The independent variable x 1 , x 2 , x 3 used for prediction of y

Question

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Answer 1

Question

Chapter 13.4, Problem 13.4E

i.

To determine

To identify:

The independent variable $x1,x2,x3$ used for prediction of $y$

i.

Expert Solution

Answer to Problem 13.4E

Independent variable $x1,x2,x3$ contribute the information to predict $y$ ,

Explanation of Solution

Given information:

$b0=1.04b1=1.29SE(b1)=0.42b2=2.72SE(b2)=0.65b3=0.41SE(b3)=0.17$

Formula used:

The student’s t- statistics is given by the following expression:

$t=bi−βtSE(bi)$

The values:

$(t≤3.0714)=0.9947(t≤4.1846)=0.9992(t≤2.4117)=0.9827$

Calculation:

To identify which independent variable $x1,x2,x3$ contribute the information to predict $y$ , need to use the significance of partial regression co-efficient.

The $F$ test with the hypothesis:

To test the null hypothesis:

$H0:βt=0$

Vs

$H0:βt≠0$

And, $df=(n−k−1)=15−3−1=11$

1. At $b1=1.29,SE(b1)=0.42$ :

The student’s t- statistics is calculated as:

$t=b1−β1SE( b 1 )t=0.129−00.42t=3.071428$

The corresponding $P$ -value is calculated as:

$2P(t≥3.0714)=2P(1−( t≤300714))2P(t≥3.0714)=2P(1−0.9947)2P(t≥3.0714)=2(0.0053)2P(t≥3.0714)=0.0106$

Reject the null hypothesis $H0$ as the $P−value=0.0106$ is less than $α=0.05$

Hence, independent variable $x1$ contribute the information to predict $y$ .

2. At $b2=2.72,SE(b2)=0.65$ :

The student’s t- statistics is calculated as:

$t=b2−β2SE( b 2 )t=2.72−00.65t=4.1846$

The corresponding $P$ -value is calculated as:

$2P(t≥4.1846)=2P(1−( t≤4.1846))2P(t≥4.1846)=2P(1−0.9992)2P(t≥4.1846)=2(0.0008)2P(t≥4.1846)=0.0016$

Reject the null hypothesis $H0$ as the $P−value=0.0016$ is less than $α=0.05$

Hence, independent variable $x2$ contribute the information to predict $y$ .

3. At $b3=0.41,SE(b3)=0.17$ :

The student’s t- statistics is calculated as:

$t=b3−β3SE( b 3 )t=0.41−00.17t=2.4117$

The corresponding $P$ -value is calculated as:

$2P(t≥2.4117)=2P(1−( t≤2.4117))2P(t≥2.4117)=2P(1−0.9827)2P(t≥2.4117)=2(0.0173)2P(t≥2.4117)=0.0346$

Reject the null hypothesis $H0$ as the $P−value=0.0346$ is less than $α=0.05$

Hence, independent variable $x3$ contribute the information to predict $y$ .

Conclusion:

Therefore, all independent variable $x1,x2,x3$ contribute the information to predict $y$ .

ii.

To determine

The least-square prediction equation

ii.

Expert Solution

Answer to Problem 13.4E

The least-square prediction equation is $Y=1.04+1.29x1+2.72x2+0.41x3$

Explanation of Solution

Given information:

$b0=1.04b1=1.29SE(b1)=0.42b2=2.72SE(b2)=0.65b3=0.41SE(b3)=0.17$

Calculation:

The line which makes the vertical distance from the data points to the regression line is known as least-square regression. This distance is as small as possible.

The least-square prediction equation is given by the following eq:

$y^=b0+b1x1+b2x2+b3x3$

Put the given values:

$y^=b0+b1x1+b2x2+b3x3y^=1.04+1.29x1+2.72x2+0.41x3$

Conclusion:

Hence, least-square prediction equation is derived as $y^=1.04+1.29x1+2.72x2+0.41x3$

iii.

To determine

To explain:

The relationship between lines shown by the graph the relationship between $x1$ and $y$ when $(x2=1,x3=0)$ and $(x2=1,x3=0.5)$

iii.

Expert Solution

Answer to Problem 13.4E

Lines shown by the graph is appearing to be parallel with each other

Explanation of Solution

Given information:

Predictor variables: $x1,x2,x3$

Calculation:

The given equation is $y^=b0+b1x1+b2x2+b3x3$ related to three predictor variables $x1,x2,x3$ .

When $(x2=1,x3=0)$ the equation becomes:

$y^=b0+b1x1+b2x2+b3x3y^=1.04+1.29x1+2.72x2+0.41x3y^=1.04+1.29x1+2.72(1)+0.41(0)y^=3.76+1.29x1$

When $(x2=1,x3=0.5)$ the equation becomes:

$y^=b0+b1x1+b2x2+b3x3y^=1.04+1.29x1+2.72x2+0.41x3y^=1.04+1.29x1+2.72(1)+0.41(0.5)y^=3.965+1.29x1$

The below graph depicting the relationship between $x1$ and $y$ when $(x2=1,x3=0)$ and $(x2=1,x3=0.5)$

EP INTRODUCTION TO PROBABILITY+STAT., Chapter 13.4, Problem 13.4E

Two lines shown by graph are seems to be parallel with each other.

Conclusion:

The above graph depicting the relationship between the two lines and they appear to be parallel to each other.

iv.

To determine

To explain:

The practical interpretation of $β1$

iv.

Expert Solution

Answer to Problem 13.4E

The measurement of changes in $x1$ when all other independent variables are held to be constant is done by $β1$ .

Explanation of Solution

Three-dimensional extension line of means is depicted by the given eq.

