Fit the model for the given data.

Question

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Answer 1

Question

Chapter 11, Problem 97SE

a.

To determine

Fit the model for the given data.

a.

Expert Solution

Answer to Problem 97SE

The fitted model for the given data is $y^=1.42857+0.5x1+0.11905x2−0.5x3$ .

Explanation of Solution

The model is as follows:

$Y=β0+β1x1+β2x2+β3x3+ε$

From the given Table, the data can be formed into matrices as follows:

$X=[1 −3 5 −11 −2 0 1 1 −1 −3 1 1 0 −4 01 1 −3 −1 1 2 0 −11 3 5 1]$ , $Y=[ 1 0 0 1 2 3 3]$

From the property of least squares, it is noted that $β^=(X′X)−1X′Y$ .

$X′=[ 1 1 1 1 1 1 1 −3 −2 −1 0 1 2 3 5 0 −3 −4 −3 0 5−1 1 1 0 −1 −1 1 ]$

Now, the matrices $X′X$ and $X′Y$ are as follows:

$X′X=[ 1 1 1 1 1 1 1 −3 −2 −1 0 1 2 3 5 0 −3 −4 −3 0 5−1 1 1 0 −1 −1 1 ][1 −3 5 −11 −2 0 1 1 −1 −3 1 1 0 −4 01 1 −3 −1 1 2 0 −11 3 5 1]=[ 7 0 0 0 0 28 0 0 0 0 84 0 0 0 0 6 ]$

$X′Y=[ 1 1 1 1 1 1 1 −3 −2 −1 0 1 2 3 5 0 −3 −4 −3 0 5−1 1 1 0 −1 −1 1 ][ 1 0 0 1 2 3 3]=[101410−3]$

Also the matrix $(X′X)−1$ is as follows:

$(X′X)−1=([ 7 0 0 0 0 28 0 0 0 0 84 0 0 0 0 6 ])−1=[1/7 0 0 00 1/28 0 00 0 1/84 00 0 0 1/6]$

The parameter $β^$ is as follows:

$β^=(X′X)−1X′Y=[1/7 0 0 00 1/28 0 00 0 1/84 00 0 0 1/6][101410−3][β^0β^1β^2β^3]=[1.428570.50.11905−0.5]$

The fitted model for the given data can be obtained as follows:

$y^=β0+β^1x1+β^2x2+β^3x3y^=1.42857+0.5x1+0.11905x2−0.5x3$

Thus, the fitted model for the given data is $y^=1.42857+0.5x1+0.11905x2−0.5x3$ .

b.

To determine

Predict Y when $x1=1,x2=−3,x3=−1$ .

Compare with the observed response in the original data.

Explain why these two are not equal.

b.

Expert Solution

Answer to Problem 97SE

The predicted value of Y when $x1=1,x2=−3,x3=−1$ is $2.07142$ .

Explanation of Solution

From Part (a), the fitted model is as follows:

$y^=1.42857+0.5x1+0.11095x2−0.5x3$ .

The predicted value of Y when $x1=1,x2=−3,x3=−1$ is as follows:

$y^=1.42857+0.5(1)+0.11905(−3)−0.5(−1)=2.07142$

Thus, the predicted value of Y when $x1=1,x2=−3,x3=−1$ is $2.07142$ .

According to the data from the table, the observed value of Y when $x1=1,x2=−3,x3=−1$ is equal to 2.

The observed and the predicted value are not equal because to predict the value one can use the fitted model based on the given data and because of that each predicted value contains some deviation from the observed value.

c.

To determine

Test the hypothesis $H0:β3=0$ against $Ha:β3≠0$ using $Ha:β3≠0$ .

c.

Expert Solution

Answer to Problem 97SE

The data provide sufficient evidence to indicate that $x3$ contributes information for the prediction of Y.

Explanation of Solution

The null and alternative hypotheses are stated as follows:

$H0:β3=0$

$Ha:β3≠0$

The significance level is 0.05.

The test statistic is as follows:

$t=β^3(sc33)$

Where,

$s=SSEn−(k+1)=Y′Y−β^′X′Yn−(k+1)$ , $β^=(X′X)−1X′Y$

Here, $k+1$ is the number of unknown $βi$ values and n is the number of data values.

Also, it is noticed that $cii$ is the element in row $(i+1)$ and column $(i+1)$ of $(X′X)−1$ .

Where, $0≤i≤k$

From the given table, it is found that the values of $n=7$ and $k=3$ .

From Part (a), the parameter $β^$ can be found as follows:

$[β^0β^1β^2β^3]=[1.428570.50.11905−0.5]$

Thus, the value of $β^3$ is –0.5.

Also, the matrix $(X′X)−1$ is as follows:

$(X′X)−1=[1/7 0 0 00 1/28 0 00 0 1/84 00 0 0 1/6]$

From the above matrix, the value of $c33=0.166666667$ .

From Part (a), the values of $Y′Y$ , $X′Y$ , and $β^′X′Y$ are as follows:

$Y′Y=[1 0 0 1 2 3 3][ 1 0 0 1 2 3 3]=24$

$X′Y=[101410−3]$

$β^′X′Y=[1.42857 0.5 0.11905 −0.5][101410−3]=23.97619$

The value of $s$ is calculated as follows:

$s=Y′Y−β^′X′Yn−(k+1)=24−23.976197−4=0.089087$

Thus, the value of $s$ is 0.089087.

The test statistic value can be obtained as follows:

$t=β^3(sc33)=−0.50.0890870.166666667=−13.748$

Thus, the test statistic value is –13.748.

Critical value:

Step-by-step procedure to obtain the t-critical value using Table 5 of Appendix 3:

Locate the degrees of freedom as 5 in the column of df.
Move left until column headed by $t0.025$ .
Select the intersection of (5, 0.025) that gives the critical value of t.

The critical value of t for the left-tailed test is $−2.571$ .

Conclusion:

The test statistic value is greater than the table value.

That is, $(|tcal|=13.748)>(|ttab|=2.571)$ .

Hence, by the rejection rule, reject the null hypothesis $H0$ at the 0.05 significance level.

Therefore, the data provide sufficient evidence to indicate that $x3$ contributes information for the prediction of Y.

d.

To determine

Find a 95% confidence interval for the expected value of Y when $x1=1$ , $x2=−3$ , and $x3=−1$ .

d.

Expert Solution

Answer to Problem 97SE

The 95% confidence interval for the expected value of Y is (1.881, 2.262).

Explanation of Solution

A 95% confidence interval for the expectedvalue of Y is as follows:

$(a′β^−t0.025.sa′(X′X)−1a,a′β^+t0.025.sa′(X′X)−1a)$

It is given that $x1=1$ , $x2=−3$ , and $x3=−1$ .

The matrix ‘ $a′$ ’ can be obtained as follows:

$a′=[1 1 −3 −1]$

From Part (a), the parameter $β^$ , $n$ and $k$ can be found as follows:

$β^=[1.428570.50.11905−0.5]$ , $n=7$ and $k=3$

The value of $a′β^$ is as follows:

$a′β^=[1 1 −3 −1][1.428570.50.11905−0.5]=2.07143$

Thus, the value of $a′β^$ is 2.07143.

The value of $a′(X′X)−1a$ is as follows:

$a′(X′X)−1a=[1 1 −3 −1][1/7 0 0 00 1/28 0 00 0 1/84 00 0 0 1/6][11−3−1]=0.45238$

From Part (c), the value of s is 0.089087.

The t-critical value using Table 5 of Appendix 3 at 3 degrees of freedom is $t0.025=3.182$ .

The 95% confidence interval for the expectedvalue of Y is as follows:

The lower limit is as follows:

$a′β^−t0.025.sa′(X′X)−1a=2.07143−3.182(0.089087)0.45238=1.881$

The upper limit is as follows:

$a′β^+t0.025.sa′(X′X)−1a=2.07143+3.182(0.089087)0.45238=2.262$

The 95% confidence interval for the expectedvalue of Y is (1.881, 2.262).

e.

To determine

Find a 95% prediction interval for the expected value of Y when $x1=1$ , $x2=−3$ , and $x3=−1$ .

e.

Expert Solution

Answer to Problem 97SE

The 95% prediction interval for the expected value of Y is (1.730, 2.413).

