: < 2.2e-16cat("\nSummary Statistics - Log-Level Model:\n")##
## Summary Statistics - Log-Level Model:summary(model_log_level)##
## Call:
## lm(formula = log(wage) ~ educ + exper + tenure, data = wage_data)
##
## Residuals:
## Min 1Q Median 3Q Max
## -2.05802 -0.29645 -0.03265 0.28788 1.42809
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 0.284360 0.104190 2.729 0.00656 **
## educ 0.092029 0.007330 12.555 < 2e-16 ***
## exper 0.004121 0.001723 2.391 0.01714 *
## tenure 0.022067 0.003094 7.133 3.29e-12 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.4409 on 522 degrees of freedom
## Multiple R-squared: 0.316, Adjusted R-squared: 0.3121
## F-statistic: 80.39 on 3 and 522 DF, p-value: < 2.2e-16#Chapter6 #3 # Load the wooldridge library
library(wooldridge)Load the dataset
data("rdchem")(i) At what point does the marginal effect of sales on rdintens become negative?
sales_negative_marginal_effect <- -coef(model)[2] / (2 * coef(model)[3])
cat("The marginal effect of sales on rdintens becomes negative when sales is greater than", sales_negative_marginal_effect, "\n")## The marginal effect of sales on rdintens becomes negative when sales is greater than -0.4742839(ii) Would you keep the quadratic term in the model? Explain.
cat("The decision to keep the quadratic term depends on the significance of the coefficient and the context of the analysis.\n")## The decision to keep the quadratic term depends on the significance of the coefficient and the context of the analysis.cat("If the quadratic term is statistically significant and improves the model fit, it may be kept.\n\n")## If the quadratic term is statistically significant and improves the model fit, it may be kept.(iii) Define salesbil as sales measured in billions of dollars
rdchem$salesbil <- rdchem$sales / 1000
model_salesbil <- lm(rdintens ~ salesbil + I(salesbil^2), data = rdchem)
summary(model_salesbil)##
## Call:
## lm(formula = rdintens ~ salesbil + I(salesbil^2), data = rdchem)
##
## Residuals:
## Min 1Q Median 3Q Max
## -2.1418 -1.3630 -0.2257 1.0688 5.5808
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 2.612512 0.429442 6.084 1.27e-06 ***
## salesbil 0.300571 0.139295 2.158 0.0394 *
## I(salesbil^2) -0.006946 0.003726 -1.864 0.0725 .
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.788 on 29 degrees of freedom
## Multiple R-squared: 0.1484, Adjusted R-squared: 0.08969
## F-statistic: 2.527 on 2 and 29 DF, p-value: 0.09733(iv) For the purpose of reporting the results, which equation do you prefer?
First Model
model1 <- lm(rdintens ~ sales + I(sales^2), data = rdchem)Second Model
model2 <- lm(rdintens ~ salesbil + I(salesbil^2), data = rdchem)Compare Adjusted R-squared
adjusted_r_squared_model1 <- summary(model1)$adj.r.squared
adjusted_r_squared_model2 <- summary(model2)$adj.r.squaredCompare Coefficients
coefficients_model1 <- coef(model1)
coefficients_model2 <- coef(model2)Compare Residuals (optional, for additional diagnostics)
residuals_model1 <- residuals(model1)
residuals_model2 <- residuals(model2)Decision
if (adjusted_r_squared_model1 == adjusted_r_squared_model2) {
preference <- "Both models have the same adjusted R-squared."
} else if (adjusted_r_squared_model1 > adjusted_r_squared_model2) {
preference <- "Model 1 is preferred based on adjusted R-squared."
} else {
preference <- "Model 2 is preferred based on adjusted R-squared."
}Report Coefficients (you can customize this based on your needs)
report_model1 <- coef(model1)
report_model2 <- coef(model2)Print results
cat("Adjusted R-squared - Model 1:", adjusted_r_squared_model1, "\n")## Adjusted R-squared - Model 1: 0.08969224cat("Adjusted R-squared - Model 2:", adjusted_r_squared_model2, "\n")## Adjusted R-squared - Model 2: 0.08969224cat("Preference:", preference, "\n")## Preference: Both models have the same adjusted R-squared.cat("\nCoefficients - Model 1:\n")##
## Coefficients - Model 1:print(report_model1)## (Intercept) sales I(sales^2)
## 2.612512e+00 3.005713e-04 -6.945939e-09# Print Coefficients - Model 2
cat("\nCoefficients - Model 2:\n")##
## Coefficients - Model 2:print(report_model2)## (Intercept) salesbil I(salesbil^2)
## 2.612512085 0.300571301 -0.006945939#10 # Load the wooldridge library
library(wooldridge)# Load the dataset
data("meapsingle")(i) If you are a policy maker trying to estimate the causal effect of per-student spending on math test performance,
explain why the first equation is more relevant than the second. What is the estimated effect of a 10% increase in expenditures per student?
cat("The first equation is more relevant because it provides a direct estimate of the effect of per-student spending on math test performance.\n")## The first equation is more relevant because it provides a direct estimate of the effect of per-student spending on math test performance.cat("In the first equation, the coefficient on lexppp represents the estimated effect of a one-unit increase in log expenditures per student.\n")## In the first equation, the coefficient on lexppp represents the estimated effect of a one-unit increase in log expenditures per student.cat("To get the estimated effect of a 10% increase, you can multiply the coefficient by 0.1.\n")## To get the estimated effect of a 10% increase, you can multiply the coefficient by 0.1.(ii) Does adding read4 to the regression have strange effects on coefficients and statistical significance other than B1exppp?
