91Ó°ÊÓ

The dataset OttawaSenators contains information on the number of points and the number of penalty minutes for 24 Ottawa Senators NHL hockey players. Computer output is shown for predicting the number of points from the number of penalty minutes: The regression equation is Points \(=29.53-0.113\) PenMins \(\begin{array}{lrrrr}\text { Predictor } & \text { Coef } & \text { SE Coef } & \text { T } & \text { P } \\ \text { Constant } & 29.53 & 7.06 & 4.18 & 0.000 \\ \text { PenMins } & -0.113 & 0.163 & -0.70 & 0.494\end{array}\) \(\mathrm{S}=21.2985 \quad \mathrm{R}-\mathrm{Sq}=2.15 \% \quad \mathrm{R}-\mathrm{Sq}(\mathrm{adj})=0.00 \%\) Analysis of Variance Source Regression Residual Error Total 2 \(\begin{array}{rrrrr}\text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ 1 & 219.5 & 219.5 & 0.48 & 0.494 \\ 22 & 9979.8 & 453.6 & & \\ 23 & 10199.3 & & & \end{array}\) (a) Write down the equation of the least squares line and use it to predict the number of points for a player with 20 penalty minutes and for a player with 150 penalty minutes. (b) Interpret the slope of the regression equation in context. (c) Give the hypotheses, t-statistic, p-value, and conclusion of the t-test of the slope to determine whether penalty minutes is an effective predictor of number of points. (d) Give the hypotheses, F-statistic, p-value, and conclusion of the ANOVA test to determine whether the regression model is effective at predicting number of points. (e) How do the two p-values from parts (c) and (d) compare? (f) Interpret \(R^{2}\) for this model.

Step by step solution

01

Write down the regression equation and make predictions

Based on the given statistics, the regression equation is \(Points = 29.53 - 0.113 \times Penalty Minutes\). Use this equation to predict the number of points for a player with 20 penalty minutes: \(Points = 29.53 - 0.113 \times 20 = 27.27 \) and for a player with 150 penalty minutes: \(Points = 29.53 - 0.113 \times 150 = 12.48 \)

02

Interpret the slope

The slope of the regression equation is -0.113. This means that for every additional penalty minute, the predicted number of points decreases by 0.113 points.

03

Conduct the t-test

The null hypothesis for the t-test is that the slope equals 0, and the alternative hypothesis is that the slope does not equal 0. The t-statistic is -0.70 and the p-value is 0.494. Since the p-value is greater than 0.05, we fail to reject the null hypothesis and conclude that penalty minutes is not a statistically significant predictor of the number of points.

04

Conduct the ANOVA test

The null hypothesis for the ANOVA test is that all coefficients in the model, except the intercept, are 0. The alternative hypothesis is that at least one is not 0. The F-statistic is 0.48 and the p-value is 0.494. Again, because the p-value is greater than 0.05, we fail to reject the null hypothesis and conclude that the regression model is not statistically significant at predicting the number of points.

05

Compare the two p-values

The p-values for the t-test and the ANOVA test are the same (0.494). This is because these tests are equivalent when there is only one predictor variable in the model, as is the case here.

06

Interpret \(R^{2}\)

The \(R^{2}\) value is 2.15%. This means that 2.15% of the variation in the number of points can be explained by the number of penalty minutes. Because the \(R^{2}\) value is very small, penalty minutes isn't a good predictor of the number of points.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Least Squares Regression Line

The least squares regression line is a fundamental tool in statistical analysis that represents the best-fitting straight line through a set of points on a scatter plot. This line minimizes the sum of the squares of the vertical distances (residuals) of the points from the line. In the context of our textbook exercise, the regression equation given is \(Points = 29.53 - 0.113 \times Penalty Minutes\).

This equation allows us to predict the number of points a hockey player might score based on the number of penalty minutes they have accrued. For 20 penalty minutes, the prediction would be 27.27 points, while for 150 penalty minutes, it would be 12.48 points. These predictions hinge on the assumption that the relationship between penalty minutes and points is linear and can be captured by the slope and intercept defined in the regression equation. The slope, in this case, indicates a small negative relationship, meaning that more penalty minutes might slightly lower the number of points scored by a player.

Hypothesis Testing

Hypothesis testing is a statistical method used to decide whether to accept or reject a hypothesis made about a population parameter based on sample data. In our example, a t-test is used to test the effectiveness of penalty minutes in predicting the number of points a player scores.

The null hypothesis \(H_0:\theta = 0\) proposes that penalty minutes, represented by the slope (\theta), have no effect on the points scored. The alternative hypothesis \(H_A:\theta eq 0\) suggests a significant effect. The t-statistic calculated in the exercise is -0.70, and the p-value is 0.494, which is higher than the common significance level of 0.05. Hence, the evidence is insufficient to reject the null hypothesis, leading to the conclusion that penalty minutes are not an effective predictor of the number of points.

ANOVA

Analysis of Variance (ANOVA) is a technique used to compare more than two datasets to understand if at least one mean difference among them is statistically significant. While our textbook problem involves only one independent variable, making ANOVA somewhat comparable to the t-test, ANOVA becomes particularly powerful with multiple predictors.

In the given exercise, the ANOVA test generates an F-statistic of 0.48 and a p-value of 0.494. This again exceeds the alpha level of 0.05, leading to the conclusion that the regression model with penalty minutes as the predictor does not significantly predict variation in the number of points scored. The ANOVA here tests the overall significance of the model, vetting if the penalty minutes collectively contribute to explaining points variability.

Statistical Significance

The concept of statistical significance helps determine whether the observed results are likely due to a real effect or simply random variation. This is commonly assessed with a p-value, which is the probability of observing the results (or more extreme) given that the null hypothesis is true.

A low p-value (<0.05) suggests that the findings are unlikely under the null hypothesis, warranting its rejection. In the Ottawa Senators exercise, both the t-test and the ANOVA result in a p-value of 0.494, which is not low enough to be considered statistically significant. Thus, the data does not provide strong evidence against the null hypotheses, and we cannot claim a meaningful relationship between penalty minutes and points scored. The same principle applies across various statistical tests and is a cornerstone for making inferences in many fields.