$E(y)=β0+β1x1+β2x2$

When $x1$ and $x2$ both are zero, the intercept-the average value of $y$ is given by $β0$

Partial slopes of the model is denoted by $β1$ and $β0$ . They are also known as partial regression coefficients.

The measurement of changes occurs in $y$ for one unit change in $x1$ when all other independent variables are held to be constant is done by $β1$ .

The slope estimated by a fit line with $x1$ alone is not same as partial regression coefficients $β1$ with $x1$ and $x2$ .

Conclusion:

Hence, the unknown constant values are estimated by using the sample data.

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Answer 2

Question

Chapter 13.4, Problem 13.4E

i.

To determine

To identify:

The independent variable $x1,x2,x3$ used for prediction of $y$

i.

Expert Solution

Answer to Problem 13.4E

Independent variable $x1,x2,x3$ contribute the information to predict $y$ ,

Explanation of Solution

Given information:

$b0=1.04b1=1.29SE(b1)=0.42b2=2.72SE(b2)=0.65b3=0.41SE(b3)=0.17$

Formula used:

The student’s t- statistics is given by the following expression:

$t=bi−βtSE(bi)$

The values:

$(t≤3.0714)=0.9947(t≤4.1846)=0.9992(t≤2.4117)=0.9827$

Calculation:

To identify which independent variable $x1,x2,x3$ contribute the information to predict $y$ , need to use the significance of partial regression co-efficient.

The $F$ test with the hypothesis:

To test the null hypothesis:

$H0:βt=0$

Vs

$H0:βt≠0$

And, $df=(n−k−1)=15−3−1=11$

1. At $b1=1.29,SE(b1)=0.42$ :

The student’s t- statistics is calculated as:

$t=b1−β1SE( b 1 )t=0.129−00.42t=3.071428$

The corresponding $P$ -value is calculated as:

$2P(t≥3.0714)=2P(1−( t≤300714))2P(t≥3.0714)=2P(1−0.9947)2P(t≥3.0714)=2(0.0053)2P(t≥3.0714)=0.0106$

Reject the null hypothesis $H0$ as the $P−value=0.0106$ is less than $α=0.05$

Hence, independent variable $x1$ contribute the information to predict $y$ .

2. At $b2=2.72,SE(b2)=0.65$ :

The student’s t- statistics is calculated as:

$t=b2−β2SE( b 2 )t=2.72−00.65t=4.1846$

The corresponding $P$ -value is calculated as:

$2P(t≥4.1846)=2P(1−( t≤4.1846))2P(t≥4.1846)=2P(1−0.9992)2P(t≥4.1846)=2(0.0008)2P(t≥4.1846)=0.0016$

Reject the null hypothesis $H0$ as the $P−value=0.0016$ is less than $α=0.05$

Hence, independent variable $x2$ contribute the information to predict $y$ .

3. At $b3=0.41,SE(b3)=0.17$ :

The student’s t- statistics is calculated as:

$t=b3−β3SE( b 3 )t=0.41−00.17t=2.4117$

The corresponding $P$ -value is calculated as:

$2P(t≥2.4117)=2P(1−( t≤2.4117))2P(t≥2.4117)=2P(1−0.9827)2P(t≥2.4117)=2(0.0173)2P(t≥2.4117)=0.0346$

Reject the null hypothesis $H0$ as the $P−value=0.0346$ is less than $α=0.05$

Hence, independent variable $x3$ contribute the information to predict $y$ .

Conclusion:

Therefore, all independent variable $x1,x2,x3$ contribute the information to predict $y$ .

ii.

To determine

The least-square prediction equation

ii.

Expert Solution

Answer to Problem 13.4E

The least-square prediction equation is $Y=1.04+1.29x1+2.72x2+0.41x3$

Explanation of Solution

Given information:

$b0=1.04b1=1.29SE(b1)=0.42b2=2.72SE(b2)=0.65b3=0.41SE(b3)=0.17$

Calculation:

The line which makes the vertical distance from the data points to the regression line is known as least-square regression. This distance is as small as possible.

The least-square prediction equation is given by the following eq:

$y^=b0+b1x1+b2x2+b3x3$

Put the given values:

$y^=b0+b1x1+b2x2+b3x3y^=1.04+1.29x1+2.72x2+0.41x3$

Conclusion:

Hence, least-square prediction equation is derived as $y^=1.04+1.29x1+2.72x2+0.41x3$

iii.

To determine

To explain:

The relationship between lines shown by the graph the relationship between $x1$ and $y$ when $(x2=1,x3=0)$ and $(x2=1,x3=0.5)$

iii.

Expert Solution

Answer to Problem 13.4E

Lines shown by the graph is appearing to be parallel with each other

Explanation of Solution

Given information:

Predictor variables: $x1,x2,x3$

Calculation:

The given equation is $y^=b0+b1x1+b2x2+b3x3$ related to three predictor variables $x1,x2,x3$ .

When $(x2=1,x3=0)$ the equation becomes:

$y^=b0+b1x1+b2x2+b3x3y^=1.04+1.29x1+2.72x2+0.41x3y^=1.04+1.29x1+2.72(1)+0.41(0)y^=3.76+1.29x1$

When $(x2=1,x3=0.5)$ the equation becomes:

$y^=b0+b1x1+b2x2+b3x3y^=1.04+1.29x1+2.72x2+0.41x3y^=1.04+1.29x1+2.72(1)+0.41(0.5)y^=3.965+1.29x1$

The below graph depicting the relationship between $x1$ and $y$ when $(x2=1,x3=0)$ and $(x2=1,x3=0.5)$

EP INTRODUCTION TO PROBABILITY+STAT., Chapter 13.4, Problem 13.4E

Two lines shown by graph are seems to be parallel with each other.

Conclusion:

The above graph depicting the relationship between the two lines and they appear to be parallel to each other.

iv.