(iii) How would you explain to someone with only basic knowledge of regression why, in this case, you prefer the equation with the smaller adjusted R-squared?
Fit the first equation
model1 <- lm(math4 ~ lexppp + free + lmedinc + pctsgle, data = meapsingle)
summary(model1)##
## Call:
## lm(formula = math4 ~ lexppp + free + lmedinc + pctsgle, data = meapsingle)
##
## Residuals:
## Min 1Q Median 3Q Max
## -33.259 -7.422 1.615 7.274 49.524
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 24.48949 59.23781 0.413 0.6797
## lexppp 9.00648 4.03530 2.232 0.0266 *
## free -0.42164 0.07064 -5.969 9.27e-09 ***
## lmedinc -0.75221 5.35816 -0.140 0.8885
## pctsgle -0.27444 0.16086 -1.706 0.0894 .
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 11.59 on 224 degrees of freedom
## Multiple R-squared: 0.4716, Adjusted R-squared: 0.4622
## F-statistic: 49.98 on 4 and 224 DF, p-value: < 2.2e-16Fit the second equation
model2 <- lm(math4 ~ lexppp + free + lmedinc + pctsgle + read4, data = meapsingle)
summary(model2)##
## Call:
## lm(formula = math4 ~ lexppp + free + lmedinc + pctsgle + read4,
## data = meapsingle)
##
## Residuals:
## Min 1Q Median 3Q Max
## -29.5690 -4.6729 -0.0349 4.3644 24.8425
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 149.37870 41.70293 3.582 0.000419 ***
## lexppp 1.93215 2.82480 0.684 0.494688
## free -0.06004 0.05399 -1.112 0.267297
## lmedinc -10.77595 3.75746 -2.868 0.004529 **
## pctsgle -0.39663 0.11143 -3.559 0.000454 ***
## read4 0.66656 0.04249 15.687 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 8.012 on 223 degrees of freedom
## Multiple R-squared: 0.7488, Adjusted R-squared: 0.7432
## F-statistic: 132.9 on 5 and 223 DF, p-value: < 2.2e-16Compare the two equations
cat("Comparison of Equations:\n")## Comparison of Equations:cat("1. Equation without read4:\n")## 1. Equation without read4:summary(model1)##
## Call:
## lm(formula = math4 ~ lexppp + free + lmedinc + pctsgle, data = meapsingle)
##
## Residuals:
## Min 1Q Median 3Q Max
## -33.259 -7.422 1.615 7.274 49.524
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 24.48949 59.23781 0.413 0.6797
## lexppp 9.00648 4.03530 2.232 0.0266 *
## free -0.42164 0.07064 -5.969 9.27e-09 ***
## lmedinc -0.75221 5.35816 -0.140 0.8885
## pctsgle -0.27444 0.16086 -1.706 0.0894 .
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 11.59 on 224 degrees of freedom
## Multiple R-squared: 0.4716, Adjusted R-squared: 0.4622
## F-statistic: 49.98 on 4 and 224 DF, p-value: < 2.2e-16cat("\n2. Equation with read4:\n")##
## 2. Equation with read4:summary(model2)##
## Call:
## lm(formula = math4 ~ lexppp + free + lmedinc + pctsgle + read4,
## data = meapsingle)
##
## Residuals:
## Min 1Q Median 3Q Max
## -29.5690 -4.6729 -0.0349 4.3644 24.8425
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 149.37870 41.70293 3.582 0.000419 ***
## lexppp 1.93215 2.82480 0.684 0.494688
## free -0.06004 0.05399 -1.112 0.267297
## lmedinc -10.77595 3.75746 -2.868 0.004529 **
## pctsgle -0.39663 0.11143 -3.559 0.000454 ***
## read4 0.66656 0.04249 15.687 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 8.012 on 223 degrees of freedom
## Multiple R-squared: 0.7488, Adjusted R-squared: 0.7432
## F-statistic: 132.9 on 5 and 223 DF, p-value: < 2.2e-16Explanation for (iii)
cat("\nExplanation for (iii):\n")##
## Explanation for (iii):cat("The adjusted R-squared value is a measure of the proportion of variance in the dependent variable that is explained by the independent variables.\n")## The adjusted R-squared value is a measure of the proportion of variance in the dependent variable that is explained by the independent variables.cat("In this case, a smaller adjusted R-squared indicates that the additional variable (read4) does not significantly improve the model's explanatory power.\n")## In this case, a smaller adjusted R-squared indicates that the additional variable (read4) does not significantly improve the model's explanatory power.cat("While the model with read4 has a higher overall R-squared, the adjusted R-squared considers the number of variables and penalizes for overfitting.\n")## While the model with read4 has a higher overall R-squared, the adjusted R-squared considers the number of variables and penalizes for overfitting.cat("Choosing the model with a smaller adjusted R-squared may be preferable if the additional variable does not contribute substantially to the model's accuracy.\n")## Choosing the model with a smaller adjusted R-squared may be preferable if the additional variable does not contribute substantially to the model's accuracy.Load the necessary library
library(tidyverse)## ── Attaching core tidyverse packages ──────────────────────── tidyverse 2.0.0 ──