Explanation of Solution

A 95% prediction interval for the expectedvalue of Y is as follows:

$(a′β^−t0.025.s1+a′(X′X)−1a,a′β^+t0.025.s1+a′(X′X)−1a)$

It is given that $x1=1$ , $x2=−3$ , and $x3=−1$

The matrix ‘ $a′$ ’ can be obtained as follows:

$a′=[1 1 −3 −1]$

From Part (d), the value of $a′β^$ is 2.07143.

From Part (d), the value of $a′(X′X)−1a$ is 0.45238.

From Part (c), the value of s is 0.089087.

The t-critical value using Table 5 of Appendix 3 at 3 degrees of freedom is $t0.025=3.182$ .

The 95% prediction interval for the expectedvalue of Y is as follows:

The lower limit is as follows:

$a′β^−t0.025.s1+a′(X′X)−1a=2.07143−3.182(0.089087)1+0.45238=1.730$

The upper limit is as follows:

$a′β^+t0.025.s1+a′(X′X)−1a=2.07143+3.182(0.089087)1+0.45238=2.413$

The 95% prediction interval for the expectedvalue of Y is (1.730, 2.413).

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Answer 2

Question

Chapter 11, Problem 97SE

a.

To determine

Fit the model for the given data.

a.

Expert Solution

Answer to Problem 97SE

The fitted model for the given data is $y^=1.42857+0.5x1+0.11905x2−0.5x3$ .

Explanation of Solution

The model is as follows:

$Y=β0+β1x1+β2x2+β3x3+ε$

From the given Table, the data can be formed into matrices as follows:

$X=[1 −3 5 −11 −2 0 1 1 −1 −3 1 1 0 −4 01 1 −3 −1 1 2 0 −11 3 5 1]$ , $Y=[ 1 0 0 1 2 3 3]$

From the property of least squares, it is noted that $β^=(X′X)−1X′Y$ .

$X′=[ 1 1 1 1 1 1 1 −3 −2 −1 0 1 2 3 5 0 −3 −4 −3 0 5−1 1 1 0 −1 −1 1 ]$

Now, the matrices $X′X$ and $X′Y$ are as follows:

$X′X=[ 1 1 1 1 1 1 1 −3 −2 −1 0 1 2 3 5 0 −3 −4 −3 0 5−1 1 1 0 −1 −1 1 ][1 −3 5 −11 −2 0 1 1 −1 −3 1 1 0 −4 01 1 −3 −1 1 2 0 −11 3 5 1]=[ 7 0 0 0 0 28 0 0 0 0 84 0 0 0 0 6 ]$

$X′Y=[ 1 1 1 1 1 1 1 −3 −2 −1 0 1 2 3 5 0 −3 −4 −3 0 5−1 1 1 0 −1 −1 1 ][ 1 0 0 1 2 3 3]=[101410−3]$

Also the matrix $(X′X)−1$ is as follows:

$(X′X)−1=([ 7 0 0 0 0 28 0 0 0 0 84 0 0 0 0 6 ])−1=[1/7 0 0 00 1/28 0 00 0 1/84 00 0 0 1/6]$

The parameter $β^$ is as follows:

$β^=(X′X)−1X′Y=[1/7 0 0 00 1/28 0 00 0 1/84 00 0 0 1/6][101410−3][β^0β^1β^2β^3]=[1.428570.50.11905−0.5]$

The fitted model for the given data can be obtained as follows:

$y^=β0+β^1x1+β^2x2+β^3x3y^=1.42857+0.5x1+0.11905x2−0.5x3$

Thus, the fitted model for the given data is $y^=1.42857+0.5x1+0.11905x2−0.5x3$ .

b.

To determine

Predict Y when $x1=1,x2=−3,x3=−1$ .

Compare with the observed response in the original data.

Explain why these two are not equal.

b.

Expert Solution

Answer to Problem 97SE

The predicted value of Y when $x1=1,x2=−3,x3=−1$ is $2.07142$ .

Explanation of Solution

From Part (a), the fitted model is as follows:

$y^=1.42857+0.5x1+0.11095x2−0.5x3$ .

The predicted value of Y when $x1=1,x2=−3,x3=−1$ is as follows:

$y^=1.42857+0.5(1)+0.11905(−3)−0.5(−1)=2.07142$

Thus, the predicted value of Y when $x1=1,x2=−3,x3=−1$ is $2.07142$ .

According to the data from the table, the observed value of Y when $x1=1,x2=−3,x3=−1$ is equal to 2.

The observed and the predicted value are not equal because to predict the value one can use the fitted model based on the given data and because of that each predicted value contains some deviation from the observed value.

c.

To determine

Test the hypothesis $H0:β3=0$ against $Ha:β3≠0$ using $Ha:β3≠0$ .

c.

Expert Solution

Answer to Problem 97SE

The data provide sufficient evidence to indicate that $x3$ contributes information for the prediction of Y.

Explanation of Solution

The null and alternative hypotheses are stated as follows:

$H0:β3=0$

$Ha:β3≠0$

The significance level is 0.05.

The test statistic is as follows:

$t=β^3(sc33)$

Where,

$s=SSEn−(k+1)=Y′Y−β^′X′Yn−(k+1)$ , $β^=(X′X)−1X′Y$

Here, $k+1$ is the number of unknown $βi$ values and n is the number of data values.

Also, it is noticed that $cii$ is the element in row $(i+1)$ and column $(i+1)$ of $(X′X)−1$ .

Where, $0≤i≤k$

From the given table, it is found that the values of $n=7$ and $k=3$ .

From Part (a), the parameter $β^$ can be found as follows:

$[β^0β^1β^2β^3]=[1.428570.50.11905−0.5]$

Thus, the value of $β^3$ is –0.5.

Also, the matrix $(X′X)−1$ is as follows:

$(X′X)−1=[1/7 0 0 00 1/28 0 00 0 1/84 00 0 0 1/6]$

From the above matrix, the value of $c33=0.166666667$ .

From Part (a), the values of $Y′Y$ , $X′Y$ , and $β^′X′Y$ are as follows:

$Y′Y=[1 0 0 1 2 3 3][ 1 0 0 1 2 3 3]=24$

$X′Y=[101410−3]$

$β^′X′Y=[1.42857 0.5 0.11905 −0.5][101410−3]=23.97619$

The value of $s$ is calculated as follows:

$s=Y′Y−β^′X′Yn−(k+1)=24−23.976197−4=0.089087$

Thus, the value of $s$ is 0.089087.

The test statistic value can be obtained as follows:

$t=β^3(sc33)=−0.50.0890870.166666667=−13.748$

Thus, the test statistic value is –13.748.

Critical value:

Step-by-step procedure to obtain the t-critical value using Table 5 of Appendix 3:

Locate the degrees of freedom as 5 in the column of df.
Move left until column headed by $t0.025$ .
Select the intersection of (5, 0.025) that gives the critical value of t.

The critical value of t for the left-tailed test is $−2.571$ .

Conclusion:

The test statistic value is greater than the table value.

That is, $(|tcal|=13.748)>(|ttab|=2.571)$ .

Hence, by the rejection rule, reject the null hypothesis $H0$ at the 0.05 significance level.

Therefore, the data provide sufficient evidence to indicate that $x3$ contributes information for the prediction of Y.

d.

To determine

Find a 95% confidence interval for the expected value of Y when $x1=1$ , $x2=−3$ , and $x3=−1$ .

d.

Expert Solution

Answer to Problem 97SE

The 95% confidence interval for the expected value of Y is (1.881, 2.262).

Explanation of Solution

A 95% confidence interval for the expectedvalue of Y is as follows:

$(a′β^−t0.025.sa′(X′X)−1a,a′β^+t0.025.sa′(X′X)−1a)$

It is given that $x1=1$ , $x2=−3$ , and $x3=−1$ .

The matrix ‘ $a′$ ’ can be obtained as follows:

$a′=[1 1 −3 −1]$

From Part (a), the parameter $β^$ , $n$ and $k$ can be found as follows:

$β^=[1.428570.50.11905−0.5]$ , $n=7$ and $k=3$

The value of $a′β^$ is as follows:

$a′β^=[1 1 −3 −1][1.428570.50.11905−0.5]=2.07143$

Thus, the value of $a′β^$ is 2.07143.

The value of $a′(X′X)−1a$ is as follows:

$a′(X′X)−1a=[1 1 −3 −1][1/7 0 0 00 1/28 0 00 0 1/84 00 0 0 1/6][11−3−1]=0.45238$

From Part (c), the value of s is 0.089087.

