Question: Improving Math SAT scores. Refer to the Chance (Winter 2001) study of students who paid a private tutor (or coach) to help them improve their Scholastic Assessment Test (SAT) scores, Exercise 2.88 (p. 113). Multiple regression was used to estimate the effect of coaching on SAT–Mathematics scores. Data on 3,492 students (573 of whom were coached) were used to fit the model $\mathbf{E}\left(\mathbf{y}\right)\mathbf{=}{\mathbf{\beta }}_{\mathbf{0}}\mathbf{+}{\mathbf{\beta }}_{\mathbf{1}}{\mathbf{x}}_{\mathbf{1}}\mathbf{+}{\mathbf{\beta }}_{\mathbf{2}}{\mathbf{x}}_{\mathbf{2}}$, where y = SAT-Math score, x₁ score on PSAT, and x₂{1 if the student was coached, 0 if not}.

1. The fitted model had an adjusted R² value of .76. Interpret this result.
2. The estimate of${\mathbf{\beta }}_{\mathbf{2}}$ in the model was 19, with a standard error of 3. Use this information to form a confidence interval for${\mathbf{\beta }}_{\mathbf{2}}$ . Interpret the interval.
3. Based on the interval, part b, what can you say about the effect of coaching on SAT–Math scores?
4. As an alternative model, the researcher added several “control” variables, including dummy variables for student ethnicity (x₃,x₄ and x₅ ), a socioeconomic status index variable (x₆) , two variables that measured high school performance (x₇ and x₈₎ , the number of math courses taken in high school (x₉) , and the overall GPA for the math courses (x₁₀) . Write the hypothesized equation for E(y) for the alternative model.
5. Give the null hypothesis for a nested model F-test comparing the initial and alternative models.
6. The nested model F-test, part e, was statistically significant at . Practically interpret this result.
7. The alternative model, part d, resulted in${\mathbf{R}}_{\mathbf{a}}^{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{79}\mathbf{,}{\hat{\mathbf{\beta }}}_{\mathbf{2}}\mathbf{=}\mathbf{14}\mathbf{}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathbf{s}{\hat{\mathbf{\beta }}}_{\mathbf{2}}\mathbf{=}\mathbf{3}$ . Interpret the value of R²_a .
8. Refer to part g. Find and interpret a confidence interval for .
9. The researcher concluded that “the estimated effect of SAT coaching decreases from the baseline model when control variables are added to the model.” Do you agree? Justify your answer.
10. As a modification to the model of part d, the researcher added all possible interactions between the coaching variable (x₂) and the other independent variables in the model. Write the equation for E(y) for this modified model.
11. Give the null hypothesis for comparing the models, parts d and j. How would you perform this test?

The value of R² represents the fraction of the sample variance

of the y-values (measured by SS_yy ) that is explained by the least squares prediction equation.

R² And R²a have similar interpretations. However, unlike $R^2,R_a^2$ takes into account (“adjusts” for) both the sample size n and the number of b parameters in the model.

The value of R² in this question is 0.76, meaning that 76% of the variation in the data is explained by the model, indicating that the model is a good fit for the data.

The confidence interval for ${\beta }_2$ _­can be written as $\left({\hat{\beta }}_2\pm t_{0.025,3491}\times s{\hat{\beta }}_2\right)$

Therefore, the confidence interval for ${\beta }_2$ is $\left(19\pm 1.96\times 3\right)$

95% confidence interval for ${\beta }_2$ is (13.12, 24.88).

Since the confidence interval is positive, it indicates that coached students have scored higher in SAT-Math.

The equation for E(y) for the alternate model can be written as

$E\left(y\right)={\beta }_0+{\beta }_1{x}_1+{\beta }_2{x}_2+{\beta }_3{x}_3+{\beta }_4{x}_4+{\beta }_5{x}_5+{\beta }_6{x}_6+{\beta }_7{x}_7+{\beta }_8{x}_8+{\beta }_9{x}_9+{\beta }_{10}{x}_{10}$

.

At least one of the parameters under test is non-zero.

$H_0\text{:}{\beta }_3={\beta }_4={\beta }_5={\beta }_6={\beta }_7={\beta }_8={\beta }_9={\beta }_{10}$

It is given in the question that the nested model F-test is statistically significant at $\alpha =0.05$ indicating that the alternate model is a better fit for the data.

The alternate model has a${\mathrm{R}}_{\mathrm{a}}^2$ value of 0.79, indicating that 79% of the variation in the data is explained by the model making the model a good fit.

The confidence interval for ${\beta }_2$ _­can be written as $\left({\hat{\beta }}_2\pm t_{0.025,3491}\times s{\hat{\beta }}_2\right)$

Therefore, the confidence interval for ${\beta }_2$ is

95% confidence interval for ${\beta }_2$ is (9.12, 20.88).

The researcher’s conclusion that the estimated effect of SAT coaching decreases from the baseline model when control variables are added to the model is not correct. The alternate model was statistically significant, indicating that the added variables fit better.

The equation for E(y) for the alternate model with interaction can be written as

$$\begin{gathered} E\left(y\right)={\beta }_0+{\beta }_1{x}_1+{\beta }_2{x}_2+{\beta }_3{x}_3+{\beta }_4{x}_4+{\beta }_5{x}_5+{\beta }_6{x}_6+{\beta }_7{x}_7+{\beta }_8{x}_8+{\beta }_9{x}_9+{\beta }_{10}{x}_{10} \\ +{\beta }_{11}{x}_2{x}_1+{\beta }_{12}{x}_2{x}_3+{\beta }_{13}{x}_2{x}_4+{\beta }_{14}{x}_2{x}_5+{\beta }_{15}{x}_2{x}_6+{\beta }_{16}{x}_2{x}_7+{\beta }_{17}{x}_2{x}_8+{\beta }_{18}{x}_2{x}_9+{\beta }_{19}{x}_2{x}_{10} \end{gathered}$$

To compare the two models, an F-test is conducted where the null and alternate hypotheses are $H_0:{\beta }_{11}={\beta }_{12}={\beta }_{13}={\beta }_{14}={\beta }_{15}={\beta }_{16}={\beta }_{17}={\beta }_{18}={\beta }_{19}=0$while

H_a: At least one of the parameters under test is non-zero.

Where, $F-\text{test} \text{statistic}=\frac{SSE}{n-\left(k+1\right)}$.