## ✔ dplyr 1.1.3 ✔ readr 2.1.4
## ✔ forcats 1.0.0 ✔ stringr 1.5.0
## ✔ lubridate 1.9.3 ✔ tibble 3.2.1
## ✔ purrr 1.0.2 ✔ tidyr 1.3.0
## ── Conflicts ────────────────────────────────────────── tidyverse_conflicts() ──
## ✖ dplyr::filter() masks stats::filter()
## ✖ dplyr::lag() masks stats::lag()
## ℹ Use the conflicted package (<http://conflicted.r-lib.org/>) to force all conflicts to become errorsLoad the dataset
data("wage2")Fit the regression model
model <- lm(log(wage) ~ educ + exper + educ * exper, data = wage2)(i) Show that the return to another year of education, holding exper fixed, is b1 + b3 * exper
coefs <- coef(model)
return_to_education <- coefs["educ"] + coefs["educ:exper"]
cat("Return to another year of education (holding exper fixed):", return_to_education, "\n")## Return to another year of education (holding exper fixed): 0.04725277(ii) State the null hypothesis that the return to education does not depend on the level of exper.
# The null hypothesis (H0) is: B3 = 0 (No interaction effect)
# Appropriate alternative: The return to education depends on the level of exper (B3 ≠ 0)
# (iii) Test the null hypothesis in (ii) against the stated alternative
summary(model)##
## Call:
## lm(formula = log(wage) ~ educ + exper + educ * exper, data = wage2)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.88558 -0.24553 0.03558 0.26171 1.28836
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 5.949455 0.240826 24.704 <2e-16 ***
## educ 0.044050 0.017391 2.533 0.0115 *
## exper -0.021496 0.019978 -1.076 0.2822
## educ:exper 0.003203 0.001529 2.095 0.0365 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.3923 on 931 degrees of freedom
## Multiple R-squared: 0.1349, Adjusted R-squared: 0.1321
## F-statistic: 48.41 on 3 and 931 DF, p-value: < 2.2e-16(iv) Return to Education When exper = 10
# Extract coefficients
B_educ <- coef(model)["educ"]
B_exper <- coef(model)["exper"]
# Calculate the return to education when exper = 10
theta <- B_educ + 10 * B_exper
# Get standard error for the estimate
se_theta <- summary(model)$coefficients["exper", "Std. Error"]
# Calculate the 95% confidence interval
confidence_interval <- c(theta - 1.96 * se_theta, theta + 1.96 * se_theta)
cat("Theta:", theta, "\n")## Theta: -0.1709096cat("95% Confidence Interval:", confidence_interval, "\n")## 95% Confidence Interval: -0.210067 -0.1317521Question 12
install.packages("wooldridge")## Installing package into '/cloud/lib/x86_64-pc-linux-gnu-library/4.3'
## (as 'lib' is unspecified)library(wooldridge)
data("k401ksubs")
subsetted_data <- k401ksubs[k401ksubs$fsize == 1, ]i.What is the youngest age of people in this sample? How many people are at that age?
youngest_age <- min(k401ksubs$age)
count_youngest_age <- sum(k401ksubs$age == youngest_age)
cat("Youngest age:", youngest_age, "\n")## Youngest age: 25cat("Number of people at the youngest age:", count_youngest_age, "\n")## Number of people at the youngest age: 211ii. What is the literal interpretation of b2? By itself, is it of much interest?
model <- lm(nettfa ~ inc + age + agesq , data = k401ksubs)
summary(model)##
## Call:
## lm(formula = nettfa ~ inc + age + agesq, data = k401ksubs)
##
## Residuals:
## Min 1Q Median 3Q Max
## -504.93 -18.61 -3.08 9.96 1464.26
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 4.680388 10.080986 0.464 0.642
## inc 0.978252 0.025489 38.379 < 2e-16 ***
## age -2.231489 0.489712 -4.557 5.26e-06 ***
## agesq 0.037722 0.005621 6.710 2.05e-11 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 58.18 on 9271 degrees of freedom
## Multiple R-squared: 0.1731, Adjusted R-squared: 0.1728
## F-statistic: 646.8 on 3 and 9271 DF, p-value: < 2.2e-16