To determine

To explain:

The practical interpretation of $β1$

iv.

Expert Solution

Answer to Problem 13.4E

The measurement of changes in $x1$ when all other independent variables are held to be constant is done by $β1$ .

Explanation of Solution

Three-dimensional extension line of means is depicted by the given eq.

$E(y)=β0+β1x1+β2x2$

When $x1$ and $x2$ both are zero, the intercept-the average value of $y$ is given by $β0$

Partial slopes of the model is denoted by $β1$ and $β0$ . They are also known as partial regression coefficients.

The measurement of changes occurs in $y$ for one unit change in $x1$ when all other independent variables are held to be constant is done by $β1$ .

The slope estimated by a fit line with $x1$ alone is not same as partial regression coefficients $β1$ with $x1$ and $x2$ .

Conclusion:

Hence, the unknown constant values are estimated by using the sample data.

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Answer 3

Question

Chapter 13.4, Problem 13.4E

i.

To determine

To identify:

The independent variable $x1,x2,x3$ used for prediction of $y$

i.

Expert Solution

Answer to Problem 13.4E

Independent variable $x1,x2,x3$ contribute the information to predict $y$ ,

Explanation of Solution

Given information:

$b0=1.04b1=1.29SE(b1)=0.42b2=2.72SE(b2)=0.65b3=0.41SE(b3)=0.17$

Formula used:

The student’s t- statistics is given by the following expression:

$t=bi−βtSE(bi)$

The values:

$(t≤3.0714)=0.9947(t≤4.1846)=0.9992(t≤2.4117)=0.9827$

Calculation:

To identify which independent variable $x1,x2,x3$ contribute the information to predict $y$ , need to use the significance of partial regression co-efficient.

The $F$ test with the hypothesis:

To test the null hypothesis:

$H0:βt=0$

Vs

$H0:βt≠0$

And, $df=(n−k−1)=15−3−1=11$

1. At $b1=1.29,SE(b1)=0.42$ :

The student’s t- statistics is calculated as:

$t=b1−β1SE( b 1 )t=0.129−00.42t=3.071428$

The corresponding $P$ -value is calculated as:

$2P(t≥3.0714)=2P(1−( t≤300714))2P(t≥3.0714)=2P(1−0.9947)2P(t≥3.0714)=2(0.0053)2P(t≥3.0714)=0.0106$

Reject the null hypothesis $H0$ as the $P−value=0.0106$ is less than $α=0.05$

Hence, independent variable $x1$ contribute the information to predict $y$ .

2. At $b2=2.72,SE(b2)=0.65$ :

The student’s t- statistics is calculated as:

$t=b2−β2SE( b 2 )t=2.72−00.65t=4.1846$

The corresponding $P$ -value is calculated as:

$2P(t≥4.1846)=2P(1−( t≤4.1846))2P(t≥4.1846)=2P(1−0.9992)2P(t≥4.1846)=2(0.0008)2P(t≥4.1846)=0.0016$

Reject the null hypothesis $H0$ as the $P−value=0.0016$ is less than $α=0.05$

Hence, independent variable $x2$ contribute the information to predict $y$ .

3. At $b3=0.41,SE(b3)=0.17$ :

The student’s t- statistics is calculated as:

$t=b3−β3SE( b 3 )t=0.41−00.17t=2.4117$

The corresponding $P$ -value is calculated as:

$2P(t≥2.4117)=2P(1−( t≤2.4117))2P(t≥2.4117)=2P(1−0.9827)2P(t≥2.4117)=2(0.0173)2P(t≥2.4117)=0.0346$

Reject the null hypothesis $H0$ as the $P−value=0.0346$ is less than $α=0.05$

Hence, independent variable $x3$ contribute the information to predict $y$ .

Conclusion:

Therefore, all independent variable $x1,x2,x3$ contribute the information to predict $y$ .

ii.

To determine

The least-square prediction equation

ii.

Expert Solution

Answer to Problem 13.4E

The least-square prediction equation is $Y=1.04+1.29x1+2.72x2+0.41x3$

Explanation of Solution

Given information:

$b0=1.04b1=1.29SE(b1)=0.42b2=2.72SE(b2)=0.65b3=0.41SE(b3)=0.17$

Calculation:

The line which makes the vertical distance from the data points to the regression line is known as least-square regression. This distance is as small as possible.

The least-square prediction equation is given by the following eq:

$y^=b0+b1x1+b2x2+b3x3$

Put the given values:

$y^=b0+b1x1+b2x2+b3x3y^=1.04+1.29x1+2.72x2+0.41x3$

Conclusion:

Hence, least-square prediction equation is derived as $y^=1.04+1.29x1+2.72x2+0.41x3$

iii.

To determine

To explain:

The relationship between lines shown by the graph the relationship between $x1$ and $y$ when $(x2=1,x3=0)$ and $(x2=1,x3=0.5)$

iii.

Expert Solution

Answer to Problem 13.4E

Lines shown by the graph is appearing to be parallel with each other

Explanation of Solution

Given information:

Predictor variables: $x1,x2,x3$

Calculation:

The given equation is $y^=b0+b1x1+b2x2+b3x3$ related to three predictor variables $x1,x2,x3$ .

When $(x2=1,x3=0)$ the equation becomes:

$y^=b0+b1x1+b2x2+b3x3y^=1.04+1.29x1+2.72x2+0.41x3y^=1.04+1.29x1+2.72(1)+0.41(0)y^=3.76+1.29x1$

When $(x2=1,x3=0.5)$ the equation becomes:

$y^=b0+b1x1+b2x2+b3x3y^=1.04+1.29x1+2.72x2+0.41x3y^=1.04+1.29x1+2.72(1)+0.41(0.5)y^=3.965+1.29x1$

The below graph depicting the relationship between $x1$ and $y$ when $(x2=1,x3=0)$ and $(x2=1,x3=0.5)$

EP INTRODUCTION TO PROBABILITY+STAT., Chapter 13.4, Problem 13.4E

Two lines shown by graph are seems to be parallel with each other.

