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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. $$ \begin{aligned} &\text { Response: }\\\ &\begin{array}{lrrrrr} & \text { Df } & \text { Sum Sq } & \text { Mean Sq } & \text { F value } & \operatorname{Pr}(>\text { F) } \\ \text { ModelB } & 1 & 10.380 & 10.380 & 2.18 & 0.141 \\ \text { Residuals } & 342 & 1630.570 & 4.768 & & \\ \text { Total } & 343 & 1640.951 & & & \end{array} \end{aligned} $$

Short Answer

Expert verified
The null hypothesis (all model coefficients are zero) and the alternative hypothesis (at least one coefficient is not zero) are tested. The F-statistic is 2.18, and the p-value is 0.141. Because the p-value is larger than 0.05, we fail to reject the null hypothesis, concluding that the regression model doesn't show statistical significance.

Step by step solution

01

State the Hypotheses

The Null Hypothesis \(H_0\) is that all coefficients of the regression model are zero, meaning the model is not significant. The Alternative Hypothesis \(H_1\) is that at least one coefficient is different from zero, implying the model is significant.
02

Identify the F-Statistic

The given F-statistic in the ANOVA table is 2.18. This value is a measure of how much variation in the response variable can be explained by the predictors in the regression model.
03

Identify the p-value

The p-value is given as 0.141. This value provides the significance level of the F-value in testing the null hypothesis. In other words, it shows what probability there is of observing a value as extreme as the test statistic, assuming the null hypothesis is true.
04

Conclusion of the Test

If the p-value is less than the chosen significance level (usually 0.05), we reject the null hypothesis. Since the p-value of 0.141 is larger than 0.05, we cannot reject the null hypothesis. Hence, we conclude that the model does not have statistically significance.

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Key Concepts

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

Null Hypothesis in ANOVA
The null hypothesis in Analysis of Variance (ANOVA) for regression is a vital component for understanding how the regression analysis is evaluated. In the context of regression, the null hypothesis (\(H_0\)) asserts that the independent variables in the model do not significantly explain the variation in the dependent variable. Essentially, it suggests that any observed correlations in the data could be due to random chance rather than a meaningful relationship.

To use our exercise as an instance, the null hypothesis proposes that the coefficients of the regression model are zero, indicating that the model doesn't significantly predict or explain the response variable's behavior. Refuting this hypothesis would mean that your regression model provides valuable insights into how the predictors affect the output.
F-Statistic
The F-statistic plays a central role in determining whether your regression model is a good fit for the data. It is calculated by dividing the model mean square (the variance explained by the model) by the residual mean square (variance unexplained by the model). The higher the value of the F-statistic, the more likely it is that the observed variation in the response variable is not due to chance.

In the example provided, the F-statistic is 2.18. This would be used to compare against a critical value from the F-distribution based on specific degrees of freedom, which correspond to the number of predictors and the number of observations in the data, respectively. A significant F-statistic supports the alternative hypothesis, hinting that the model imparts a significant explanatory power over the variation of the response variable.
p-value in ANOVA
The p-value is intimately connected with the F-statistic and helps us make empirical decisions about our hypotheses. It is the probability of finding the observed results when the null hypothesis is true. In other words, it shows how extreme the obtained results are if we assume there was no effect.

The given p-value in our exercise example is 0.141. When the p-value is less than the conventional threshold of 0.05, we say that the results are 'statistically significant,' and it would be reasonable to reject the null hypothesis. However, with a p-value higher than 0.05, as in this case, we do not have sufficient evidence to dismiss the null hypothesis, and we would conclude that the model may not be a significant predictor of the response variable.
Statistical Significance
Statistical significance is a determination of whether the results obtained from a test, such as ANOVA, reflect a true effect or if they are likely due to random variance. It is often represented through a p-value. If a p-value is lower than our pre-defined level (usually 0.05), the results are deemed statistically significant, suggesting strong evidence against the null hypothesis.

In ANOVA for regression, as articulated in our example, reaching statistical significance means we can confidently state that our model provides a better explanation for variation than mere chance. Since the p-value exceeded the 0.05 threshold, our results do not reach statistical significance, and thus, the variable(s) in question do not provide a significant predictive value for the dependent variable.

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Most popular questions from this chapter

Show some computer output for fitting simple linear models. State the value of the sample slope for each model and give the null and alternative hypotheses for testing if the slope in the population is different from zero. Identify the p-value and use it (and a \(5 \%\) significance level) to make a clear conclusion about the effectiveness of the model.$$ \begin{array}{lrrrr} \text { Coefficients: } & \text { Estimate } & \text { Std.Error } & \mathrm{t} \text { value } & \operatorname{Pr}(>|\mathrm{t}|) \\ \text { (Intercept) } & 7.277 & 1.167 & 6.24 & 0.000 \\ \text { Dose } & -0.3560 & 0.2007 & -1.77 & 0.087 \end{array} $$

Test the correlation, as indicated. Show all details of the test. Test for evidence of a linear association; \(r=0.28 ; n=10\)

Show some computer output for fitting simple linear models. State the value of the sample slope for each model and give the null and alternative hypotheses for testing if the slope in the population is different from zero. Identify the p-value and use it (and a \(5 \%\) significance level) to make a clear conclusion about the effectiveness of the model.$$ \begin{array}{lrrrr} \text { Coefficients: } & \text { Estimate } & \text { Std.Error } & \mathrm{t} \text { value } & \operatorname{Pr}(>|\mathrm{t}|) \\ \text { (Intercept) } & 807.79 & 87.78 & 9.30 & 0.000 \\ \mathrm{~A} & -3.659 & 1.199 & -3.05 & 0.006 \end{array} $$

in Fish In Exercise 9.21 , we see that the conditions are met for using the \(\mathrm{pH}\) of a lake in Florida to predict the mercury level of fish in the lake. The data are given in FloridaLakes. Computer output is shown for the linear model with several values missing: The regression equation is AvgMercury \(=1.53-0.152 \mathrm{pH}\) Predictor Constant \(\mathrm{pH}\) \(\begin{array}{rrrr}\text { Coef } & \text { SE Coef } & \text { T } & \text { P } \\ 1.5309 & 0.2035 & 7.52 & 0.000 \\ -0.15230 & \text { * (c) }^{* *} & -5.02 & 0.000\end{array}\) stant \(S=* *(b)^{* *} \quad R-S q=* *(a)^{* x}\) Analysis of Variance Source Regression Residual Erri Total \(\begin{array}{rrrrrr} & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { ion } & 1 & 2.0024 & 2.0024 & 25.24 & 0.000 \\ \text { Error } & 51 & 4.0455 & 0.0793 & & \\ & 52 & 6.0479 & & & \end{array}\) (a) Use the information in the ANOVA table to compute and interpret the value of \(R^{2}\). (b) Show how to estimate the standard deviation of the error term, \(s_{\epsilon}\). (c) Use the result from part (b) and the summary statistics below to compute the standard error of the slope, \(S E,\) for this model: $$ \begin{array}{lrrrrr} \text { Variable } & \text { N } & \text { Mean } & \text { StDev } & \text { Minimum } & \text { Maximum } \\ \text { pH } & 53 & 6.591 & 1.288 & 3.600 & 9.100 \\ \text { AvgMercury } & 53 & 0.5272 & 0.3410 & 0.0400 & 1.3300 \end{array} $$

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.

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