Elementary Statistics (Text Only)
Elementary Statistics (Text Only)
2nd Edition
ISBN: 9780077836351
Author: Author
Publisher: McGraw Hill
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 13.3, Problem 21E

a.

To determine

To find:The regression equation for the data.

a.

Expert Solution
Check Mark

Answer to Problem 21E

The regression equation for the data is y=8.8720.0068x1+0.7163x2+2.03x3 .

Explanation of Solution

Given information: The data is shown below.

    y(x1)(x2)(x3)y(x1)(x2)(x3)y(x1)(x2)(x3)y(x1)(x2)(x3)
    31.3 50 194.037.46030 5.0 24.060244.051.080347.5
    56.990388.043.360267.036.2 50 217.034.360222.5
    43.170286.536.370257.526.550172.031.56024 5.0
    41.57025 5.5 38.47031 5.5 47.18034 8.5 33.260234.0
    39.060266.541.560277.538.170275.539.260296.5
    40.970295.036.l60236.033.560242.546.770277.5
    35.96023 5.5 38.560236.043.6702710.030.470324.0
    43.57028 5.5 42.460249.041.08032 6.5 43.26025 5.5
    47.9 80 346.546.57031 5.5 50.2 50 269.030.660263.5
    33.870264.543.180326.034.450224.043.370287.5
    41.170268.050.8602610.047.980346.532.460225.5
    38.770268.044.270284.546.2502110.035.560224.5

Calculation:

The MINITAB is shown below,

  Elementary Statistics (Text Only), Chapter 13.3, Problem 21E , additional homework tip  1

  Figure-1

From Figure-1 it is clear that the regression equation is y=8.8720.0068x1+0.7163x2+2.03x3

b.

To determine

To find: The value of variable y .

b.

Expert Solution
Check Mark

Answer to Problem 21E

The value of variable y is 42.201 .

Explanation of Solution

Given information: The data is shown below.

    y(x1)(x2)(x3)y(x1)(x2)(x3)y(x1)(x2)(x3)y(x1)(x2)(x3)
    31.3 50 194.037.46030 5.0 24.060244.051.080347.5
    56.990388.043.360267.036.2 50 217.034.360222.5
    43.170286.536.370257.526.550172.031.56024 5.0
    41.57025 5.5 38.47031 5.5 47.18034 8.5 33.260234.0
    39.060266.541.560277.538.170275.539.260296.5
    40.970295.036.l60236.033.560242.546.770277.5
    35.96023 5.5 38.560236.043.6702710.030.470324.0
    43.57028 5.5 42.460249.041.08032 6.5 43.26025 5.5
    47.9 80 346.546.57031 5.5 50.2 50 269.030.660263.5
    33.870264.543.180326.034.450224.043.370287.5
    41.170268.050.8602610.047.980346.532.460225.5
    38.770268.044.270284.546.2502110.035.560224.5

Calculation:

From Figure-1 it is clear that the regression equation is y=8.8720.0068x1+0.7163x2+2.03x3

Substitute the values in above equation.

  y=8.8720.0068(50)+0.7163(30)+2.03(6)=42.201

Thus, the value of variable y is 42.201 .

c.

To determine

To find: The confidence interval.

c.

Expert Solution
Check Mark

Answer to Problem 21E

The confidence intervalis (36.934,47.47) .

Explanation of Solution

Given information: The data is shown below.

    y(x1)(x2)(x3)y(x1)(x2)(x3)y(x1)(x2)(x3)y(x1)(x2)(x3)
    31.3 50 194.037.46030 5.0 24.060244.051.080347.5
    56.990388.043.360267.036.2 50 217.034.360222.5
    43.170286.536.370257.526.550172.031.56024 5.0
    41.57025 5.5 38.47031 5.5 47.18034 8.5 33.260234.0
    39.060266.541.560277.538.170275.539.260296.5
    40.970295.036.l60236.033.560242.546.770277.5
    35.96023 5.5 38.560236.043.6702710.030.470324.0
    43.57028 5.5 42.460249.041.08032 6.5 43.26025 5.5
    47.9 80 346.546.57031 5.5 50.2 50 269.030.660263.5
    33.870264.543.180326.034.450224.043.370287.5
    41.170268.050.8602610.047.980346.532.460225.5
    38.770268.044.270284.546.2502110.035.560224.5

Calculation:

The MINITAB output is shown below.