Conclusion:

The above graph depicting the relationship between the two lines and they appear to be parallel to each other.

iv.

To determine

To explain:

The practical interpretation of $β1$

iv.

Expert Solution

Answer to Problem 13.4E

The measurement of changes in $x1$ when all other independent variables are held to be constant is done by $β1$ .

Explanation of Solution

Three-dimensional extension line of means is depicted by the given eq.

$E(y)=β0+β1x1+β2x2$

When $x1$ and $x2$ both are zero, the intercept-the average value of $y$ is given by $β0$

Partial slopes of the model is denoted by $β1$ and $β0$ . They are also known as partial regression coefficients.

The measurement of changes occurs in $y$ for one unit change in $x1$ when all other independent variables are held to be constant is done by $β1$ .

The slope estimated by a fit line with $x1$ alone is not same as partial regression coefficients $β1$ with $x1$ and $x2$ .

Conclusion:

Hence, the unknown constant values are estimated by using the sample data.

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Answer 4

Question

Chapter 13.4, Problem 13.4E

i.

To determine

To identify:

The independent variable $x1,x2,x3$ used for prediction of $y$

i.

Expert Solution

Answer to Problem 13.4E

Independent variable $x1,x2,x3$ contribute the information to predict $y$ ,

Explanation of Solution

Given information:

$b0=1.04b1=1.29SE(b1)=0.42b2=2.72SE(b2)=0.65b3=0.41SE(b3)=0.17$

Formula used:

The student’s t- statistics is given by the following expression:

$t=bi−βtSE(bi)$

The values:

$(t≤3.0714)=0.9947(t≤4.1846)=0.9992(t≤2.4117)=0.9827$

Calculation:

To identify which independent variable $x1,x2,x3$ contribute the information to predict $y$ , need to use the significance of partial regression co-efficient.

The $F$ test with the hypothesis:

To test the null hypothesis:

$H0:βt=0$

Vs

$H0:βt≠0$

And, $df=(n−k−1)=15−3−1=11$

1. At $b1=1.29,SE(b1)=0.42$ :

The student’s t- statistics is calculated as:

$t=b1−β1SE( b 1 )t=0.129−00.42t=3.071428$

The corresponding $P$ -value is calculated as:

$2P(t≥3.0714)=2P(1−( t≤300714))2P(t≥3.0714)=2P(1−0.9947)2P(t≥3.0714)=2(0.0053)2P(t≥3.0714)=0.0106$

Reject the null hypothesis $H0$ as the $P−value=0.0106$ is less than $α=0.05$

Hence, independent variable $x1$ contribute the information to predict $y$ .

2. At $b2=2.72,SE(b2)=0.65$ :

The student’s t- statistics is calculated as:

$t=b2−β2SE( b 2 )t=2.72−00.65t=4.1846$

The corresponding $P$ -value is calculated as:

$2P(t≥4.1846)=2P(1−( t≤4.1846))2P(t≥4.1846)=2P(1−0.9992)2P(t≥4.1846)=2(0.0008)2P(t≥4.1846)=0.0016$

Reject the null hypothesis $H0$ as the $P−value=0.0016$ is less than $α=0.05$

Hence, independent variable $x2$ contribute the information to predict $y$ .

3. At $b3=0.41,SE(b3)=0.17$ :

The student’s t- statistics is calculated as:

$t=b3−β3SE( b 3 )t=0.41−00.17t=2.4117$

The corresponding $P$ -value is calculated as:

$2P(t≥2.4117)=2P(1−( t≤2.4117))2P(t≥2.4117)=2P(1−0.9827)2P(t≥2.4117)=2(0.0173)2P(t≥2.4117)=0.0346$

Reject the null hypothesis $H0$ as the $P−value=0.0346$ is less than $α=0.05$

Hence, independent variable $x3$ contribute the information to predict $y$ .

Conclusion:

Therefore, all independent variable $x1,x2,x3$ contribute the information to predict $y$ .

ii.

To determine

The least-square prediction equation

ii.

Expert Solution

Answer to Problem 13.4E

The least-square prediction equation is $Y=1.04+1.29x1+2.72x2+0.41x3$

Explanation of Solution

Given information:

$b0=1.04b1=1.29SE(b1)=0.42b2=2.72SE(b2)=0.65b3=0.41SE(b3)=0.17$

Calculation:

The line which makes the vertical distance from the data points to the regression line is known as least-square regression. This distance is as small as possible.

The least-square prediction equation is given by the following eq:

$y^=b0+b1x1+b2x2+b3x3$

Put the given values:

$y^=b0+b1x1+b2x2+b3x3y^=1.04+1.29x1+2.72x2+0.41x3$

Conclusion:

Hence, least-square prediction equation is derived as $y^=1.04+1.29x1+2.72x2+0.41x3$

iii.

To determine

To explain:

The relationship between lines shown by the graph the relationship between $x1$ and $y$ when $(x2=1,x3=0)$ and $(x2=1,x3=0.5)$

iii.

Expert Solution

Answer to Problem 13.4E

Lines shown by the graph is appearing to be parallel with each other

Explanation of Solution

Given information:

Predictor variables: $x1,x2,x3$

Calculation:

The given equation is $y^=b0+b1x1+b2x2+b3x3$ related to three predictor variables $x1,x2,x3$ .

