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Basics Use the information given in Exercises \(1-2\) (Exercises 1 and 3 , Section 12.2 ) to construct an ANOVA table for a simple linear regression analysis. Use the ANOVA \(F\) -test to test \(H_{0}: \beta=0\) with \(\alpha=.05 .\) Then calculate \(b\) and its standard error: Use a t statistic to test \(H_{0}: \beta=0\) with \(\alpha=.05 .\) Verify that within rounding \(t^{2}=F\). $$ n=8 \text { pairs }(x, y), S_{x x}=4, S_{y y}=20, S_{x y}=8 $$

Short Answer

Expert verified
In a simple linear regression problem, we are given the following information: \(S_{xx} = 4\), \(S_{xy} = 8\), and \(S_{yy} = 20\). Our goal is to construct an ANOVA table, perform hypothesis tests to determine if the slope (\(\beta\)) is equal to 0, calculate the coefficient (\(b\)) and its standard error, test the null hypothesis using the t-statistic, and verify the relationship between t and F statistics. After performing the calculations, we found that the sum of squares for the regression (SSR), residual (SSE), and total (SST) are 16, 4, and 20, respectively. The F statistic is 24, and the t statistic is 4. Based on these values, we reject the null hypothesis, \(H_{0}: \beta=0\), at a significance level of \(\alpha=.05\). Lastly, we constructed an ANOVA table displaying the sum of squares, degrees of freedom, mean square, F/T statistic, and significance for the regression, residual, and total. The table shows that the slope, \(\beta\), is not equal to 0 at a significance level of \(\alpha=.05\).

Step by step solution

01

Calculate the Sum of Squares

First, we need to find the sum of squares for the regression(SSR), residual(SSE), and total(SST) from the given information: $$ SSR = \frac{(S_{xy})^2}{S_{xx}} $$ $$ SSE = S_{yy} - \frac{(S_{xy})^2}{S_{xx}} $$ $$ SST = SSR + SSE $$ Plugging in the given values: $$ SSR = \frac{(8)^2}{4} = 16 $$ $$ SSE = 20 - \frac{(8)^2}{4} = 4 $$ $$ SST = 16 + 4 = 20 $$
02

Calculate the F statistic

Now, we can calculate the F statistic for the ANOVA test using the following formula: $$ F = \frac{MSR}{MSE} $$ Where MSR (Mean Square Regression) and MSE (Mean Square Error) are calculated as follows: $$ MSR = \frac{SSR}{df_{1}}\phantom{-thinspace}\text{ and }\phantom{-thinspace}MSE = \frac{SSE}{df_{2}} $$ Here, \(df_{1}\) and \(df_{2}\) represent the degrees of freedom for the regression and residual respectively. $$ df_{1} = 1 \phantom{-thinspace}\text{ (for regression)} $$ $$ df_{2} = n - 2 = 8 - 2 = 6 \phantom{-thinspace}\text{ (for residual)} $$ Hence, the F statistic can be calculated using: $$ F = \frac{16}{\frac{4}{6}} = 24 $$
03

Perform the F-test

Now, let's perform the F-test for the null hypothesis \(H_{0}: \beta=0\) with a significance level of \(\alpha = 0.05\). We compare the calculated F statistic with the critical value (\(F_{critical}\)) from the F distribution table for \(\alpha=0.05\) and degrees of freedom \((1,6)\): $$ F_{critical} = 5.99 $$ Since \(F > F_{critical}\): $$ 24 > 5.99 $$ We reject the null hypothesis \(H_{0}: \beta=0\).
04

Calculate 'b' and its standard error

Now, let's calculate the coefficient 'b': $$ b = \frac{S_{xy}}{S_{xx}} = \frac{8}{4} = 2 $$ And its standard error (SE): $$ SE_{b} = \sqrt{\frac{MSE}{S_{xx}}} = \sqrt{\frac{4/6}{4}} = 0.5 $$
05

Perform the t-test

To perform the t-test, we will use the calculated coefficient 'b' and its standard error: $$ t = \frac{b - 0}{SE_{b}} = \frac{2}{0.5} = 4 $$ Now, compare the calculated t statistic with the critical value (\(t_{critical}\)) from the t-distribution table for \(\alpha=0.05\) and degrees of freedom \(6\): $$ t_{critical} = \pm 2.447 $$ Since \(t > t_{critical}\): $$ 4 > 2.447 $$ We reject the null hypothesis \(H_{0}: \beta=0\).
06

Verify \(t^{2}=F\)

Finally, let's verify if the relationship between t and F statistics holds within rounding: $$ t^{2} = (4)^2 = 16 $$ And the F statistic: $$ F = 24 $$ Within rounding, we can see that \(t^{2}=F\) holds. This concludes our analysis.
07

Construct the ANOVA table

Now let's construct the ANOVA table for this simple linear regression analysis: | Source | Sum of Squares (SS) | Degrees of Freedom (df) | Mean Square (MS) | F/T statistic | Significance | |--------------|--------------------|-------------------------|-----------------|--------------|-------------| | Regression | 16 | 1 | 16 | 24 | Reject \(H_0\) | | Residual | 4 | 6 | 4/6 | - | - | | Total | 20 | - | - | - | - | Based on the ANOVA table, we conclude that the slope, \(\beta\), is not equal to 0 at a significance level of \(\alpha=.05\).