  Elementary Statistics (Text Only), Chapter 13.3, Problem 21E , additional homework tip  2

  Figure-2

From Figure-2 it is clear that the confidence interval is (36.934,47.47) .

d.

To determine

To find: The prediction interval.

d.

Expert Solution
Check Mark

Answer to Problem 21E

The prediction intervalis (32.888,51.517) .

Explanation of Solution

Given information: The data is shown below.

    y(x1)(x2)(x3)y(x1)(x2)(x3)y(x1)(x2)(x3)y(x1)(x2)(x3)
    31.3 50 194.037.46030 5.0 24.060244.051.080347.5
    56.990388.043.360267.036.2 50 217.034.360222.5
    43.170286.536.370257.526.550172.031.56024 5.0
    41.57025 5.5 38.47031 5.5 47.18034 8.5 33.260234.0
    39.060266.541.560277.538.170275.539.260296.5
    40.970295.036.l60236.033.560242.546.770277.5
    35.96023 5.5 38.560236.043.6702710.030.470324.0
    43.57028 5.5 42.460249.041.08032 6.5 43.26025 5.5
    47.9 80 346.546.57031 5.5 50.2 50 269.030.660263.5
    33.870264.543.180326.034.450224.043.370287.5
    41.170268.050.8602610.047.980346.532.460225.5
    38.770268.044.270284.546.2502110.035.560224.5

Calculation:

From Figure-12 it is clear that the (32.888,51.517)

e.

To determine

To find: The percentage of variation in variable y .

e.

Expert Solution
Check Mark

Answer to Problem 21E

The percentage of variation in variable y is 70.2% .

Explanation of Solution

Given information: The data is shown below.

    y(x1)(x2)(x3)y(x1)(x2)(x3)y(x1)(x2)(x3)y(x1)(x2)(x3)
    31.3 50 194.037.46030 5.0 24.060244.051.080347.5
    56.990388.043.360267.036.2 50 217.034.360222.5
    43.170286.536.370257.526.550172.031.56024 5.0
    41.57025 5.5 38.47031 5.5 47.18034 8.5 33.260234.0
    39.060266.541.560277.538.170275.539.260296.5
    40.970295.036.l60236.033.560242.546.770277.5
    35.96023 5.5 38.560236.043.6702710.030.470324.0
    43.57028 5.5 42.460249.041.08032 6.5 43.26025 5.5
    47.9 80 346.546.57031 5.5 50.2 50 269.030.660263.5
    33.870264.543.180326.034.450224.043.370287.5
    41.170268.050.8602610.047.980346.532.460225.5
    38.770268.044.270284.546.2502110.035.560224.5

Calculation:

From Figure-1 it is clear that the percentage of variation in variable y is 70.2% .

f.

To determine

To find:Whether the given model is useful for prediction.

f.

Expert Solution
Check Mark

Answer to Problem 21E

The model is useful in prediction.

Explanation of Solution

Given information: The data is shown below.

    y(x1)(x2)(x3)y(x1)(x2)(x3)y(x1)(x2)(x3)y(x1)(x2)(x3)
    31.3 50 194.037.46030 5.0 24.060244.051.080347.5
    56.990388.043.360267.036.2 50 217.034.360222.5
    43.170286.536.370257.526.550172.031.56024 5.0
    41.57025 5.5 38.47031 5.5 47.18034 8.5 33.260234.0
    39.060266.541.560277.538.170275.539.260296.5
    40.970295.036.l60236.033.560242.546.770277.5
    35.96023 5.5 38.560236.043.6702710.030.470324.0
    43.57028 5.5 42.460249.041.08032 6.5 43.26025 5.5
    47.9 80 346.546.57031 5.5 50.2 50 269.030.660263.5
    33.870264.543.180326.034.450224.043.370287.5
    41.170268.050.8602610.047.980346.532.460225.5
    38.770268.044.270284.546.2502110.035.560224.5

Calculation:

The null hypothesis is, the model is not useful for prediction and the alternative hypothesis is, the model is useful in prediction.

From Figure-1 it is clear that the p value is less than the level of significance of 0.01 .

Hence, the null hypothesis is rejected.

