1. **State the problem:** We want to test if there is a linear relationship between GPA and starting salary using the given data. 2. **Formula used:** The test statistic for the linear relationship is based on the correlation coefficient $r$. The test statistic $t$ is given by: $$t = \frac{r\sqrt{n-2}}{\sqrt{1-r^2}}$$ where $n$ is the number of data points. 3. **Calculate the correlation coefficient $r$:** Given data points: GPA: $[2.3, 3.9, 3.2, 3.5, 3.8, 2.5, 3.3, 3.9, 2.2, 2.8]$ Salary: $[44.1, 33.8, 42.6, 28.4, 34.3, 31.3, 40.8, 41.2, 26.2, 28.9]$ Calculate means: $$\bar{x} = \frac{\sum x_i}{n} = \frac{2.3+3.9+3.2+3.5+3.8+2.5+3.3+3.9+2.2+2.8}{10} = 3.14$$ $$\bar{y} = \frac{\sum y_i}{n} = \frac{44.1+33.8+42.6+28.4+34.3+31.3+40.8+41.2+26.2+28.9}{10} = 35.16$$ Calculate covariance and variances: $$S_{xy} = \sum (x_i - \bar{x})(y_i - \bar{y}) = -10.462$$ $$S_{xx} = \sum (x_i - \bar{x})^2 = 4.004$$ $$S_{yy} = \sum (y_i - \bar{y})^2 = 292.424$$ Calculate correlation coefficient: $$r = \frac{S_{xy}}{\sqrt{S_{xx} S_{yy}}} = \frac{-10.462}{\sqrt{4.004 \times 292.424}} = \frac{-10.462}{34.25} = -0.31$$ 4. **Calculate the test statistic $t$:** $$n = 10$$ $$t = \frac{-0.31 \times \sqrt{10-2}}{\sqrt{1 - (-0.31)^2}} = \frac{-0.31 \times \sqrt{8}}{\sqrt{1 - 0.0961}} = \frac{-0.31 \times 2.828}{\sqrt{0.9039}} = \frac{-0.877}{0.951} = -0.92$$ 5. **Interpretation:** The test statistic value is approximately $-0.92$. **Final answer:** The value of the test statistic is $-0.92$.