When $(x2=1,x3=0)$ the equation becomes:

$y^=b0+b1x1+b2x2+b3x3y^=1.04+1.29x1+2.72x2+0.41x3y^=1.04+1.29x1+2.72(1)+0.41(0)y^=3.76+1.29x1$

When $(x2=1,x3=0.5)$ the equation becomes:

$y^=b0+b1x1+b2x2+b3x3y^=1.04+1.29x1+2.72x2+0.41x3y^=1.04+1.29x1+2.72(1)+0.41(0.5)y^=3.965+1.29x1$

The below graph depicting the relationship between $x1$ and $y$ when $(x2=1,x3=0)$ and $(x2=1,x3=0.5)$

EP INTRODUCTION TO PROBABILITY+STAT., Chapter 13.4, Problem 13.4E

Two lines shown by graph are seems to be parallel with each other.

Conclusion:

The above graph depicting the relationship between the two lines and they appear to be parallel to each other.

iv.

To determine

To explain:

The practical interpretation of $β1$

iv.

Expert Solution

Answer to Problem 13.4E

The measurement of changes in $x1$ when all other independent variables are held to be constant is done by $β1$ .

Explanation of Solution

Three-dimensional extension line of means is depicted by the given eq.

$E(y)=β0+β1x1+β2x2$

When $x1$ and $x2$ both are zero, the intercept-the average value of $y$ is given by $β0$

Partial slopes of the model is denoted by $β1$ and $β0$ . They are also known as partial regression coefficients.

The measurement of changes occurs in $y$ for one unit change in $x1$ when all other independent variables are held to be constant is done by $β1$ .

The slope estimated by a fit line with $x1$ alone is not same as partial regression coefficients $β1$ with $x1$ and $x2$ .

Conclusion:

Hence, the unknown constant values are estimated by using the sample data.

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Answer 5

Chapter 13.4, Problem 13.4E

i.

To determine

To identify:

The independent variable $x1,x2,x3$ used for prediction of $y$

i.

Expert Solution

Answer to Problem 13.4E

Independent variable $x1,x2,x3$ contribute the information to predict $y$ ,

Explanation of Solution

Given information:

$b0=1.04b1=1.29SE(b1)=0.42b2=2.72SE(b2)=0.65b3=0.41SE(b3)=0.17$

Formula used:

The student’s t- statistics is given by the following expression:

$t=bi−βtSE(bi)$

The values:

$(t≤3.0714)=0.9947(t≤4.1846)=0.9992(t≤2.4117)=0.9827$

Calculation:

To identify which independent variable $x1,x2,x3$ contribute the information to predict $y$ , need to use the significance of partial regression co-efficient.

The $F$ test with the hypothesis:

To test the null hypothesis:

$H0:βt=0$

Vs

$H0:βt≠0$

And, $df=(n−k−1)=15−3−1=11$

1. At $b1=1.29,SE(b1)=0.42$ :

The student’s t- statistics is calculated as:

$t=b1−β1SE( b 1 )t=0.129−00.42t=3.071428$

The corresponding $P$ -value is calculated as:

$2P(t≥3.0714)=2P(1−( t≤300714))2P(t≥3.0714)=2P(1−0.9947)2P(t≥3.0714)=2(0.0053)2P(t≥3.0714)=0.0106$

Reject the null hypothesis $H0$ as the $P−value=0.0106$ is less than $α=0.05$

Hence, independent variable $x1$ contribute the information to predict $y$ .

2. At $b2=2.72,SE(b2)=0.65$ :

The student’s t- statistics is calculated as:

$t=b2−β2SE( b 2 )t=2.72−00.65t=4.1846$

The corresponding $P$ -value is calculated as:

$2P(t≥4.1846)=2P(1−( t≤4.1846))2P(t≥4.1846)=2P(1−0.9992)2P(t≥4.1846)=2(0.0008)2P(t≥4.1846)=0.0016$

Reject the null hypothesis $H0$ as the $P−value=0.0016$ is less than $α=0.05$

Hence, independent variable $x2$ contribute the information to predict $y$ .

3. At $b3=0.41,SE(b3)=0.17$ :

The student’s t- statistics is calculated as:

$t=b3−β3SE( b 3 )t=0.41−00.17t=2.4117$

The corresponding $P$ -value is calculated as:

$2P(t≥2.4117)=2P(1−( t≤2.4117))2P(t≥2.4117)=2P(1−0.9827)2P(t≥2.4117)=2(0.0173)2P(t≥2.4117)=0.0346$

Reject the null hypothesis $H0$ as the $P−value=0.0346$ is less than $α=0.05$

Hence, independent variable $x3$ contribute the information to predict $y$ .

Conclusion:

Therefore, all independent variable $x1,x2,x3$ contribute the information to predict $y$ .

ii.

To determine

The least-square prediction equation

ii.

Expert Solution

Answer to Problem 13.4E

The least-square prediction equation is $Y=1.04+1.29x1+2.72x2+0.41x3$

Explanation of Solution

Given information:

$b0=1.04b1=1.29SE(b1)=0.42b2=2.72SE(b2)=0.65b3=0.41SE(b3)=0.17$

Calculation:

The line which makes the vertical distance from the data points to the regression line is known as least-square regression. This distance is as small as possible.

The least-square prediction equation is given by the following eq:

$y^=b0+b1x1+b2x2+b3x3$

Put the given values:

$y^=b0+b1x1+b2x2+b3x3y^=1.04+1.29x1+2.72x2+0.41x3$

Conclusion:

Hence, least-square prediction equation is derived as $y^=1.04+1.29x1+2.72x2+0.41x3$

iii.

To determine

To explain:

The relationship between lines shown by the graph the relationship between $x1$ and $y$ when $(x2=1,x3=0)$ and $(x2=1,x3=0.5)$

iii.

