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Chapter 9: Problem 40

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 \(n=25\).

Short Answer

Expert verified
The complete ANOVA table would be as follows: Model df=1, SS=8.5, MS=8.5, F-statistic calculated as per step 4, and p-value output from software; Error df=23, SS=247.2, MS calculated as per step 3; Total df=24, SS calculated as per step 2. Note: The F-statistic and p-value require additional computation/interpretation.

Step by step solution

01

Fill SS and df fields

Let's start by filling out the SS for Model and Error, and filling out the df using the sample size. In this instance, SSModel is given as \(8.5\) and SSError as \(247.2\). Degrees of freedom for Model (dfModel) is always 1 in simple regression, and degrees of freedom for Error (dfError) is calculated as \(n - 2\), where n is the sample size (25 in this case).
02

Calculate SSTotal and dfTotal

Next we'll calculate SSTotal and dfTotal. Sum of Squares Total (SSTotal) is the sum of SSModel and SSError, and Degrees of Freedom Total (dfTotal) is the sum of dfModel and dfError.
03

Calculate MSModel and MSError

Mean Square (MS) is calculated as SS divided by df. So, we can calculate MSModel as \(SSModel/dfModel\), and MSError as \(SSError/dfError\).
04

Calculate F-statistic

Now we'll compute F-statistic. In Analysis of Variance, F-statistic is calculated as \(MSModel/MSError\).
05

Determine P-value

P-value can be found from the F-distribution table with dfModel and dfError, given F-statistic. Alternatively, it can be computed with statistical software. This step involves interpretation and can't be computed analytically.

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Key Concepts

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

Degrees of Freedom
Degrees of freedom (df) is a crucial concept when working with statistical models, particularly in regression analysis. It refers to the number of independent values or quantities that can vary in analysis without breaking any constraints. In the context of an Analysis of Variance (ANOVA) for regression, degrees of freedom are used to allocate responsibility for the observed variability.

In a simple linear regression, the model degrees of freedom (dfModel) is always set to 1. This is because you are estimating one parameter, the slope of the line. The degrees of freedom for error (dfError) is calculated based on the sample size and the regression model. It is determined by the formula:
  • dfError = n - 2
where n is the total number of observations.

For calculating the total degrees of freedom in your analysis, sum the dfModel and dfError. Thus:
  • dfTotal = dfModel + dfError
Understanding degrees of freedom is fundamental because it influences other statistical measures like the Mean Square and the F-statistic.
Mean Square
Mean Square (MS) is another key measure in regression analysis, particularly in the ANOVA context. It's a way of estimating the variance and provides insight into the spread of the data points around the regression line.

The Mean Square for Model (MSModel) is obtained by dividing the sum of squares for the model (SSModel) by its degrees of freedom (dfModel). The equation is as follows:
  • MSModel = SSModel / dfModel
Similarly, the Mean Square Error (MSError) is obtained by dividing the sum of squares for error (SSError) by its degrees of freedom (dfError):
  • MSError = SSError / dfError
These calculations of Mean Squares allow statisticians to compare variance due to the regression model to the variance that remains unexplained by the model. These comparisons are then used to calculate the F-statistic, a measure of the overall significance of the regression.
F-statistic
The F-statistic is a ratio that allows us to assess whether the overall regression model is a good fit for the data. It is a powerful tool in validation because it makes sense of the calculated Mean Squares.

To calculate the F-statistic within the context of ANOVA for regression, you divide the Mean Square for the Model (MSModel) by the Mean Square Error (MSError):
  • F = MSModel / MSError
This test statistic follows an F-distribution and is used to test the null hypothesis that all the regression coefficients are equal to zero, indicating no relationship between the independent and dependent variables.

A larger F-statistic suggests that the model has more explanatory power in relation to the error. The p-value, usually found using statistical software or an F-distribution table, provides a probability score for obtaining an F-statistic as extreme as, or more than, the observed value, which helps in decision-making pertaining to the model's validity and significance.

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Most popular questions from this chapter

Exercises 9.5 to 9.8 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. $$ \begin{array}{lrrrr} \text { The regression equation is } \mathrm{Y}=89.4 & -8.20 \mathrm{X} & \\ \text { Predictor } & \text { Coef } & \text { SE Coef } & \mathrm{T} & \mathrm{P} \\ \text { Constant } & 89.406 & 4.535 & 19.71 & 0.000 \\ \mathrm{X} & -8.1952 & 0.9563 & -8.57 & 0.000 \end{array} $$

Exercises 9.66 and 9.67 refer to the regression line (given in Exercise 9.46 ): Cognition \(=102.3-3.34 \cdot\) Years using years playing football to predict the score on a cognition test. In each exercise, (a) Find the predicted cognition score for that case. (b) Two intervals are shown: one is a \(95 \%\) confidence interval for the mean response and the other is a \(95 \%\) prediction interval for the response. Which is which? A person who has played football for 8 years. \(\begin{array}{ll}\text { Interval I: }(22.7,128.5) & \text { Interval II: }(63.4,87.8)\end{array}\)

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

Exercise 2.143 on page 102 introduces a study examining years playing football, brain size, and percentile score on a cognitive skills test. We show computer output below for a model to predict Cognition score based on Years playing football. (The scatterplot given in Exercise 2.143 allows us to proceed without serious concerns about the conditions.) Pearson correlation of Years and Cognition \(=-0.366\) P-Value \(=0.015\) Regression Equation Cognition \(=102.3-3.34\) Years Coefficients \(\begin{array}{lrrrr}\text { Term } & \text { Coef } & \text { SE Coef } & \text { T-Value } & \text { P-Value } \\ \text { Constant } & 102.3 & 15.6 & 6.56 & 0.000 \\ \text { Years } & -3.34 & 1.31 & -2.55 & 0.015 \\ & & & & \\ & \text { S } & \text { R-sq } & \text { R-sq(adj) } & \text { R-sq(pred) } \\ 25.4993 & 13.39 \% & 11.33 \% & 5.75 \%\end{array}\) Analysis of Variance \(\begin{array}{lrrrrr}\text { Source } & \text { DF } & \text { Adj SS } & \text { Adj MS } & \text { F-Value } & \text { P-Value } \\\ \text { Regression } & 1 & 4223 & 4223.2 & 6.50 & 0.015 \\ \text { Error } & 42 & 27309 & 650.2 & & \\ \text { Total } & 43 & 31532 & & & \\ & \-- & & & \end{array}\) (a) What is the correlation between these two variables? What is the p-value for testing the correlation? (b) What is the slope of the regression line to predict cognition score based on years playing football? What is the t-statistic for testing the slope? What is the p-value for the test? (c) The ANOVA table is given for testing the effectiveness of this model. What is the F-statistic for the test? What is the p-value? (d) What do you notice about the three p-values for the three tests in parts \((\mathrm{a}),(\mathrm{b}),\) and \((\mathrm{c}) ?\) (e) In every case, at a \(5 \%\) level, what is the conclusion of the test in terms of football and cognition?

Hantavirus is carried by wild rodents and causes severe lung disease in humans. A study \(^{5}\) on the California Channel Islands found that increased prevalence of the virus was linked with greater precipitation. The study adds "Precipitation accounted for \(79 \%\) of the variation in prevalence." (a) What notation or terminology do we use for the value \(79 \%\) in this context? (b) What is the response variable? What is the explanatory variable? (c) What is the correlation between the two variables?
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