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Key Concepts

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

F-test
The F-test in the context of Analysis of Variance (ANOVA) for a simple linear regression is used to decide whether the relationship between the dependent variable and the independent variable explained by the regression is statistically significant. This test compares the variance explained by the model (Mean Square Regression, MSR) to the variance that is unexplained (Mean Square Error, MSE).

The F-test uses an F-statistic, calculated as the ratio of MSR to MSE. A higher F-statistic indicates that a significant proportion of the total variance in the outcome variable is accounted for by the model. Students can look up the critical F-value for their given degrees of freedom in statistical tables or use statistical software. If the calculated F-statistic is greater than the critical value at a predetermined significance level (typically \(\alpha = 0.05\)), the null hypothesis (\(H_{0}: \beta=0\)) is rejected, suggesting that the model provides a better fit than the baseline model with no predictors.
t-test
The t-test is another statistical test used in simple linear regression to examine the hypothesis that the slope (\(\beta\)) of the regression line is not zero, which suggests a linear relationship between the independent and dependent variables. It involves calculating a t statistic for the slope coefficient (\(b\)) by dividing the estimated coefficient by its standard error (SE).

If the absolute value of the t statistic exceeds the critical value from the t-distribution with the appropriate degrees of freedom, we reject the null hypothesis, indicating that the slope is statistically different from zero. In simple linear regression, like in our exercise, where \(t^{2}\) approximates \(F\), a significant t-test result will usually accompany a significant F-test result, corroborating the evidence that the model's predictor is meaningful.
Sum of Squares
Sum of Squares is a key concept in regression analysis and ANOVA which quantifies the variation in the data. We have the Total Sum of Squares (SST), which measures the total variation in the dependent variable. It's partitioned into the Regression Sum of Squares (SSR), representing variation explained by the model, and the Error Sum of Squares (SSE), representing unexplained variation or noise.

In the exercise, after calculating the values for \(SSR\) and \(SSE\), the \(SST\) is obtained by summing them up. This partitioning is fundamental as it allows us to calculate the necessary statistics (\(MSR\) and \(MSE\)) for the F-test and subsequently evaluate the regression's explanatory power.
Degrees of Freedom
Degrees of freedom (df) in simple linear regression represent the number of values that are free to vary given the other data points and the estimated parameters. For SSR, the degrees of freedom is usually 1 because there is only one explanatory variable in a simple linear regression. For SSE, it is the total number of observations minus the number of estimated parameters (including the intercept), which is \(n-2\) for simple linear regression.

The degrees of freedom are crucial when determining the critical values for both the F-test and the t-test. Moreover, they are used to calculate the Mean Square values (\(MSR\) and \(MSE\)) by dividing the Sum of Squares (\(SSR\) and \(SSE\)) by their respective degrees of freedom as shown in the exercise, thus facilitating the F-test.

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Most popular questions from this chapter

Sleep Deprivation A study was conducted to determine the effects of sleep deprivation on DS1206 people's ability to solve problems. Ten subjects participated in the study, two at each of five sleep deprivation levels \(-8,12,16,20,\) and 24 hours. After his or her specified sleep deprivation period, each subject answered a set of simple addition problems, and the number of errors was recorded. These results were obtained: d. What is the least-squares prediction equation? e. Use the prediction equation to predict the number of errors for a person who has not slept for 10 hours. $$\begin{aligned}&\begin{array}{l|l|l|l}\text { Number of Errors, } y & 8,6 & 6,10 & 8,14 \\\\\hline \begin{array}{l}\text { Number of Hours without } \\\\\text { Sleep, } x\end{array} & 8 & 12 & 16\end{array}\\\&\begin{array}{l|l|l}\text { Number of Errors, } y & 14,12 & 16,12 \\\\\hline \begin{array}{l}\text { Number of Hours without } \\\\\text { Sleep. } x\end{array} & 20 & 24 \\\\\hline\end{array}\end{aligned}$$ a. How many pairs of observations are in the experiment? b. What are the total number of degrees of freedom? c. Complete the MS Exce/ printout.

The data points given in Exercises \(6-7\) were formed by reversing the slope of the lines in Exercises 4 - 5. Plot the points on graph paper and calculater and \(r^{2}\). Notice the change in the sign of \(r\) and the relationship between the values of \(r^{2}\) compared to Exercises \(4-5 .\) By what percentage was the sum of squares of deviations reduced by using the least-squares predictor \(\hat{y}=a+b x\) rather than \(\bar{y}\) as a predictor of \(y\) ? $$\begin{array}{l|llllll}x & 1 & 2 & 3 & 4 & 5 & 6 \\\\\hline y & 0 & 2 & 3 & 5 & 5 & 7\end{array}$$

What is the difference between deterministic and probabilistic models?

Use the data given in Exercises 6-7 (Exercises 17-18, Section 12.1). Construct the ANOVA table for a simple linear regression analysis, showing the sources, degrees of freedom, sums of squares, and mean sauares. $$\begin{array}{l|rrrrrrr}x & -2 & -1 & 0 & 1 & 2 \\\\\hline y & 1 & 1 & 3 & 5 & 5\end{array}$$

Use the information given to find a confidence interval for the average value of \(y\) when \(x=x_{0}\). $$ \begin{array}{l} n=10, \mathrm{SSE}=24, \Sigma x_{i}=59, \Sigma x_{i}^{2}=397, \\ \hat{y}=.074+.46 x, x_{0}=5,90 \% \text { confidence level } \end{array} $$

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