Expert Solution

Answer to Problem 13.4E

Lines shown by the graph is appearing to be parallel with each other

Explanation of Solution

Given information:

Predictor variables: $x1,x2,x3$

Calculation:

The given equation is $y^=b0+b1x1+b2x2+b3x3$ related to three predictor variables $x1,x2,x3$ .

When $(x2=1,x3=0)$ the equation becomes:

$y^=b0+b1x1+b2x2+b3x3y^=1.04+1.29x1+2.72x2+0.41x3y^=1.04+1.29x1+2.72(1)+0.41(0)y^=3.76+1.29x1$

When $(x2=1,x3=0.5)$ the equation becomes:

$y^=b0+b1x1+b2x2+b3x3y^=1.04+1.29x1+2.72x2+0.41x3y^=1.04+1.29x1+2.72(1)+0.41(0.5)y^=3.965+1.29x1$

The below graph depicting the relationship between $x1$ and $y$ when $(x2=1,x3=0)$ and $(x2=1,x3=0.5)$

EP INTRODUCTION TO PROBABILITY+STAT., Chapter 13.4, Problem 13.4E

Two lines shown by graph are seems to be parallel with each other.

Conclusion:

The above graph depicting the relationship between the two lines and they appear to be parallel to each other.

iv.

To determine

To explain:

The practical interpretation of $β1$

iv.

Expert Solution

Answer to Problem 13.4E

The measurement of changes in $x1$ when all other independent variables are held to be constant is done by $β1$ .

Explanation of Solution

Three-dimensional extension line of means is depicted by the given eq.

$E(y)=β0+β1x1+β2x2$

When $x1$ and $x2$ both are zero, the intercept-the average value of $y$ is given by $β0$

Partial slopes of the model is denoted by $β1$ and $β0$ . They are also known as partial regression coefficients.

The measurement of changes occurs in $y$ for one unit change in $x1$ when all other independent variables are held to be constant is done by $β1$ .

The slope estimated by a fit line with $x1$ alone is not same as partial regression coefficients $β1$ with $x1$ and $x2$ .

Conclusion:

Hence, the unknown constant values are estimated by using the sample data.

Answer 6

Chapter 13.4, Problem 13.4E

i.

To determine

To identify:

The independent variable $x1,x2,x3$ used for prediction of $y$

i.

Expert Solution

Answer to Problem 13.4E

Independent variable $x1,x2,x3$ contribute the information to predict $y$ ,

Explanation of Solution

Given information:

$b0=1.04b1=1.29SE(b1)=0.42b2=2.72SE(b2)=0.65b3=0.41SE(b3)=0.17$

Formula used:

The student’s t- statistics is given by the following expression:

$t=bi−βtSE(bi)$

The values:

$(t≤3.0714)=0.9947(t≤4.1846)=0.9992(t≤2.4117)=0.9827$

Calculation:

To identify which independent variable $x1,x2,x3$ contribute the information to predict $y$ , need to use the significance of partial regression co-efficient.

The $F$ test with the hypothesis:

To test the null hypothesis:

$H0:βt=0$

Vs

$H0:βt≠0$

And, $df=(n−k−1)=15−3−1=11$

1. At $b1=1.29,SE(b1)=0.42$ :

The student’s t- statistics is calculated as:

$t=b1−β1SE( b 1 )t=0.129−00.42t=3.071428$

The corresponding $P$ -value is calculated as:

$2P(t≥3.0714)=2P(1−( t≤300714))2P(t≥3.0714)=2P(1−0.9947)2P(t≥3.0714)=2(0.0053)2P(t≥3.0714)=0.0106$

Reject the null hypothesis $H0$ as the $P−value=0.0106$ is less than $α=0.05$

Hence, independent variable $x1$ contribute the information to predict $y$ .

2. At $b2=2.72,SE(b2)=0.65$ :

The student’s t- statistics is calculated as:

$t=b2−β2SE( b 2 )t=2.72−00.65t=4.1846$

The corresponding $P$ -value is calculated as:

$2P(t≥4.1846)=2P(1−( t≤4.1846))2P(t≥4.1846)=2P(1−0.9992)2P(t≥4.1846)=2(0.0008)2P(t≥4.1846)=0.0016$

Reject the null hypothesis $H0$ as the $P−value=0.0016$ is less than $α=0.05$

Hence, independent variable $x2$ contribute the information to predict $y$ .

3. At $b3=0.41,SE(b3)=0.17$ :

The student’s t- statistics is calculated as:

$t=b3−β3SE( b 3 )t=0.41−00.17t=2.4117$

The corresponding $P$ -value is calculated as:

$2P(t≥2.4117)=2P(1−( t≤2.4117))2P(t≥2.4117)=2P(1−0.9827)2P(t≥2.4117)=2(0.0173)2P(t≥2.4117)=0.0346$

Reject the null hypothesis $H0$ as the $P−value=0.0346$ is less than $α=0.05$

Hence, independent variable $x3$ contribute the information to predict $y$ .

Conclusion:

Therefore, all independent variable $x1,x2,x3$ contribute the information to predict $y$ .

Answer 7

Chapter 13.4, Problem 13.4E

i.

To determine

To identify:

The independent variable $x1,x2,x3$ used for prediction of $y$

Answer 8

i.

Expert Solution

Answer to Problem 13.4E

Independent variable $x1,x2,x3$ contribute the information to predict $y$ ,

Answer 9

i.

Expert Solution

Answer 10

i.

