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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. $$ \begin{array}{lrrrr} \text { Analysis of Variance } & & & & \\ \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Regression } & 1 & 3396.8 & 3396.8 & 21.85 & 0.000 \\ \text { Residual Error } & 174 & 27053.7 & 155.5 & & \\ \text { Total } & 175 & 30450.5 & & & \end{array} $$

Short Answer

Expert verified
The null hypothesis that there's no significant relationship in the model is rejected based on the F-statistic of 21.85 and the corresponding p-value of nearly 0.

Step by step solution

01

State the Hypotheses

In a test for regression, we aim to establish whether there is a significant relationship between our independent and dependent variables. The Null Hypothesis \(H_0\) states that there is no significant relationship (i.e. all model coefficients equal to zero), while the Alternative Hypothesis \(H_1\) contends that at least one coefficient does not equal zero, implying there is some relationship.
02

Calculate the F-statistic

The F-statistic is given in the table as 21.85. This value represents the ratio of the Mean Square Regression (MSR) to the Mean Square Error (MSE). In this case, it was calculated as \(21.85 = \frac{MSR}{MSE} = \frac{3396.8}{155.5}\), comparing the explained variance to the unexplained variance.
03

Determine the p-value

The p-value is given in the table as 0.000, meaning that the probability of getting an F-statistic as extreme as 21.85, given that the null hypothesis is true, is extremely low, essentially zero.
04

State the Conclusion

As the p-value is less than the typical threshold of 0.05, we reject the null hypothesis in favor of the alternative. This indicates that there is a significant relationship between the independent and dependent variables.

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Key Concepts

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

Null Hypothesis
The null hypothesis (\(H_0\)) in the context of ANOVA for regression is a foundational concept that serves as the starting point for any hypothesis testing. It posits that there is no effect or no significant relationship between the independent variables (predictors) and the dependent variable (outcome). In simpler terms, it assumes that any observed differences in means across groups are due to random chance alone.

In regression analysis, the null hypothesis specifically states that the coefficients of the independent variables are all equal to zero, indicating that the variables have no influence on the predicted outcome. The null hypothesis is what we assume to be true until evidence suggests otherwise. Therefore, in our case, the null hypothesis would assert that the independent variable does not significantly predict the dependent variable.
F-statistic
The F-statistic is a crucial value when conducting ANOVA for regression as it assesses whether the group means are significantly different from each other. It is calculated based on the ratio of variation between the groups (Mean Square Regression, or MSR) to the variation within the groups (Mean Square Error, or MSE). The larger the F-statistic, the more likely it is that the observed variance is due to the independent variable having a genuine effect, rather than random variation.

In our exercise, the F-statistic is calculated as 21.85, suggesting a relatively high ratio of explained to unexplained variance, giving the first indication that our independent variable might indeed have an effect on our dependent variable. Essentially, a high F-statistic prompts us to suspect that the null hypothesis may not be true.
p-value
The p-value is a critical concept often misunderstood by students. It represents the probability of obtaining a test statistic (like the F-statistic) as extreme as the observed one, under the assumption that the null hypothesis is true. The p-value helps us make decisions about our hypotheses.

Commonly, if the p-value is less than 0.05 (5%), the result is considered statistically significant, and we have reason to reject the null hypothesis. In our ANOVA table, the p-value is 0.000, essentially zero, which strongly suggests that the null hypothesis is false and that there is a significant relationship between the variables in our regression model.
Mean Square Regression
Mean Square Regression (MSR) is an essential element of the ANOVA table that represents the average amount of variance in the dependent variable explained by the independent variable(s) in the regression model. It is calculated by dividing the sum of squares due to regression (SSR) by the degrees of freedom associated with the model's regression (DF).

The MSR in our example is 3396.8, indicating that, on average, each unit of the independent variable accounts for 3396.8 units of variance in the dependent variable. A higher MSR means more of the variance is explained by the regression model, pointing to a potentially significant relationship.
Alternative Hypothesis
The alternative hypothesis (\(H_1\) or (\(H_a\) is the hypothesis that contradicts the null hypothesis. In ANOVA for regression, the alternative hypothesis posits that there is a significant effect or relationship between the independent variable(s) and the dependent variable. It implies that the observed differences are too large to be due to random chance alone.

In our specific case, the alternative hypothesis states that at least one of the independent variables' coefficients is not equal to zero, suggesting a real effect on the dependent variable. The evidence we gather from our p-value and F-statistic leads us to reject the null hypothesis and support the alternative hypothesis, affirming that there is a significant relationship present.

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Most popular questions from this chapter

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

Use the computer output (from different computer packages) to estimate the intercept \(\beta_{0},\) the slope \(\beta_{1},\) and to give the equation for the least squares line for the sample. Assume the response variable is \(Y\) in each case. $$ \begin{array}{lrrrr} \text { Coefficients: } & \text { Estimate } & \text { Std.Error } & \mathrm{t} \text { value } & \operatorname{Pr}(>|\mathrm{t}|) \\ \text { (Intercept) } & 77.44 & 14.43 & 5.37 & 0.000 \\ \text { Score } & -15.904 & 5.721 & -2.78 & 0.012 \end{array} $$

Data 9.1 on page 577 introduces the dataset InkjetPrinters, which includes information on all-in-one printers. Two of the variables are Price (the price of the printer in dollars) and CostColor (average cost per page in cents for printing in color). Computer output for predicting the price from the cost of printing in color is shown: $$ \begin{aligned} &\text { The regression equation is Price }=378-18.6 \text { CostColor }\\\ &\begin{array}{lrrrrr} \text { Analysis of Variance } & & & & \\ \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Regression } & 1 & 57604 & 57604 & 13.19 & 0.002 \\ \text { Residual Error } & 18 & 78633 & 4369 & & \\ \text { Total } & 19 & 136237 & & & \end{array} \end{aligned} $$ (a) What is the predicted price of a printer that costs 10 cents a page for color printing? (b) According to the model, does it tend to cost more or less (per page) to do color printing on a cheaper printer? (c) Use the information in the ANOVA table to determine the number of printers included in the dataset. (d) Use the information in the ANOVA table to compute and interpret \(R^{2}\). (e) Is the linear model effective at predicting the price of a printer? Use information from the computer output and state the conclusion in context.

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.$$ \begin{array}{lrrrr} \text { Coefficients: } & \text { Estimate } & \text { Std.Error } & \mathrm{t} \text { value } & \operatorname{Pr}(>|\mathrm{t}|) \\ \text { (Intercept) } & 807.79 & 87.78 & 9.30 & 0.000 \\ \mathrm{~A} & -3.659 & 1.199 & -3.05 & 0.006 \end{array} $$

Test the correlation, as indicated. Show all details of the test. Test for evidence of a linear association; \(r=0.28 ; n=100\).

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