The t-critical value using Table 5 of Appendix 3 at 3 degrees of freedom is $t0.025=3.182$ .

The 95% confidence interval for the expectedvalue of Y is as follows:

The lower limit is as follows:

$a′β^−t0.025.sa′(X′X)−1a=2.07143−3.182(0.089087)0.45238=1.881$

The upper limit is as follows:

$a′β^+t0.025.sa′(X′X)−1a=2.07143+3.182(0.089087)0.45238=2.262$

The 95% confidence interval for the expectedvalue of Y is (1.881, 2.262).

e.

To determine

Find a 95% prediction interval for the expected value of Y when $x1=1$ , $x2=−3$ , and $x3=−1$ .

e.

Expert Solution

Answer to Problem 97SE

The 95% prediction interval for the expected value of Y is (1.730, 2.413).

Explanation of Solution

A 95% prediction interval for the expectedvalue of Y is as follows:

$(a′β^−t0.025.s1+a′(X′X)−1a,a′β^+t0.025.s1+a′(X′X)−1a)$

It is given that $x1=1$ , $x2=−3$ , and $x3=−1$

The matrix ‘ $a′$ ’ can be obtained as follows:

$a′=[1 1 −3 −1]$

From Part (d), the value of $a′β^$ is 2.07143.

From Part (d), the value of $a′(X′X)−1a$ is 0.45238.

From Part (c), the value of s is 0.089087.

The t-critical value using Table 5 of Appendix 3 at 3 degrees of freedom is $t0.025=3.182$ .

The 95% prediction interval for the expectedvalue of Y is as follows:

The lower limit is as follows:

$a′β^−t0.025.s1+a′(X′X)−1a=2.07143−3.182(0.089087)1+0.45238=1.730$

The upper limit is as follows:

$a′β^+t0.025.s1+a′(X′X)−1a=2.07143+3.182(0.089087)1+0.45238=2.413$

The 95% prediction interval for the expectedvalue of Y is (1.730, 2.413).

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Answer 3

Question

Chapter 11, Problem 97SE

a.

To determine

Fit the model for the given data.

a.

Expert Solution

Answer to Problem 97SE

The fitted model for the given data is $y^=1.42857+0.5x1+0.11905x2−0.5x3$ .

Explanation of Solution

The model is as follows:

$Y=β0+β1x1+β2x2+β3x3+ε$

From the given Table, the data can be formed into matrices as follows:

$X=[1 −3 5 −11 −2 0 1 1 −1 −3 1 1 0 −4 01 1 −3 −1 1 2 0 −11 3 5 1]$ , $Y=[ 1 0 0 1 2 3 3]$

From the property of least squares, it is noted that $β^=(X′X)−1X′Y$ .

$X′=[ 1 1 1 1 1 1 1 −3 −2 −1 0 1 2 3 5 0 −3 −4 −3 0 5−1 1 1 0 −1 −1 1 ]$

Now, the matrices $X′X$ and $X′Y$ are as follows:

$X′X=[ 1 1 1 1 1 1 1 −3 −2 −1 0 1 2 3 5 0 −3 −4 −3 0 5−1 1 1 0 −1 −1 1 ][1 −3 5 −11 −2 0 1 1 −1 −3 1 1 0 −4 01 1 −3 −1 1 2 0 −11 3 5 1]=[ 7 0 0 0 0 28 0 0 0 0 84 0 0 0 0 6 ]$

$X′Y=[ 1 1 1 1 1 1 1 −3 −2 −1 0 1 2 3 5 0 −3 −4 −3 0 5−1 1 1 0 −1 −1 1 ][ 1 0 0 1 2 3 3]=[101410−3]$

Also the matrix $(X′X)−1$ is as follows:

$(X′X)−1=([ 7 0 0 0 0 28 0 0 0 0 84 0 0 0 0 6 ])−1=[1/7 0 0 00 1/28 0 00 0 1/84 00 0 0 1/6]$

The parameter $β^$ is as follows:

$β^=(X′X)−1X′Y=[1/7 0 0 00 1/28 0 00 0 1/84 00 0 0 1/6][101410−3][β^0β^1β^2β^3]=[1.428570.50.11905−0.5]$

The fitted model for the given data can be obtained as follows:

$y^=β0+β^1x1+β^2x2+β^3x3y^=1.42857+0.5x1+0.11905x2−0.5x3$

Thus, the fitted model for the given data is $y^=1.42857+0.5x1+0.11905x2−0.5x3$ .

b.

To determine

Predict Y when $x1=1,x2=−3,x3=−1$ .

Compare with the observed response in the original data.

Explain why these two are not equal.

b.

Expert Solution

Answer to Problem 97SE

The predicted value of Y when $x1=1,x2=−3,x3=−1$ is $2.07142$ .

Explanation of Solution

From Part (a), the fitted model is as follows:

$y^=1.42857+0.5x1+0.11095x2−0.5x3$ .

The predicted value of Y when $x1=1,x2=−3,x3=−1$ is as follows:

$y^=1.42857+0.5(1)+0.11905(−3)−0.5(−1)=2.07142$

Thus, the predicted value of Y when $x1=1,x2=−3,x3=−1$ is $2.07142$ .

According to the data from the table, the observed value of Y when $x1=1,x2=−3,x3=−1$ is equal to 2.

The observed and the predicted value are not equal because to predict the value one can use the fitted model based on the given data and because of that each predicted value contains some deviation from the observed value.

c.

To determine

Test the hypothesis $H0:β3=0$ against $Ha:β3≠0$ using $Ha:β3≠0$ .

c.

Expert Solution

Answer to Problem 97SE

The data provide sufficient evidence to indicate that $x3$ contributes information for the prediction of Y.

Explanation of Solution

The null and alternative hypotheses are stated as follows:

$H0:β3=0$

$Ha:β3≠0$

The significance level is 0.05.

The test statistic is as follows:

$t=β^3(sc33)$

Where,

$s=SSEn−(k+1)=Y′Y−β^′X′Yn−(k+1)$ , $β^=(X′X)−1X′Y$

Here, $k+1$ is the number of unknown $βi$ values and n is the number of data values.

Also, it is noticed that $cii$ is the element in row $(i+1)$ and column $(i+1)$ of $(X′X)−1$ .

Where, $0≤i≤k$

From the given table, it is found that the values of $n=7$ and $k=3$ .

From Part (a), the parameter $β^$ can be found as follows:

$[β^0β^1β^2β^3]=[1.428570.50.11905−0.5]$

Thus, the value of $β^3$ is –0.5.

Also, the matrix $(X′X)−1$ is as follows:

$(X′X)−1=[1/7 0 0 00 1/28 0 00 0 1/84 00 0 0 1/6]$

From the above matrix, the value of $c33=0.166666667$ .

From Part (a), the values of $Y′Y$ , $X′Y$ , and $β^′X′Y$ are as follows:

$Y′Y=[1 0 0 1 2 3 3][ 1 0 0 1 2 3 3]=24$

$X′Y=[101410−3]$

$β^′X′Y=[1.42857 0.5 0.11905 −0.5][101410−3]=23.97619$

The value of $s$ is calculated as follows:

$s=Y′Y−β^′X′Yn−(k+1)=24−23.976197−4=0.089087$

Thus, the value of $s$ is 0.089087.

The test statistic value can be obtained as follows:

$t=β^3(sc33)=−0.50.0890870.166666667=−13.748$

Thus, the test statistic value is –13.748.

Critical value:

Step-by-step procedure to obtain the t-critical value using Table 5 of Appendix 3:

Locate the degrees of freedom as 5 in the column of df.
Move left until column headed by $t0.025$ .
Select the intersection of (5, 0.025) that gives the critical value of t.

The critical value of t for the left-tailed test is $−2.571$ .

Conclusion:

The test statistic value is greater than the table value.

That is, $(|tcal|=13.748)>(|ttab|=2.571)$ .

Hence, by the rejection rule, reject the null hypothesis $H0$ at the 0.05 significance level.

Therefore, the data provide sufficient evidence to indicate that $x3$ contributes information for the prediction of Y.

d.

To determine

Find a 95% confidence interval for the expected value of Y when $x1=1$ , $x2=−3$ , and $x3=−1$ .

d.

Expert Solution

Answer to Problem 97SE

The 95% confidence interval for the expected value of Y is (1.881, 2.262).