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given information:

$b0=1.04b1=1.29SE(b1)=0.42b2=2.72SE(b2)=0.65b3=0.41SE(b3)=0.17$

Formula used:

The student’s t- statistics is given by the following expression:

$t=bi−βtSE(bi)$

The values:

$(t≤3.0714)=0.9947(t≤4.1846)=0.9992(t≤2.4117)=0.9827$

Calculation:

To identify which independent variable $x1,x2,x3$ contribute the information to predict $y$ , need to use the significance of partial regression co-efficient.

The $F$ test with the hypothesis:

To test the null hypothesis:

$H0:βt=0$

Vs

$H0:βt≠0$

And, $df=(n−k−1)=15−3−1=11$

1. At $b1=1.29,SE(b1)=0.42$ :

The student’s t- statistics is calculated as:

$t=b1−β1SE( b 1 )t=0.129−00.42t=3.071428$

The corresponding $P$ -value is calculated as:

$2P(t≥3.0714)=2P(1−( t≤300714))2P(t≥3.0714)=2P(1−0.9947)2P(t≥3.0714)=2(0.0053)2P(t≥3.0714)=0.0106$

Reject the null hypothesis $H0$ as the $P−value=0.0106$ is less than $α=0.05$

Hence, independent variable $x1$ contribute the information to predict $y$ .

2. At $b2=2.72,SE(b2)=0.65$ :

The student’s t- statistics is calculated as:

$t=b2−β2SE( b 2 )t=2.72−00.65t=4.1846$

The corresponding $P$ -value is calculated as:

$2P(t≥4.1846)=2P(1−( t≤4.1846))2P(t≥4.1846)=2P(1−0.9992)2P(t≥4.1846)=2(0.0008)2P(t≥4.1846)=0.0016$

Reject the null hypothesis $H0$ as the $P−value=0.0016$ is less than $α=0.05$

Hence, independent variable $x2$ contribute the information to predict $y$ .

3. At $b3=0.41,SE(b3)=0.17$ :

The student’s t- statistics is calculated as:

$t=b3−β3SE( b 3 )t=0.41−00.17t=2.4117$

The corresponding $P$ -value is calculated as:

$2P(t≥2.4117)=2P(1−( t≤2.4117))2P(t≥2.4117)=2P(1−0.9827)2P(t≥2.4117)=2(0.0173)2P(t≥2.4117)=0.0346$

Reject the null hypothesis $H0$ as the $P−value=0.0346$ is less than $α=0.05$

Hence, independent variable $x3$ contribute the information to predict $y$ .

Conclusion:

Therefore, all independent variable $x1,x2,x3$ contribute the information to predict $y$ .

Answer 13

ii.

To determine

The least-square prediction equation

ii.

Expert Solution

Answer to Problem 13.4E

The least-square prediction equation is $Y=1.04+1.29x1+2.72x2+0.41x3$

Explanation of Solution

Given information:

$b0=1.04b1=1.29SE(b1)=0.42b2=2.72SE(b2)=0.65b3=0.41SE(b3)=0.17$

Calculation:

The line which makes the vertical distance from the data points to the regression line is known as least-square regression. This distance is as small as possible.

The least-square prediction equation is given by the following eq:

$y^=b0+b1x1+b2x2+b3x3$

Put the given values:

$y^=b0+b1x1+b2x2+b3x3y^=1.04+1.29x1+2.72x2+0.41x3$

Conclusion:

Hence, least-square prediction equation is derived as $y^=1.04+1.29x1+2.72x2+0.41x3$

Answer 14

ii.

To determine

The least-square prediction equation

Answer 15

ii.

Expert Solution

Answer to Problem 13.4E

The least-square prediction equation is $Y=1.04+1.29x1+2.72x2+0.41x3$

Answer 16

ii.

Expert Solution

Answer 17

ii.

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given information:

$b0=1.04b1=1.29SE(b1)=0.42b2=2.72SE(b2)=0.65b3=0.41SE(b3)=0.17$

Calculation:

The line which makes the vertical distance from the data points to the regression line is known as least-square regression. This distance is as small as possible.

The least-square prediction equation is given by the following eq:

$y^=b0+b1x1+b2x2+b3x3$

Put the given values:

$y^=b0+b1x1+b2x2+b3x3y^=1.04+1.29x1+2.72x2+0.41x3$

Conclusion:

Hence, least-square prediction equation is derived as $y^=1.04+1.29x1+2.72x2+0.41x3$

Answer 20

iii.

To determine

To explain:

The relationship between lines shown by the graph the relationship between $x1$ and $y$ when $(x2=1,x3=0)$ and $(x2=1,x3=0.5)$

iii.

Expert Solution

Answer to Problem 13.4E

Lines shown by the graph is appearing to be parallel with each other

Explanation of Solution

Given information:

Predictor variables: $x1,x2,x3$

Calculation:

The given equation is $y^=b0+b1x1+b2x2+b3x3$ related to three predictor variables $x1,x2,x3$ .

When $(x2=1,x3=0)$ the equation becomes:

$y^=b0+b1x1+b2x2+b3x3y^=1.04+1.29x1+2.72x2+0.41x3y^=1.04+1.29x1+2.72(1)+0.41(0)y^=3.76+1.29x1$

When $(x2=1,x3=0.5)$ the equation becomes:

$y^=b0+b1x1+b2x2+b3x3y^=1.04+1.29x1+2.72x2+0.41x3y^=1.04+1.29x1+2.72(1)+0.41(0.5)y^=3.965+1.29x1$

The below graph depicting the relationship between $x1$ and $y$ when $(x2=1,x3=0)$ and $(x2=1,x3=0.5)$

EP INTRODUCTION TO PROBABILITY+STAT., Chapter 13.4, Problem 13.4E

Two lines shown by graph are seems to be parallel with each other.

Conclusion:

The above graph depicting the relationship between the two lines and they appear to be parallel to each other.

Answer 21

iii.