Thus, the model is useful in prediction.

g.

To determine

To explain:The test for the hypothesis H0:β1=0 versus H1:β10 .

g.

Expert Solution
Check Mark

Explanation of Solution

Given information: The data is shown below.

    y(x1)(x2)(x3)y(x1)(x2)(x3)y(x1)(x2)(x3)y(x1)(x2)(x3)
    31.3 50 194.037.46030 5.0 24.060244.051.080347.5
    56.990388.043.360267.036.2 50 217.034.360222.5
    43.170286.536.370257.526.550172.031.56024 5.0
    41.57025 5.5 38.47031 5.5 47.18034 8.5 33.260234.0
    39.060266.541.560277.538.170275.539.260296.5
    40.970295.036.l60236.033.560242.546.770277.5
    35.96023 5.5 38.560236.043.6702710.030.470324.0
    43.57028 5.5 42.460249.041.08032 6.5 43.26025 5.5
    47.9 80 346.546.57031 5.5 50.2 50 269.030.660263.5
    33.870264.543.180326.034.450224.043.370287.5
    41.170268.050.8602610.047.980346.532.460225.5
    38.770268.044.270284.546.2502110.035.560224.5

The null hypothesis is, there is no relationship between y and x1,x2,x3 and the alternative hypothesis is, there is relationship between y and x1,x2,x3 .

For the variable x1 ,

From Figure-1 it is clear that the p value is 0.954 which is greater than the level of significance of 0.05 .

Hence, the null hypothesis is not rejected.

Thus, there is no linear relationship between y and x1 .

For the variable x2 ,

From Figure-1 it is clear that the p value is 0.008 which is less than the level of significance of 0.05 .

Hence, the null hypothesis is rejected.

Thus, there is alinear relationship between y and x2 .

For the variable x3 .

From Figure-1 it is clear that the p value is 0 which is less than the level of significance of 0.05 .

Hence, the null hypothesis is rejected.

Thus, there is a linear relationship between y and x3 .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
Heights (in centimeters) and weights (in kilograms) of 7 supermodels are given below. Find the regression equation, letting the first variable be the independent (x) variable, and predict the weight of a supermodel who is 167 cm tall.   \begin{array}{c|ccccccc} \mbox{Height} & 178 & 176 & 166 & 178 & 174 & 172 & 174 \cr \hline \mbox{Weight} & 58 & 54 & 47 & 57 & 55 & 53 & 54 \cr \end{array} The regression equation is \hat{y} =   +   x . The best predicted weight of a supermodel who is 167 cm tall is   .
Find the regression line of the points: (1, 9),   (2, 8),   (8, 4),   (9, 2)
The table shows the amounts of crude oil​ (in thousands of barrels per​ day) produced by a certain country and the amounts of crude oil​ (in thousands of barrels per​ day) imported by the same country for seven years. The equation of the regression line is y=−1.326x+16,981.04. Complete parts​ (a) and​ (b) below. Produced,_x  5,789    5,679, 5,626, 5,407, 5,234, 5,138, 5,094 Imported,_y   9,309   9,130, 9,681, 10,090, 10,151, 10,152, 10,017 (a) Find the coefficient of determination and interpret the result.   r2= ​(Round to three decimal places as​ needed.)   ​(b) Find the standard error of estimate se and interpret the result.   Se= ​(Round to three decimal places as​ needed.)

Chapter 13 Solutions

Elementary Statistics (Text Only)

