Ch 3 and 12 Test Review: Regression and Inference

Least-Squares Regression Line (LSRL) Equation: $\hat{AR} = 5.81 - 0.19(\text{Hours})$ .
Residual Calculation: For the point $(3, 10)$ , the predicted value is $\hat{y} = 5.81 - 0.19(3) = 5.24$ . The residual is $\text{observed} - \text{predicted} = 10 - 5.24 = 4.76$ .
Impact of Removing Points: * Large Residual Point: Removing a point with a large residual (e.g., $(3, 10)$ ) results in the slope ( $b$ ) remaining about the same, the correlation ( $r$ ) increasing, and the typical residual size ( $s$ ) decreasing. * Influential Outlier: Removing an outlier in the explanatory variable (e.g., $(15, 2)$ ) causes the slope ( $b$ ) to change significantly and the correlation ( $r$ ) to increase.

LSRL Context: $\hat{GPA} = 2.6476 + 0.10176(\text{SLEEP HOURS})$ .
Slope Interpretation: For each additional hour of sleep, a student's $\text{GPA}$ is predicted to increase by $0.10176$ points.
Standard Deviation of Residuals ( $s$ ): The value $s = 0.1804$ represents the typical prediction error or the typical residual in $\text{GPA}$ points.
Hypothesis Test for Slope: * Hypotheses: $H_0: \beta_1 = 0$ vs. H_a: \beta_1 > 0. * Degrees of Freedom: $df = n - 2 = 14 - 2 = 12$ . * Test Statistic: $t(12) = \frac{0.10176 - 0}{0.04347} = 2.341$ .
* P-value: $P(t(12) \geq 2.341) = 0.0187$ . * Conclusion: Since 0.0187 < 0.05, reject $H_0$ . There is convincing evidence of a positive linear relationship between hours of sleep and $\text{GPA}$ .
95% Confidence Interval for Slope: * Formula: $b_1 \pm t(12)^\times \times SE_{b_1}$ . * Calculation: $0.10176 \pm 2.179 \times 0.04347 \rightarrow (0.007, 0.196)$ . * Interpretation: We are $95\%$ confident that the true slope of the LSRL for $\text{GPA}$ vs Hours of Sleep is between $0.007$ and $0.196$ points/hour.

Linear Model: $\hat{\text{Area}} = -4.133 + 3.0333(\text{seconds})$ . Prediction at $5.5\,\text{seconds}$ is $12.55\,cm^2$ .
Model Selection: The Exponential Model (Model 1) is superior because it has the highest $r^2$ and the residual plot lacks a clearly curved pattern.
Exponential Model Equation: $\log(\hat{\text{Area}}) = 0.34057 + 0.115582(\text{seconds})$ .
Transformed Prediction: To predict area at $5.5\,\text{seconds}$ , calculate $\log(\hat{\text{Area}}) = 0.34057 + 0.115582(5.5) = 0.976271$ . Solving for $\text{Area}$ gives $\text{Area} = 10^{0.976271} = 9.468\,cm^2$ .