![[The Homework Assignments]](_images/icon-assignments.svg "homework"){kind=icon}
Part I: Ordinary Least Squares Regression
In this part of the homework, let us be completely in the realm of the classical linear model (CLM) on a simple linear model. Thus, the model is
\[ y = \beta_0 + \beta_1 x + \varepsilon \]
and the assumption is
\[ \varepsilon \stackrel{\text{iid}}{\sim} \text{Normal}(0;\ \sigma^2) \]
In this part, you will show that MSE is an unbiased estimator of \(\sigma^2\). The extra credit problem at the end will have you do the same, but in a different way.
Problem 1: MSE Unbiasedness, the Start
Show that the sum of squared errors, SSE, can be written as
\[ \text{SSE} = S_{yy} - \hat{\beta}_1\ S_{xy} \]
Problem 2: MSE Unbiasedness, the Sequel
Show that the sum of squared errors, SSE, can be written as
\[ \mathbb{E}\left[\text{SSE}\right] = (n-2) \sigma^2 \]
Problem 3: MSE Unbiasedness, the Conclusion
Finally, in one line, show that
\[ \text{MSE} := \frac{ \text{SSE} }{n-2} \]
is an unbiased estimator of 2.
Part II: Applied OLS Regression
In this part, you need to use the crime dataset to answer these questions. The crime data file can be found in the expected place:
https://rfs.kvasaheim.com/data/crime.csv
Problem 4: Modeling
Fit a model for the average school enrollment in 2000 (enroll00) using the gross state product (GSP) per capita in 1990 (gspcap90). Check the assumptions of the regression. Even if a requirement is violated, pretend it is not and report the p-value for the Shapiro-Wilk test.
Problem 5: Effects
Report a point estimate and a 95% confidence interval for the effect of GSP per capita in 1990 on the enrollment rate.
Problem 6: Estimation
Estimate the expected enrollment rate for a state with a GSP per capita of $60,000. Also, provide a 95% confidence interval.
Problem 7: Predictons
Predict the enrollment rate for a state with a GSP per capita of $60,000. Also, provide a 95% prediction interval.
The Extra Credit Problem: There is Another…
Above, you proved that the MSE was an unbiased estimator of \(\sigma^2\). Here, you will do it a different way.
Under the CLM assumptions, prove
\[ \frac{(n-2)\ \text{MSE}}{\sigma^2} \sim \chi^2_{n-2} \]
Then use the fact that the expected value of a Chi-square random variable is its degrees of freedom.