Explanation of Solution

A 95% confidence interval for the expectedvalue of Y is as follows:

$(a′β^−t0.025.sa′(X′X)−1a,a′β^+t0.025.sa′(X′X)−1a)$

It is given that $x1=1$ , $x2=−3$ , and $x3=−1$ .

The matrix ‘ $a′$ ’ can be obtained as follows:

$a′=[1 1 −3 −1]$

From Part (a), the parameter $β^$ , $n$ and $k$ can be found as follows:

$β^=[1.428570.50.11905−0.5]$ , $n=7$ and $k=3$

The value of $a′β^$ is as follows:

$a′β^=[1 1 −3 −1][1.428570.50.11905−0.5]=2.07143$

Thus, the value of $a′β^$ is 2.07143.

The value of $a′(X′X)−1a$ is as follows:

$a′(X′X)−1a=[1 1 −3 −1][1/7 0 0 00 1/28 0 00 0 1/84 00 0 0 1/6][11−3−1]=0.45238$

From Part (c), the value of s is 0.089087.

The t-critical value using Table 5 of Appendix 3 at 3 degrees of freedom is $t0.025=3.182$ .

The 95% confidence interval for the expectedvalue of Y is as follows:

The lower limit is as follows:

$a′β^−t0.025.sa′(X′X)−1a=2.07143−3.182(0.089087)0.45238=1.881$

The upper limit is as follows:

$a′β^+t0.025.sa′(X′X)−1a=2.07143+3.182(0.089087)0.45238=2.262$

The 95% confidence interval for the expectedvalue of Y is (1.881, 2.262).

e.

To determine

Find a 95% prediction interval for the expected value of Y when $x1=1$ , $x2=−3$ , and $x3=−1$ .

e.

Expert Solution

Answer to Problem 97SE

The 95% prediction interval for the expected value of Y is (1.730, 2.413).

Explanation of Solution

A 95% prediction interval for the expectedvalue of Y is as follows:

$(a′β^−t0.025.s1+a′(X′X)−1a,a′β^+t0.025.s1+a′(X′X)−1a)$

It is given that $x1=1$ , $x2=−3$ , and $x3=−1$

The matrix ‘ $a′$ ’ can be obtained as follows:

$a′=[1 1 −3 −1]$

From Part (d), the value of $a′β^$ is 2.07143.

From Part (d), the value of $a′(X′X)−1a$ is 0.45238.

From Part (c), the value of s is 0.089087.

The t-critical value using Table 5 of Appendix 3 at 3 degrees of freedom is $t0.025=3.182$ .

The 95% prediction interval for the expectedvalue of Y is as follows:

The lower limit is as follows:

$a′β^−t0.025.s1+a′(X′X)−1a=2.07143−3.182(0.089087)1+0.45238=1.730$

The upper limit is as follows:

$a′β^+t0.025.s1+a′(X′X)−1a=2.07143+3.182(0.089087)1+0.45238=2.413$

The 95% prediction interval for the expectedvalue of Y is (1.730, 2.413).

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Answer 4

Question

Chapter 11, Problem 97SE

a.

To determine

Fit the model for the given data.

a.

Expert Solution

Answer to Problem 97SE

The fitted model for the given data is $y^=1.42857+0.5x1+0.11905x2−0.5x3$ .

Explanation of Solution

The model is as follows:

$Y=β0+β1x1+β2x2+β3x3+ε$

From the given Table, the data can be formed into matrices as follows:

$X=[1 −3 5 −11 −2 0 1 1 −1 −3 1 1 0 −4 01 1 −3 −1 1 2 0 −11 3 5 1]$ , $Y=[ 1 0 0 1 2 3 3]$

From the property of least squares, it is noted that $β^=(X′X)−1X′Y$ .

$X′=[ 1 1 1 1 1 1 1 −3 −2 −1 0 1 2 3 5 0 −3 −4 −3 0 5−1 1 1 0 −1 −1 1 ]$

Now, the matrices $X′X$ and $X′Y$ are as follows:

$X′X=[ 1 1 1 1 1 1 1 −3 −2 −1 0 1 2 3 5 0 −3 −4 −3 0 5−1 1 1 0 −1 −1 1 ][1 −3 5 −11 −2 0 1 1 −1 −3 1 1 0 −4 01 1 −3 −1 1 2 0 −11 3 5 1]=[ 7 0 0 0 0 28 0 0 0 0 84 0 0 0 0 6 ]$

$X′Y=[ 1 1 1 1 1 1 1 −3 −2 −1 0 1 2 3 5 0 −3 −4 −3 0 5−1 1 1 0 −1 −1 1 ][ 1 0 0 1 2 3 3]=[101410−3]$

Also the matrix $(X′X)−1$ is as follows:

$(X′X)−1=([ 7 0 0 0 0 28 0 0 0 0 84 0 0 0 0 6 ])−1=[1/7 0 0 00 1/28 0 00 0 1/84 00 0 0 1/6]$

The parameter $β^$ is as follows:

$β^=(X′X)−1X′Y=[1/7 0 0 00 1/28 0 00 0 1/84 00 0 0 1/6][101410−3][β^0β^1β^2β^3]=[1.428570.50.11905−0.5]$

The fitted model for the given data can be obtained as follows:

$y^=β0+β^1x1+β^2x2+β^3x3y^=1.42857+0.5x1+0.11905x2−0.5x3$

Thus, the fitted model for the given data is $y^=1.42857+0.5x1+0.11905x2−0.5x3$ .

b.

To determine

Predict Y when $x1=1,x2=−3,x3=−1$ .

Compare with the observed response in the original data.

Explain why these two are not equal.

b.

Expert Solution

Answer to Problem 97SE

The predicted value of Y when $x1=1,x2=−3,x3=−1$ is $2.07142$ .

Explanation of Solution

From Part (a), the fitted model is as follows:

$y^=1.42857+0.5x1+0.11095x2−0.5x3$ .

The predicted value of Y when $x1=1,x2=−3,x3=−1$ is as follows:

$y^=1.42857+0.5(1)+0.11905(−3)−0.5(−1)=2.07142$

Thus, the predicted value of Y when $x1=1,x2=−3,x3=−1$ is $2.07142$ .

According to the data from the table, the observed value of Y when $x1=1,x2=−3,x3=−1$ is equal to 2.

The observed and the predicted value are not equal because to predict the value one can use the fitted model based on the given data and because of that each predicted value contains some deviation from the observed value.

c.

To determine

Test the hypothesis $H0:β3=0$ against $Ha:β3≠0$ using $Ha:β3≠0$ .

c.

Expert Solution

Answer to Problem 97SE

The data provide sufficient evidence to indicate that $x3$ contributes information for the prediction of Y.

Explanation of Solution

The null and alternative hypotheses are stated as follows:

$H0:β3=0$

$Ha:β3≠0$

The significance level is 0.05.

The test statistic is as follows:

$t=β^3(sc33)$

Where,

$s=SSEn−(k+1)=Y′Y−β^′X′Yn−(k+1)$ , $β^=(X′X)−1X′Y$

Here, $k+1$ is the number of unknown $βi$ values and n is the number of data values.

Also, it is noticed that $cii$ is the element in row $(i+1)$ and column $(i+1)$ of $(X′X)−1$ .

Where, $0≤i≤k$

From the given table, it is found that the values of $n=7$ and $k=3$ .

From Part (a), the parameter $β^$ can be found as follows:

$[β^0β^1β^2β^3]=[1.428570.50.11905−0.5]$

Thus, the value of $β^3$ is –0.5.

Also, the matrix $(X′X)−1$ is as follows:

$(X′X)−1=[1/7 0 0 00 1/28 0 00 0 1/84 00 0 0 1/6]$

From the above matrix, the value of $c33=0.166666667$ .

From Part (a), the values of $Y′Y$ , $X′Y$ , and $β^′X′Y$ are as follows:

$Y′Y=[1 0 0 1 2 3 3][ 1 0 0 1 2 3 3]=24$

$X′Y=[101410−3]$

$β^′X′Y=[1.42857 0.5 0.11905 −0.5][101410−3]=23.97619$

The value of $s$ is calculated as follows:

$s=Y′Y−β^′X′Yn−(k+1)=24−23.976197−4=0.089087$

Thus, the value of $s$ is 0.089087.

The test statistic value can be obtained as follows:

$t=β^3(sc33)=−0.50.0890870.166666667=−13.748$

Thus, the test statistic value is –13.748.