Ch. 13.1 - Prob. 17ECh. 13.1 - Prob. 18ECh. 13.1 - Prob. 19ECh. 13.1 - Prob. 20ECh. 13.1 - Prob. 21ECh. 13.1 - Prob. 22ECh. 13.1 - Prob. 23ECh. 13.1 - Prob. 24ECh. 13.1 - Prob. 25ECh. 13.1 - Prob. 26ECh. 13.1 - Calculator display: The following TI-84 Plus...Ch. 13.1 - Prob. 28ECh. 13.1 - Prob. 29ECh. 13.1 - Prob. 30ECh. 13.1 - Confidence interval for the conditional mean: In...Ch. 13.2 - Prob. 3ECh. 13.2 - Prob. 4ECh. 13.2 - Prob. 5ECh. 13.2 - Prob. 6ECh. 13.2 - Prob. 7ECh. 13.2 - Prob. 8ECh. 13.2 - Prob. 9ECh. 13.2 - Prob. 10ECh. 13.2 - Prob. 11ECh. 13.2 - Prob. 12ECh. 13.2 - Prob. 13ECh. 13.2 - Prob. 14ECh. 13.2 - Prob. 15ECh. 13.2 - Prob. 16ECh. 13.2 - Prob. 17ECh. 13.2 - Dry up: Use the data in Exercise 26 in Section...Ch. 13.2 - Prob. 19ECh. 13.2 - Prob. 20ECh. 13.2 - Prob. 21ECh. 13.3 - Prob. 7ECh. 13.3 - Prob. 8ECh. 13.3 - Prob. 9ECh. 13.3 - In Exercises 9 and 10, determine whether the...Ch. 13.3 - Prob. 11ECh. 13.3 - Prob. 12ECh. 13.3 - Prob. 13ECh. 13.3 - For the following data set: Construct the multiple...Ch. 13.3 - Engine emissions: In a laboratory test of a new...Ch. 13.3 - Prob. 16ECh. 13.3 - Prob. 17ECh. 13.3 - Prob. 18ECh. 13.3 - Prob. 19ECh. 13.3 - Prob. 20ECh. 13.3 - Prob. 21ECh. 13.3 - Prob. 22ECh. 13.3 - Prob. 23ECh. 13 - A confidence interval for 1 is to be constructed...Ch. 13 - A confidence interval for a mean response and a...Ch. 13 - Prob. 3CQCh. 13 - Prob. 4CQCh. 13 - Prob. 5CQCh. 13 - Prob. 6CQCh. 13 - Construct a 95% confidence interval for 1.Ch. 13 - Prob. 8CQCh. 13 - Prob. 9CQCh. 13 - Prob. 10CQCh. 13 - Prob. 11CQCh. 13 - Prob. 12CQCh. 13 - Prob. 13CQCh. 13 - Prob. 14CQCh. 13 - Prob. 15CQCh. 13 - Prob. 1RECh. 13 - Prob. 2RECh. 13 - Prob. 3RECh. 13 - Prob. 4RECh. 13 - Prob. 5RECh. 13 - Prob. 6RECh. 13 - Prob. 7RECh. 13 - Prob. 8RECh. 13 - Prob. 9RECh. 13 - Prob. 10RECh. 13 - Air pollution: Following are measurements of...Ch. 13 - Icy lakes: Following are data on maximum ice...Ch. 13 - Prob. 13RECh. 13 - Prob. 14RECh. 13 - Prob. 15RECh. 13 - Prob. 1WAICh. 13 - Prob. 2WAICh. 13 - Prob. 1CSCh. 13 - Prob. 2CSCh. 13 - Prob. 3CSCh. 13 - Prob. 4CSCh. 13 - Prob. 5CSCh. 13 - Prob. 6CSCh. 13 - Prob. 7CS
Knowledge Booster
Background pattern image
Statistics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, statistics and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Text book image
Glencoe Algebra 1, Student Edition, 9780079039897...
Algebra
ISBN:9780079039897
Author:Carter
Publisher:McGraw Hill
Text book image
Trigonometry (MindTap Course List)
Trigonometry
ISBN:9781305652224
Author:Charles P. McKeague, Mark D. Turner
Publisher:Cengage Learning
Text book image
College Algebra
Algebra
ISBN:9781938168383
Author:Jay Abramson
Publisher:OpenStax
Text book image
Algebra: Structure And Method, Book 1
Algebra
ISBN:9780395977224
Author:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Publisher:McDougal Littell
Text book image
Algebra & Trigonometry with Analytic Geometry
Algebra
ISBN:9781133382119
Author:Swokowski
Publisher:Cengage
Correlation Vs Regression: Difference Between them with definition & Comparison Chart; Author: Key Differences;https://www.youtube.com/watch?v=Ou2QGSJVd0U;License: Standard YouTube License, CC-BY
Correlation and Regression: Concepts with Illustrative examples; Author: LEARN & APPLY : Lean and Six Sigma;https://www.youtube.com/watch?v=xTpHD5WLuoA;License: Standard YouTube License, CC-BY