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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. $$ \begin{array}{l} \text { Analysis of Variance } \\ \begin{array}{lrrrr} \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Regression } & 1 & 303.7 & 303.7 & 1.75 & 0.187 \\ \text { Residual Error } & 174 & 30146.8 & 173.3 & & \\ \text { Total } & 175 & 30450.5 & & & \end{array} \end{array} $$

Short Answer

Expert verified
The hypotheses for the test is that the Null Hypothesis \(H_0\) posits no relation between variables, while the Alternative Hypothesis \(H_1\) asserts there is a relation. The given F-statistic is 1.75, the p-value is 0.187. As the p-value > 0.05, we do not reject the null hypothesis, concluding that there's no significant evidence to claim a relationship between variables in the regression test.

Step by step solution

01

Identify Hypothesis

From the given table, since we're looking at a regression test, our Null Hypothesis \(H_0\) is that there is no relation between the variables (i.e., the coefficients are equal to zero). The Alternative Hypothesis \(H_1\) is that there is a relation (i.e., the coefficients are not zero).
02

Obtain F-statistic

The F-statistic value is presented directly in the ANOVA table under the 'F' column in the 'Regression' row. From the table, fe can see that the F-statistic is 1.75.
03

Obtain P-value and perform test conclusion

The p-value for the F-statistic is reported in the 'P' column of the 'Regression' row. Here it is 0.187. Since the p-value is greater than the threshold (usual threshold is 0.05), we cannot reject the null hypothesis. That means, based on the data in the ANOVA table, there is no significant evidence to claim that there is a relation between the variables in the regression test.

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Key Concepts

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

Null Hypothesis
Understanding the null hypothesis is foundational to any statistical analysis. It's a presumption that there is no effect or no association between variables until evidence suggests otherwise. In the context of ANOVA regression analysis, the null hypothesis (\( H_0 \)) posits that the predictive factors being tested have no impact on the dependent variable—that is, any observed differences are merely due to random chance.

In the exercise provided, the null hypothesis specifically states that there is no relationship between the independent variable(s) and the dependent variable, implying that the coefficients in the regression model are equal to zero. Rejection of this hypothesis would suggest that there is at least one coefficient in the model that is significantly different from zero, indicating a relationship worth consideration.
F-statistic
The F-statistic is a crucial measure in regression analysis, serving as the test statistic for ANOVA. It is calculated as a ratio of two variances: the variance explained by the model (between-group variance) and the variance that is unexplained by the model (within-group variance).

In simpler terms, the F-statistic compares the collective effect of all the variables included in the model relative to the variation not explained by the model. A higher F-statistic indicates a greater degree of explained variance by the variables in the model. In the given exercise, the F-statistic is reported as 1.75. This number needs to be evaluated against a critical value or assessed through its p-value to understand if the observed ratio of variances is statistically significant.
p-value
The p-value tells us about the likelihood of the observed data given that the null hypothesis is true. It provides the probability of obtaining test results at least as extreme as the ones observed during the test, assuming that the null hypothesis is correct.

In the context of ANOVA and regression analysis, the p-value helps determine whether to reject or fail to reject the null hypothesis. A low p-value (typically <0.05) indicates strong evidence against the null hypothesis, thus favoring the alternative. However, in our exercise, the p-value is 0.187, which is above the common alpha level of 0.05. This means there isn't enough statistical evidence to reject the null hypothesis, and we conclude that the model does not significantly predict the outcome variable.
Regression Test
A regression test in statistics examines the relationship between a dependent variable and one or more independent variables. The aim is to find out if changes in the predictors are significantly associated with changes in the response variable.

When performing an ANOVA regression analysis, as seen in the given exercise, the regression test involves comparing the calculated F-statistic against a critical value or assessing the corresponding p-value to determine if the overall regression model is significant. If the evidence suggests that the null hypothesis can be rejected, the test indicates that the model explains a significant portion of the variability in the dependent variable. In our exercise, however, the conclusion is that the regression model does not significantly predict the outcome since the p-value did not reach the commonly accepted level of significance.

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Most popular questions from this chapter

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. $$ \begin{aligned} &\text { The regression equation is }\\\ &Y=808-3.66 \mathrm{~A}\\\ &\begin{array}{lrrrr} \text { Predictor } & \text { Coef } & \text { SE Coef } & \text { T } & \text { P } \\ \text { Constant } & 807.79 & 87.78 & 9.20 & 0.000 \\ \text { A } & -3.659 & 1.199 & -3.05 & 0.006 \end{array} \end{aligned} $$

FIBER IN CEREALS AS A PREDICTOR OF CALORIES In Example 9.10 on page \(592,\) we look at a model to predict the number of calories in a cup of breakfast cereal using the number of grams of sugars. In Exercises 9.64 and 9.65 , we give computer output with two regression intervals and information about a specific amount of sugar. Interpret each of the intervals in the context of this data situation. (a) The \(95 \%\) confidence interval for the mean response (b) The \(95 \%\) prediction interval for the response The intervals given are for cereals with 16 grams of sugars: Sugars 95 \(\mathrm{P}\) \(\begin{array}{rrr}\text { rs Fit } & \text { SE Fit } \\\ 6 & 157.88 & 7.10 & \text { (143.3 }\end{array}\) \(95 \% \mathrm{Cl}\) 35,172.42) \(9 \%\) \(\begin{array}{lllll}16 & 15788 & 7.10 & (143.35,172.42) & (101.46\end{array}\) 214.31)

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

In Exercises 9.62 and 9.63 , we give computer output with two regression intervals and information about the percent of calories eaten during the day. Interpret each of the intervals in the context of this data situation. (a) The \(95 \%\) confidence interval for the mean response (b) The \(95 \%\) prediction interval for the response The intervals given are for mice that eat \(50 \%\) of their calories during the day: \(\begin{array}{rrrrr}\text { DayPct } & \text { Fit } & \text { SE Fit } & 95 \% \mathrm{Cl} & 95 \% \mathrm{PI} \\ 50.0 & 7.476 & 0.457 & (6.535,8.417) & (2.786,12.166)\end{array}\)

Use this information to fill in all values in an analysis of variance for regression table as shown. $$ \begin{array}{|l|l|l|l|l|l|} \hline \text { Source } & \text { df } & \text { SS } & \text { MS } & \text { F-statistic } & \text { p-value } \\ \hline \text { Model } & & & & & \\ \hline \text { Error } & & & & & \\ \hline \text { Total } & & & & & \\ \hline \end{array} $$ SSModel \(=8.5\) with SSError \(=247.2\) and a sample size of \(n=25\).

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