Critical value:

Step-by-step procedure to obtain the t-critical value using Table 5 of Appendix 3:

Locate the degrees of freedom as 5 in the column of df.
Move left until column headed by $t0.025$ .
Select the intersection of (5, 0.025) that gives the critical value of t.

The critical value of t for the left-tailed test is $−2.571$ .

Conclusion:

The test statistic value is greater than the table value.

That is, $(|tcal|=13.748)>(|ttab|=2.571)$ .

Hence, by the rejection rule, reject the null hypothesis $H0$ at the 0.05 significance level.

Therefore, the data provide sufficient evidence to indicate that $x3$ contributes information for the prediction of Y.

d.

To determine

Find a 95% confidence interval for the expected value of Y when $x1=1$ , $x2=−3$ , and $x3=−1$ .

d.

Expert Solution

Answer to Problem 97SE

The 95% confidence interval for the expected value of Y is (1.881, 2.262).

Explanation of Solution

A 95% confidence interval for the expectedvalue of Y is as follows:

$(a′β^−t0.025.sa′(X′X)−1a,a′β^+t0.025.sa′(X′X)−1a)$

It is given that $x1=1$ , $x2=−3$ , and $x3=−1$ .

The matrix ‘ $a′$ ’ can be obtained as follows:

$a′=[1 1 −3 −1]$

From Part (a), the parameter $β^$ , $n$ and $k$ can be found as follows:

$β^=[1.428570.50.11905−0.5]$ , $n=7$ and $k=3$

The value of $a′β^$ is as follows:

$a′β^=[1 1 −3 −1][1.428570.50.11905−0.5]=2.07143$

Thus, the value of $a′β^$ is 2.07143.

The value of $a′(X′X)−1a$ is as follows:

$a′(X′X)−1a=[1 1 −3 −1][1/7 0 0 00 1/28 0 00 0 1/84 00 0 0 1/6][11−3−1]=0.45238$

From Part (c), the value of s is 0.089087.

The t-critical value using Table 5 of Appendix 3 at 3 degrees of freedom is $t0.025=3.182$ .

The 95% confidence interval for the expectedvalue of Y is as follows:

The lower limit is as follows:

$a′β^−t0.025.sa′(X′X)−1a=2.07143−3.182(0.089087)0.45238=1.881$

The upper limit is as follows:

$a′β^+t0.025.sa′(X′X)−1a=2.07143+3.182(0.089087)0.45238=2.262$

The 95% confidence interval for the expectedvalue of Y is (1.881, 2.262).

e.

To determine

Find a 95% prediction interval for the expected value of Y when $x1=1$ , $x2=−3$ , and $x3=−1$ .

e.

Expert Solution

Answer to Problem 97SE

The 95% prediction interval for the expected value of Y is (1.730, 2.413).

Explanation of Solution

A 95% prediction interval for the expectedvalue of Y is as follows:

$(a′β^−t0.025.s1+a′(X′X)−1a,a′β^+t0.025.s1+a′(X′X)−1a)$

It is given that $x1=1$ , $x2=−3$ , and $x3=−1$

The matrix ‘ $a′$ ’ can be obtained as follows:

$a′=[1 1 −3 −1]$

From Part (d), the value of $a′β^$ is 2.07143.

From Part (d), the value of $a′(X′X)−1a$ is 0.45238.

From Part (c), the value of s is 0.089087.

The t-critical value using Table 5 of Appendix 3 at 3 degrees of freedom is $t0.025=3.182$ .

The 95% prediction interval for the expectedvalue of Y is as follows:

The lower limit is as follows:

$a′β^−t0.025.s1+a′(X′X)−1a=2.07143−3.182(0.089087)1+0.45238=1.730$

The upper limit is as follows:

$a′β^+t0.025.s1+a′(X′X)−1a=2.07143+3.182(0.089087)1+0.45238=2.413$

The 95% prediction interval for the expectedvalue of Y is (1.730, 2.413).

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Answer 5

Chapter 11, Problem 97SE

a.

To determine

Fit the model for the given data.

a.

Expert Solution

Answer to Problem 97SE

The fitted model for the given data is $y^=1.42857+0.5x1+0.11905x2−0.5x3$ .

Explanation of Solution

The model is as follows:

$Y=β0+β1x1+β2x2+β3x3+ε$

From the given Table, the data can be formed into matrices as follows:

$X=[1 −3 5 −11 −2 0 1 1 −1 −3 1 1 0 −4 01 1 −3 −1 1 2 0 −11 3 5 1]$ , $Y=[ 1 0 0 1 2 3 3]$

From the property of least squares, it is noted that $β^=(X′X)−1X′Y$ .

$X′=[ 1 1 1 1 1 1 1 −3 −2 −1 0 1 2 3 5 0 −3 −4 −3 0 5−1 1 1 0 −1 −1 1 ]$

Now, the matrices $X′X$ and $X′Y$ are as follows:

$X′X=[ 1 1 1 1 1 1 1 −3 −2 −1 0 1 2 3 5 0 −3 −4 −3 0 5−1 1 1 0 −1 −1 1 ][1 −3 5 −11 −2 0 1 1 −1 −3 1 1 0 −4 01 1 −3 −1 1 2 0 −11 3 5 1]=[ 7 0 0 0 0 28 0 0 0 0 84 0 0 0 0 6 ]$

$X′Y=[ 1 1 1 1 1 1 1 −3 −2 −1 0 1 2 3 5 0 −3 −4 −3 0 5−1 1 1 0 −1 −1 1 ][ 1 0 0 1 2 3 3]=[101410−3]$

Also the matrix $(X′X)−1$ is as follows:

$(X′X)−1=([ 7 0 0 0 0 28 0 0 0 0 84 0 0 0 0 6 ])−1=[1/7 0 0 00 1/28 0 00 0 1/84 00 0 0 1/6]$

The parameter $β^$ is as follows:

$β^=(X′X)−1X′Y=[1/7 0 0 00 1/28 0 00 0 1/84 00 0 0 1/6][101410−3][β^0β^1β^2β^3]=[1.428570.50.11905−0.5]$

The fitted model for the given data can be obtained as follows:

$y^=β0+β^1x1+β^2x2+β^3x3y^=1.42857+0.5x1+0.11905x2−0.5x3$

Thus, the fitted model for the given data is $y^=1.42857+0.5x1+0.11905x2−0.5x3$ .

b.

To determine

Predict Y when $x1=1,x2=−3,x3=−1$ .

Compare with the observed response in the original data.

Explain why these two are not equal.

b.

Expert Solution

Answer to Problem 97SE

The predicted value of Y when $x1=1,x2=−3,x3=−1$ is $2.07142$ .

Explanation of Solution

From Part (a), the fitted model is as follows:

$y^=1.42857+0.5x1+0.11095x2−0.5x3$ .

The predicted value of Y when $x1=1,x2=−3,x3=−1$ is as follows:

$y^=1.42857+0.5(1)+0.11905(−3)−0.5(−1)=2.07142$

Thus, the predicted value of Y when $x1=1,x2=−3,x3=−1$ is $2.07142$ .

According to the data from the table, the observed value of Y when $x1=1,x2=−3,x3=−1$ is equal to 2.

The observed and the predicted value are not equal because to predict the value one can use the fitted model based on the given data and because of that each predicted value contains some deviation from the observed value.

c.

To determine

Test the hypothesis $H0:β3=0$ against $Ha:β3≠0$ using $Ha:β3≠0$ .

c.

Expert Solution

Answer to Problem 97SE

The data provide sufficient evidence to indicate that $x3$ contributes information for the prediction of Y.

Explanation of Solution

The null and alternative hypotheses are stated as follows:

$H0:β3=0$

$Ha:β3≠0$

The significance level is 0.05.

The test statistic is as follows:

$t=β^3(sc33)$

Where,

$s=SSEn−(k+1)=Y′Y−β^′X′Yn−(k+1)$ , $β^=(X′X)−1X′Y$

Here, $k+1$ is the number of unknown $βi$ values and n is the number of data values.

Also, it is noticed that $cii$ is the element in row $(i+1)$ and column $(i+1)$ of $(X′X)−1$ .

Where, $0≤i≤k$

From the given table, it is found that the values of $n=7$ and $k=3$ .

From Part (a), the parameter $β^$ can be found as follows:

$[β^0β^1β^2β^3]=[1.428570.50.11905−0.5]$

Thus, the value of $β^3$ is –0.5.