To determine

To explain:

The relationship between lines shown by the graph the relationship between $x1$ and $y$ when $(x2=1,x3=0)$ and $(x2=1,x3=0.5)$

Answer 22

iii.

Expert Solution

Answer to Problem 13.4E

Lines shown by the graph is appearing to be parallel with each other

Answer 23

iii.

Expert Solution

Answer 24

iii.

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given information:

Predictor variables: $x1,x2,x3$

Calculation:

The given equation is $y^=b0+b1x1+b2x2+b3x3$ related to three predictor variables $x1,x2,x3$ .

When $(x2=1,x3=0)$ the equation becomes:

$y^=b0+b1x1+b2x2+b3x3y^=1.04+1.29x1+2.72x2+0.41x3y^=1.04+1.29x1+2.72(1)+0.41(0)y^=3.76+1.29x1$

When $(x2=1,x3=0.5)$ the equation becomes:

$y^=b0+b1x1+b2x2+b3x3y^=1.04+1.29x1+2.72x2+0.41x3y^=1.04+1.29x1+2.72(1)+0.41(0.5)y^=3.965+1.29x1$

The below graph depicting the relationship between $x1$ and $y$ when $(x2=1,x3=0)$ and $(x2=1,x3=0.5)$

EP INTRODUCTION TO PROBABILITY+STAT., Chapter 13.4, Problem 13.4E

Two lines shown by graph are seems to be parallel with each other.

Conclusion:

The above graph depicting the relationship between the two lines and they appear to be parallel to each other.

Answer 27

iv.

To determine

To explain:

The practical interpretation of $β1$

iv.

Expert Solution

Answer to Problem 13.4E

The measurement of changes in $x1$ when all other independent variables are held to be constant is done by $β1$ .

Explanation of Solution

Three-dimensional extension line of means is depicted by the given eq.

$E(y)=β0+β1x1+β2x2$

When $x1$ and $x2$ both are zero, the intercept-the average value of $y$ is given by $β0$

Partial slopes of the model is denoted by $β1$ and $β0$ . They are also known as partial regression coefficients.

The measurement of changes occurs in $y$ for one unit change in $x1$ when all other independent variables are held to be constant is done by $β1$ .

The slope estimated by a fit line with $x1$ alone is not same as partial regression coefficients $β1$ with $x1$ and $x2$ .

Conclusion:

Hence, the unknown constant values are estimated by using the sample data.

Answer 28

iv.

To determine

To explain:

The practical interpretation of $β1$

Answer 29

iv.

Expert Solution

Answer to Problem 13.4E

The measurement of changes in $x1$ when all other independent variables are held to be constant is done by $β1$ .

Answer 30

iv.

Expert Solution

Answer 31

iv.

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Three-dimensional extension line of means is depicted by the given eq.

$E(y)=β0+β1x1+β2x2$

When $x1$ and $x2$ both are zero, the intercept-the average value of $y$ is given by $β0$

Partial slopes of the model is denoted by $β1$ and $β0$ . They are also known as partial regression coefficients.

The measurement of changes occurs in $y$ for one unit change in $x1$ when all other independent variables are held to be constant is done by $β1$ .

The slope estimated by a fit line with $x1$ alone is not same as partial regression coefficients $β1$ with $x1$ and $x2$ .

Conclusion:

Hence, the unknown constant values are estimated by using the sample data.

Answer 34

Ch. 13.4 - Prob. 13.1E Ch. 13.4 - Prob. 13.2E Ch. 13.4 - Suppose that you fit the model E(y)=0+1x1+2x2+3x3...Ch. 13.4 - Prob. 13.4E Ch. 13.4 - Prob. 13.5E Ch. 13.4 - Prob. 13.6E Ch. 13.4 - Prob. 13.7E Ch. 13.4 - Prob. 13.8E Ch. 13.4 - Prob. 13.9E Ch. 13.4 - College Textbooks A publisher of college textbooks...

The independent variable x 1 , x 2 , x 3 used for prediction of y

Concept explainers

Videos

To identify:

The independent variable $x1,x2,x3$ used for prediction of $y$

Answer to Problem 13.4E

Explanation of Solution

The least-square prediction equation

Answer to Problem 13.4E

Explanation of Solution

To explain:

The relationship between lines shown by the graph the relationship between $x1$ and $y$ when $(x2=1,x3=0)$ and $(x2=1,x3=0.5)$

Answer to Problem 13.4E

Explanation of Solution

To explain:

The practical interpretation of $β1$

Answer to Problem 13.4E

Explanation of Solution

Want to see more full solutions like this?

Chapter 13 Solutions

The independent variable x 1 , x 2 , x 3 used for prediction of y

Concept explainers

Videos

To identify: The independent variable x1,x2,x3<m:math><m:mrow><m:msub><m:mi>x</m:mi><m:mn>1</m:mn></m:msub><m:mo>,</m:mo><m:msub><m:mi>x</m:mi><m:mn>2</m:mn></m:msub><m:mo>,</m:mo><m:msub><m:mi>x</m:mi><m:mn>3</m:mn></m:msub></m:mrow></m:math> used for prediction of y<m:math><m:mi>y</m:mi></m:math>

Answer to Problem 13.4E

Explanation of Solution

The least-square prediction equation

Answer to Problem 13.4E

Explanation of Solution

Answer to Problem 13.4E

Explanation of Solution

To explain: The practical interpretation of β1<m:math><m:mrow><m:msub><m:mi>β</m:mi><m:mn>1</m:mn></m:msub></m:mrow></m:math>

Answer to Problem 13.4E

Explanation of Solution

Want to see more full solutions like this?

Chapter 13 Solutions

To identify:

The independent variable $x1,x2,x3$ used for prediction of $y$

To explain:

The practical interpretation of $β1$