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We show an ANOVA table for regression. State the hypotheses of the test, give the F-statistic and the p-value, and state the conclusion of the test. $$ \begin{aligned} &\text { Response: } Y\\\ &\begin{array}{lrrrrr} & \text { Df } & \text { Sum Sq } & \text { Mean Sq } & \mathrm{F} \text { value } & \operatorname{Pr}(>\mathrm{F}) \\ \text { ModelA } & 1 & 352.97 & 352.97 & 11.01 & 0.001 * * \\ \text { Residuals } & 359 & 11511.22 & 32.06 & & \\ \text { Total } & 360 & 11864.20 & & & \end{array} \end{aligned} $$

Short Answer

Expert verified
The hypotheses for the test were \(H_0\): All coefficients of ModelA are equal to zero, and \(H_A\): At least one coefficient of ModelA is different from zero. The F-Statistic was 11.01. The P-Value calculated was 0.001, which is less than the significance level of 0.05. As a result, the null hypothesis can be rejected in favor of the alternative hypothesis, providing evidence that at least one coefficient in ModelA is non-zero.

Step by step solution

01

Formulate the Hypotheses

For any hypothesis test, we will always have a null hypothesis (\(H_0\)) and an alternative hypothesis (\(H_1\) or \(H_A\)). In this context, the null hypothesis (\(H_0\)) is that the coefficients associated with ModelA are equal to zero. The alternative hypothesis (\(H_A\)) is that at least one coefficient associated with ModelA is different from zero.
02

Calculate the F-Statistics

From the ANOVA table, we know that the F-statistic is a ratio of the Model Mean Square to the Residual Mean Square. The Model Mean Square is 352.97 and the Residual Mean Square is 32.06. The F-statistic is given directly in the ANOVA table as 11.01.
03

Analyze the P-Value

In hypothesis testing, the P-value helps us decide whether our sample gives strong evidence against the null hypothesis. From the ANOVA table, the P-value is 0.001.
04

State the Conclusion Based on the P-Value and F-Statistic

If our P-value is less than our significance level (assume the significance level, \(\alpha\), is 0.05), we reject the null hypothesis. In our example, the P-value is 0.001, which is less than the \(\alpha\), so we reject the null hypothesis and conclude that there's evidence that at least one coefficient of ModelA is different from zero.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a core concept in statistics, which allows researchers to make inferences about populations based on sample data. The process involves making an assumption called the null hypothesis, represented as \(H_0\), which suggests that there is no effect or no difference, and then determining if the data collected provides sufficient evidence to reject this assumption in favor of an alternative hypothesis, represented as \(H_A\) or \(H_1\).
In the case of ANOVA for regression, the hypothesis test is used to determine whether there is a statistically significant relationship between the independent and dependent variables. Very simply put, we want to know if our regression model explains the variability in the data better than if we had no model at all. If our analysis leads us to reject the null hypothesis, we can conclude that the model provides useful information about the relationship between the variables.
F-statistic
The F-statistic is a key output of ANOVA tests and forms the basis for deciding if the observed relationships in the data are statistically significant. It is calculated by dividing the variance explained by the model (Mean Square of the Model) by the variance unexplained by the model (Mean Square of the Residuals). This ratio forms the F-statistic, which follows an F-distribution under the null hypothesis.
In our example, the F-statistic is \(11.01\), derived from dividing the Model Mean Square \(352.97\) by the Residual Mean Square \(32.06\), indicating how many times the explained variance is greater than the unexplained variance. A higher F-statistic often implies that the model explains a significant amount of the variance in the dependent variable.
P-value
The p-value represents the probability that the observed data, or something more extreme, would occur if the null hypothesis were true. In other words, it quantifies how likely it is to obtain at least as strong evidence against \(H_0\) if, in fact, \(H_0\) were true. The smaller the p-value, the greater the statistical evidence against the null hypothesis.
In hypothesis testing, we compare the p-value to a predetermined significance level, often \(\alpha = 0.05\), to decide whether to reject \(H_0\). If the p-value is less than \(\alpha\), we reject the null hypothesis. With a p-value of \(0.001\) in the provided ANOVA table, we have very strong evidence against the null hypothesis, since \(0.001 < 0.05\), leading us to believe that our model is indeed capturing a true relationship in the population.
Null and Alternative Hypotheses
The null and alternative hypotheses are the two mutually exclusive statements that hypothesis testing seeks to evaluate. The null hypothesis \(H_0\) often postulates a default state of no effect or no difference; for instance, it might state that the regression coefficients are zero (implying no relationship between the variables). The alternative hypothesis \(H_A\) or \(H_1\) is what you suspect might be true instead—in this case, that at least one regression coefficient is not zero (indicating a potential relationship).
The hypothesis test does not prove an alternative hypothesis; rather, it simply assesses whether there is enough evidence to reject the null hypothesis. When we reject \(H_0\), as in our example, we are saying that the data are inconsistent with \(H_0\) being true, and thus we have evidence to support \(H_A\).

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Most popular questions from this chapter

We show an ANOVA table for regression. State the hypotheses of the test, give the F-statistic and the p-value, and state the conclusion of the test. $$ \begin{array}{lrrrr} \text { Analysis of Variance } & & & & \\ \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Regression } & 1 & 3396.8 & 3396.8 & 21.85 & 0.000 \\ \text { Residual Error } & 174 & 27053.7 & 155.5 & & \\ \text { Total } & 175 & 30450.5 & & & \end{array} $$

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A common (and hotly debated) saying among sports fans is "Defense wins championships." Is offensive scoring ability or defensive stinginess a better indicator of a team's success? To investigate this question we'll use data from the \(2015-2016\) National Basketball Association (NBA) regular season. The data \(^{6}\) stored in NBAStandings2016 include each team's record (wins, losses, and winning percentage) along with the average number of points the team scored per game (PtsFor) and average number of points scored against them ( PtsAgainst). (a) Examine scatterplots for predicting \(\operatorname{WinPct}\) using PtsFor and predicting WinPct using PtsAgainst. In each case, discuss whether conditions for fitting a linear model appear to be met. (b) Fit a model to predict winning percentage (WinPct) using offensive ability (PtsFor). Write down the prediction equation and comment on whether PtsFor is an effective predictor. (c) Repeat the process of part (b) using PtsAgainst as the predictor. (d) Compare and interpret \(R^{2}\) for both models. (e) The Golden State Warriors set an NBA record by winning 73 games in the regular season and only losing 9 (WinPct \(=0.890\) ). They scored an average of 114.9 points per game while giving up an average of 104.1 points against. Find the predicted winning percentage for the Warriors using each of the models in (b) and (c). (f) Overall, does one of the predictors, PtsFor or PtsAgainst, appear to be more effective at explaining winning percentages for NBA teams? Give some justification for your answer.

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