Also, the matrix $(X′X)−1$ is as follows:

$(X′X)−1=[1/7 0 0 00 1/28 0 00 0 1/84 00 0 0 1/6]$

From the above matrix, the value of $c33=0.166666667$ .

From Part (a), the values of $Y′Y$ , $X′Y$ , and $β^′X′Y$ are as follows:

$Y′Y=[1 0 0 1 2 3 3][ 1 0 0 1 2 3 3]=24$

$X′Y=[101410−3]$

$β^′X′Y=[1.42857 0.5 0.11905 −0.5][101410−3]=23.97619$

The value of $s$ is calculated as follows:

$s=Y′Y−β^′X′Yn−(k+1)=24−23.976197−4=0.089087$

Thus, the value of $s$ is 0.089087.

The test statistic value can be obtained as follows:

$t=β^3(sc33)=−0.50.0890870.166666667=−13.748$

Thus, the test statistic value is –13.748.

Critical value:

Step-by-step procedure to obtain the t-critical value using Table 5 of Appendix 3:

Locate the degrees of freedom as 5 in the column of df.
Move left until column headed by $t0.025$ .
Select the intersection of (5, 0.025) that gives the critical value of t.

The critical value of t for the left-tailed test is $−2.571$ .

Conclusion:

The test statistic value is greater than the table value.

That is, $(|tcal|=13.748)>(|ttab|=2.571)$ .

Hence, by the rejection rule, reject the null hypothesis $H0$ at the 0.05 significance level.

Therefore, the data provide sufficient evidence to indicate that $x3$ contributes information for the prediction of Y.

d.

To determine

Find a 95% confidence interval for the expected value of Y when $x1=1$ , $x2=−3$ , and $x3=−1$ .

d.

Expert Solution

Answer to Problem 97SE

The 95% confidence interval for the expected value of Y is (1.881, 2.262).

Explanation of Solution

A 95% confidence interval for the expectedvalue of Y is as follows:

$(a′β^−t0.025.sa′(X′X)−1a,a′β^+t0.025.sa′(X′X)−1a)$

It is given that $x1=1$ , $x2=−3$ , and $x3=−1$ .

The matrix ‘ $a′$ ’ can be obtained as follows:

$a′=[1 1 −3 −1]$

From Part (a), the parameter $β^$ , $n$ and $k$ can be found as follows:

$β^=[1.428570.50.11905−0.5]$ , $n=7$ and $k=3$

The value of $a′β^$ is as follows:

$a′β^=[1 1 −3 −1][1.428570.50.11905−0.5]=2.07143$

Thus, the value of $a′β^$ is 2.07143.

The value of $a′(X′X)−1a$ is as follows:

$a′(X′X)−1a=[1 1 −3 −1][1/7 0 0 00 1/28 0 00 0 1/84 00 0 0 1/6][11−3−1]=0.45238$

From Part (c), the value of s is 0.089087.

The t-critical value using Table 5 of Appendix 3 at 3 degrees of freedom is $t0.025=3.182$ .

The 95% confidence interval for the expectedvalue of Y is as follows:

The lower limit is as follows:

$a′β^−t0.025.sa′(X′X)−1a=2.07143−3.182(0.089087)0.45238=1.881$

The upper limit is as follows:

$a′β^+t0.025.sa′(X′X)−1a=2.07143+3.182(0.089087)0.45238=2.262$

The 95% confidence interval for the expectedvalue of Y is (1.881, 2.262).

e.

To determine

Find a 95% prediction interval for the expected value of Y when $x1=1$ , $x2=−3$ , and $x3=−1$ .

e.

Expert Solution

Answer to Problem 97SE

The 95% prediction interval for the expected value of Y is (1.730, 2.413).

Explanation of Solution

A 95% prediction interval for the expectedvalue of Y is as follows:

$(a′β^−t0.025.s1+a′(X′X)−1a,a′β^+t0.025.s1+a′(X′X)−1a)$

It is given that $x1=1$ , $x2=−3$ , and $x3=−1$

The matrix ‘ $a′$ ’ can be obtained as follows:

$a′=[1 1 −3 −1]$

From Part (d), the value of $a′β^$ is 2.07143.

From Part (d), the value of $a′(X′X)−1a$ is 0.45238.

From Part (c), the value of s is 0.089087.

The t-critical value using Table 5 of Appendix 3 at 3 degrees of freedom is $t0.025=3.182$ .

The 95% prediction interval for the expectedvalue of Y is as follows:

The lower limit is as follows:

$a′β^−t0.025.s1+a′(X′X)−1a=2.07143−3.182(0.089087)1+0.45238=1.730$

The upper limit is as follows:

$a′β^+t0.025.s1+a′(X′X)−1a=2.07143+3.182(0.089087)1+0.45238=2.413$

The 95% prediction interval for the expectedvalue of Y is (1.730, 2.413).

Answer 6

Chapter 11, Problem 97SE

a.

To determine

Fit the model for the given data.

a.

Expert Solution

Answer to Problem 97SE

The fitted model for the given data is $y^=1.42857+0.5x1+0.11905x2−0.5x3$ .

Explanation of Solution

The model is as follows:

$Y=β0+β1x1+β2x2+β3x3+ε$

From the given Table, the data can be formed into matrices as follows:

$X=[1 −3 5 −11 −2 0 1 1 −1 −3 1 1 0 −4 01 1 −3 −1 1 2 0 −11 3 5 1]$ , $Y=[ 1 0 0 1 2 3 3]$

From the property of least squares, it is noted that $β^=(X′X)−1X′Y$ .

$X′=[ 1 1 1 1 1 1 1 −3 −2 −1 0 1 2 3 5 0 −3 −4 −3 0 5−1 1 1 0 −1 −1 1 ]$

Now, the matrices $X′X$ and $X′Y$ are as follows:

$X′X=[ 1 1 1 1 1 1 1 −3 −2 −1 0 1 2 3 5 0 −3 −4 −3 0 5−1 1 1 0 −1 −1 1 ][1 −3 5 −11 −2 0 1 1 −1 −3 1 1 0 −4 01 1 −3 −1 1 2 0 −11 3 5 1]=[ 7 0 0 0 0 28 0 0 0 0 84 0 0 0 0 6 ]$

$X′Y=[ 1 1 1 1 1 1 1 −3 −2 −1 0 1 2 3 5 0 −3 −4 −3 0 5−1 1 1 0 −1 −1 1 ][ 1 0 0 1 2 3 3]=[101410−3]$

Also the matrix $(X′X)−1$ is as follows:

$(X′X)−1=([ 7 0 0 0 0 28 0 0 0 0 84 0 0 0 0 6 ])−1=[1/7 0 0 00 1/28 0 00 0 1/84 00 0 0 1/6]$

The parameter $β^$ is as follows:

$β^=(X′X)−1X′Y=[1/7 0 0 00 1/28 0 00 0 1/84 00 0 0 1/6][101410−3][β^0β^1β^2β^3]=[1.428570.50.11905−0.5]$

The fitted model for the given data can be obtained as follows:

$y^=β0+β^1x1+β^2x2+β^3x3y^=1.42857+0.5x1+0.11905x2−0.5x3$

Thus, the fitted model for the given data is $y^=1.42857+0.5x1+0.11905x2−0.5x3$ .

Answer 7

Chapter 11, Problem 97SE

a.

To determine

Fit the model for the given data.

Answer 8

a.

Expert Solution

Answer to Problem 97SE

The fitted model for the given data is $y^=1.42857+0.5x1+0.11905x2−0.5x3$ .

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

The model is as follows:

$Y=β0+β1x1+β2x2+β3x3+ε$

From the given Table, the data can be formed into matrices as follows:

$X=[1 −3 5 −11 −2 0 1 1 −1 −3 1 1 0 −4 01 1 −3 −1 1 2 0 −11 3 5 1]$ , $Y=[ 1 0 0 1 2 3 3]$

From the property of least squares, it is noted that $β^=(X′X)−1X′Y$ .

$X′=[ 1 1 1 1 1 1 1 −3 −2 −1 0 1 2 3 5 0 −3 −4 −3 0 5−1 1 1 0 −1 −1 1 ]$

Now, the matrices $X′X$ and $X′Y$ are as follows:

$X′X=[ 1 1 1 1 1 1 1 −3 −2 −1 0 1 2 3 5 0 −3 −4 −3 0 5−1 1 1 0 −1 −1 1 ][1 −3 5 −11 −2 0 1 1 −1 −3 1 1 0 −4 01 1 −3 −1 1 2 0 −11 3 5 1]=[ 7 0 0 0 0 28 0 0 0 0 84 0 0 0 0 6 ]$

$X′Y=[ 1 1 1 1 1 1 1 −3 −2 −1 0 1 2 3 5 0 −3 −4 −3 0 5−1 1 1 0 −1 −1 1 ][ 1 0 0 1 2 3 3]=[101410−3]$

Also the matrix $(X′X)−1$ is as follows:

$(X′X)−1=([ 7 0 0 0 0 28 0 0 0 0 84 0 0 0 0 6 ])−1=[1/7 0 0 00 1/28 0 00 0 1/84 00 0 0 1/6]$

The parameter $β^$ is as follows:

$β^=(X′X)−1X′Y=[1/7 0 0 00 1/28 0 00 0 1/84 00 0 0 1/6][101410−3][β^0β^1β^2β^3]=[1.428570.50.11905−0.5]$

The fitted model for the given data can be obtained as follows:

$y^=β0+β^1x1+β^2x2+β^3x3y^=1.42857+0.5x1+0.11905x2−0.5x3$

Thus, the fitted model for the given data is $y^=1.42857+0.5x1+0.11905x2−0.5x3$ .

Answer 13

b.

To determine

Predict Y when $x1=1,x2=−3,x3=−1$ .

Compare with the observed response in the original data.

Explain why these two are not equal.

b.

Expert Solution

Answer to Problem 97SE

The predicted value of Y when $x1=1,x2=−3,x3=−1$ is $2.07142$ .

Explanation of Solution

From Part (a), the fitted model is as follows:

$y^=1.42857+0.5x1+0.11095x2−0.5x3$ .

The predicted value of Y when $x1=1,x2=−3,x3=−1$ is as follows:

$y^=1.42857+0.5(1)+0.11905(−3)−0.5(−1)=2.07142$

Thus, the predicted value of Y when $x1=1,x2=−3,x3=−1$ is $2.07142$ .

According to the data from the table, the observed value of Y when $x1=1,x2=−3,x3=−1$ is equal to 2.

The observed and the predicted value are not equal because to predict the value one can use the fitted model based on the given data and because of that each predicted value contains some deviation from the observed value.

Answer 14

b.

To determine

Predict Y when $x1=1,x2=−3,x3=−1$ .

Compare with the observed response in the original data.

Explain why these two are not equal.

Answer 15

b.

Expert Solution

Answer to Problem 97SE

The predicted value of Y when $x1=1,x2=−3,x3=−1$ is $2.07142$ .

Answer 16

b.

Expert Solution

Answer 17

b.

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

From Part (a), the fitted model is as follows:

$y^=1.42857+0.5x1+0.11095x2−0.5x3$ .

The predicted value of Y when $x1=1,x2=−3,x3=−1$ is as follows:

$y^=1.42857+0.5(1)+0.11905(−3)−0.5(−1)=2.07142$

Thus, the predicted value of Y when $x1=1,x2=−3,x3=−1$ is $2.07142$ .

According to the data from the table, the observed value of Y when $x1=1,x2=−3,x3=−1$ is equal to 2.

The observed and the predicted value are not equal because to predict the value one can use the fitted model based on the given data and because of that each predicted value contains some deviation from the observed value.

Answer 20

c.

To determine

Test the hypothesis $H0:β3=0$ against $Ha:β3≠0$ using $Ha:β3≠0$ .

c.

Expert Solution

Answer to Problem 97SE

The data provide sufficient evidence to indicate that $x3$ contributes information for the prediction of Y.

Explanation of Solution

The null and alternative hypotheses are stated as follows:

$H0:β3=0$

$Ha:β3≠0$

The significance level is 0.05.

The test statistic is as follows:

$t=β^3(sc33)$

Where,

$s=SSEn−(k+1)=Y′Y−β^′X′Yn−(k+1)$ , $β^=(X′X)−1X′Y$

Here, $k+1$ is the number of unknown $βi$ values and n is the number of data values.

Also, it is noticed that $cii$ is the element in row $(i+1)$ and column $(i+1)$ of $(X′X)−1$ .

Where, $0≤i≤k$

From the given table, it is found that the values of $n=7$ and $k=3$ .

From Part (a), the parameter $β^$ can be found as follows:

$[β^0β^1β^2β^3]=[1.428570.50.11905−0.5]$

Thus, the value of $β^3$ is –0.5.

Also, the matrix $(X′X)−1$ is as follows:

$(X′X)−1=[1/7 0 0 00 1/28 0 00 0 1/84 00 0 0 1/6]$

From the above matrix, the value of $c33=0.166666667$ .

From Part (a), the values of $Y′Y$ , $X′Y$ , and $β^′X′Y$ are as follows:

$Y′Y=[1 0 0 1 2 3 3][ 1 0 0 1 2 3 3]=24$

$X′Y=[101410−3]$

$β^′X′Y=[1.42857 0.5 0.11905 −0.5][101410−3]=23.97619$

The value of $s$ is calculated as follows:

$s=Y′Y−β^′X′Yn−(k+1)=24−23.976197−4=0.089087$

Thus, the value of $s$ is 0.089087.

The test statistic value can be obtained as follows:

$t=β^3(sc33)=−0.50.0890870.166666667=−13.748$

Thus, the test statistic value is –13.748.

Critical value:

Step-by-step procedure to obtain the t-critical value using Table 5 of Appendix 3:

Locate the degrees of freedom as 5 in the column of df.
Move left until column headed by $t0.025$ .
Select the intersection of (5, 0.025) that gives the critical value of t.

The critical value of t for the left-tailed test is $−2.571$ .

Conclusion:

The test statistic value is greater than the table value.

That is, $(|tcal|=13.748)>(|ttab|=2.571)$ .

Hence, by the rejection rule, reject the null hypothesis $H0$ at the 0.05 significance level.

Therefore, the data provide sufficient evidence to indicate that $x3$ contributes information for the prediction of Y.

Answer 21

c.

To determine

Test the hypothesis $H0:β3=0$ against $Ha:β3≠0$ using $Ha:β3≠0$ .

Answer 22

c.

Expert Solution

Answer to Problem 97SE

The data provide sufficient evidence to indicate that $x3$ contributes information for the prediction of Y.

Answer 23

c.

Expert Solution

Answer 24

c.

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

The null and alternative hypotheses are stated as follows:

$H0:β3=0$

$Ha:β3≠0$

The significance level is 0.05.

The test statistic is as follows:

$t=β^3(sc33)$

Where,

$s=SSEn−(k+1)=Y′Y−β^′X′Yn−(k+1)$ , $β^=(X′X)−1X′Y$

Here, $k+1$ is the number of unknown $βi$ values and n is the number of data values.

Also, it is noticed that $cii$ is the element in row $(i+1)$ and column $(i+1)$ of $(X′X)−1$ .

Where, $0≤i≤k$

From the given table, it is found that the values of $n=7$ and $k=3$ .

From Part (a), the parameter $β^$ can be found as follows:

$[β^0β^1β^2β^3]=[1.428570.50.11905−0.5]$

Thus, the value of $β^3$ is –0.5.

Also, the matrix $(X′X)−1$ is as follows:

$(X′X)−1=[1/7 0 0 00 1/28 0 00 0 1/84 00 0 0 1/6]$

From the above matrix, the value of $c33=0.166666667$ .

From Part (a), the values of $Y′Y$ , $X′Y$ , and $β^′X′Y$ are as follows:

$Y′Y=[1 0 0 1 2 3 3][ 1 0 0 1 2 3 3]=24$

$X′Y=[101410−3]$

$β^′X′Y=[1.42857 0.5 0.11905 −0.5][101410−3]=23.97619$

The value of $s$ is calculated as follows:

$s=Y′Y−β^′X′Yn−(k+1)=24−23.976197−4=0.089087$

Thus, the value of $s$ is 0.089087.

The test statistic value can be obtained as follows:

$t=β^3(sc33)=−0.50.0890870.166666667=−13.748$

Thus, the test statistic value is –13.748.

Critical value:

Step-by-step procedure to obtain the t-critical value using Table 5 of Appendix 3:

Locate the degrees of freedom as 5 in the column of df.
Move left until column headed by $t0.025$ .
Select the intersection of (5, 0.025) that gives the critical value of t.

The critical value of t for the left-tailed test is $−2.571$ .

Conclusion:

The test statistic value is greater than the table value.

That is, $(|tcal|=13.748)>(|ttab|=2.571)$ .

Hence, by the rejection rule, reject the null hypothesis $H0$ at the 0.05 significance level.

Therefore, the data provide sufficient evidence to indicate that $x3$ contributes information for the prediction of Y.

Answer 27

d.

To determine

Find a 95% confidence interval for the expected value of Y when $x1=1$ , $x2=−3$ , and $x3=−1$ .

d.

Expert Solution

Answer to Problem 97SE

The 95% confidence interval for the expected value of Y is (1.881, 2.262).

Explanation of Solution

A 95% confidence interval for the expectedvalue of Y is as follows:

$(a′β^−t0.025.sa′(X′X)−1a,a′β^+t0.025.sa′(X′X)−1a)$

It is given that $x1=1$ , $x2=−3$ , and $x3=−1$ .

The matrix ‘ $a′$ ’ can be obtained as follows:

$a′=[1 1 −3 −1]$

From Part (a), the parameter $β^$ , $n$ and $k$ can be found as follows:

$β^=[1.428570.50.11905−0.5]$ , $n=7$ and $k=3$

The value of $a′β^$ is as follows:

$a′β^=[1 1 −3 −1][1.428570.50.11905−0.5]=2.07143$

Thus, the value of $a′β^$ is 2.07143.

The value of $a′(X′X)−1a$ is as follows:

$a′(X′X)−1a=[1 1 −3 −1][1/7 0 0 00 1/28 0 00 0 1/84 00 0 0 1/6][11−3−1]=0.45238$

From Part (c), the value of s is 0.089087.

The t-critical value using Table 5 of Appendix 3 at 3 degrees of freedom is $t0.025=3.182$ .

The 95% confidence interval for the expectedvalue of Y is as follows:

The lower limit is as follows:

$a′β^−t0.025.sa′(X′X)−1a=2.07143−3.182(0.089087)0.45238=1.881$

The upper limit is as follows:

$a′β^+t0.025.sa′(X′X)−1a=2.07143+3.182(0.089087)0.45238=2.262$

The 95% confidence interval for the expectedvalue of Y is (1.881, 2.262).

Answer 28

d.

To determine

Find a 95% confidence interval for the expected value of Y when $x1=1$ , $x2=−3$ , and $x3=−1$ .

Answer 29

d.

Expert Solution

Answer to Problem 97SE

The 95% confidence interval for the expected value of Y is (1.881, 2.262).

Answer 30

d.

Expert Solution

Answer 31

d.

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

A 95% confidence interval for the expectedvalue of Y is as follows:

$(a′β^−t0.025.sa′(X′X)−1a,a′β^+t0.025.sa′(X′X)−1a)$

It is given that $x1=1$ , $x2=−3$ , and $x3=−1$ .

The matrix ‘ $a′$ ’ can be obtained as follows:

$a′=[1 1 −3 −1]$

From Part (a), the parameter $β^$ , $n$ and $k$ can be found as follows:

$β^=[1.428570.50.11905−0.5]$ , $n=7$ and $k=3$

The value of $a′β^$ is as follows:

$a′β^=[1 1 −3 −1][1.428570.50.11905−0.5]=2.07143$

Thus, the value of $a′β^$ is 2.07143.

The value of $a′(X′X)−1a$ is as follows:

$a′(X′X)−1a=[1 1 −3 −1][1/7 0 0 00 1/28 0 00 0 1/84 00 0 0 1/6][11−3−1]=0.45238$

From Part (c), the value of s is 0.089087.

The t-critical value using Table 5 of Appendix 3 at 3 degrees of freedom is $t0.025=3.182$ .

The 95% confidence interval for the expectedvalue of Y is as follows:

The lower limit is as follows:

$a′β^−t0.025.sa′(X′X)−1a=2.07143−3.182(0.089087)0.45238=1.881$

The upper limit is as follows:

$a′β^+t0.025.sa′(X′X)−1a=2.07143+3.182(0.089087)0.45238=2.262$

The 95% confidence interval for the expectedvalue of Y is (1.881, 2.262).

Answer 34

e.

To determine

Find a 95% prediction interval for the expected value of Y when $x1=1$ , $x2=−3$ , and $x3=−1$ .

e.

Expert Solution

Answer to Problem 97SE

The 95% prediction interval for the expected value of Y is (1.730, 2.413).

Explanation of Solution

A 95% prediction interval for the expectedvalue of Y is as follows:

$(a′β^−t0.025.s1+a′(X′X)−1a,a′β^+t0.025.s1+a′(X′X)−1a)$

It is given that $x1=1$ , $x2=−3$ , and $x3=−1$

The matrix ‘ $a′$ ’ can be obtained as follows:

$a′=[1 1 −3 −1]$

From Part (d), the value of $a′β^$ is 2.07143.

From Part (d), the value of $a′(X′X)−1a$ is 0.45238.

From Part (c), the value of s is 0.089087.

The t-critical value using Table 5 of Appendix 3 at 3 degrees of freedom is $t0.025=3.182$ .

The 95% prediction interval for the expectedvalue of Y is as follows:

The lower limit is as follows:

$a′β^−t0.025.s1+a′(X′X)−1a=2.07143−3.182(0.089087)1+0.45238=1.730$

The upper limit is as follows:

$a′β^+t0.025.s1+a′(X′X)−1a=2.07143+3.182(0.089087)1+0.45238=2.413$

The 95% prediction interval for the expectedvalue of Y is (1.730, 2.413).

Answer 35

e.

To determine

Find a 95% prediction interval for the expected value of Y when $x1=1$ , $x2=−3$ , and $x3=−1$ .

Answer 36

e.

Expert Solution

Answer to Problem 97SE

The 95% prediction interval for the expected value of Y is (1.730, 2.413).

Answer 37

e.

Expert Solution

Answer 38

e.

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

A 95% prediction interval for the expectedvalue of Y is as follows:

$(a′β^−t0.025.s1+a′(X′X)−1a,a′β^+t0.025.s1+a′(X′X)−1a)$

It is given that $x1=1$ , $x2=−3$ , and $x3=−1$

The matrix ‘ $a′$ ’ can be obtained as follows:

$a′=[1 1 −3 −1]$

From Part (d), the value of $a′β^$ is 2.07143.

From Part (d), the value of $a′(X′X)−1a$ is 0.45238.

From Part (c), the value of s is 0.089087.

The t-critical value using Table 5 of Appendix 3 at 3 degrees of freedom is $t0.025=3.182$ .

The 95% prediction interval for the expectedvalue of Y is as follows:

The lower limit is as follows:

$a′β^−t0.025.s1+a′(X′X)−1a=2.07143−3.182(0.089087)1+0.45238=1.730$

The upper limit is as follows:

$a′β^+t0.025.s1+a′(X′X)−1a=2.07143+3.182(0.089087)1+0.45238=2.413$

The 95% prediction interval for the expectedvalue of Y is (1.730, 2.413).

Answer 41

Ch. 11.3 - If 0 and 1 are the least-squares estimates for the...Ch. 11.3 - Prob. 2E Ch. 11.3 - Fit a straight line to the five data points in the...Ch. 11.3 - Auditors are often required to compare the audited...Ch. 11.3 - Prob. 5E Ch. 11.3 - Applet Exercise Refer to Exercises 11.2 and 11.5....Ch. 11.3 - Prob. 7E Ch. 11.3 - Laboratory experiments designed to measure LC50...Ch. 11.3 - Prob. 9E Ch. 11.3 - Suppose that we have postulated the model...

Fit the model for the given data.

Videos

Fit the model for the given data.

Answer to Problem 97SE

Explanation of Solution

Predict Y when $x1=1,x2=−3,x3=−1$ .

Compare with the observed response in the original data.

Explain why these two are not equal.

Answer to Problem 97SE

Explanation of Solution

Test the hypothesis $H0:β3=0$ against $Ha:β3≠0$ using $Ha:β3≠0$ .

Answer to Problem 97SE

Explanation of Solution

Find a 95% confidence interval for the expected value of Y when $x1=1$ , $x2=−3$ , and $x3=−1$ .

Answer to Problem 97SE

Explanation of Solution

Find a 95% prediction interval for the expected value of Y when $x1=1$ , $x2=−3$ , and $x3=−1$ .

Answer to Problem 97SE

Explanation of Solution

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