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					Chapter 11: Simple Linear Regression Analysis


                 CHAPTER 11 – Simple Linear Regression Analysis
   11.1   When there appears to be a linear relationship between y and x

   11.2   y   y x    the observed value of the dependent variable
           y x   0  1 x  the mean value of the y when the value of the independent variable is x
           = error term
   11.3   1 : the change in the mean value of the dependent variable that is associated with a one-unit
          increase in the value of the independent variable
           0 : the mean value of the dependent variable when the value of the independent variable is
          zero

   11.4   When data is observed in time sequence, the data is called time series data. Cross-sectional
          data is observed at a single point in time.

   11.5   The straight line appearance of this data plot suggests that the simple linear regression model
          with a positive slope might be appropriate.

   11.6   a.    y x  4.00   0  1 (4.00) is the mean of the starting salaries of all marketing graduates
               having a 4.00 GPA.

          b.    y x  2.50  0  1 (2. 50) is the mean of the starting salaries of all marketing graduates
               having a 2.50 GPA.

          c.   1  the change in mean starting salary associated with a one point increase in the grade
               point average.

          d.    0  the mean starting salary for marketing graduates with a grade point average of
               0.00.
               The interpretation of  0 fails to make practical sense because it requires that a marketing
               graduate have a grade point average of 0.00 in which case, of course, the marketing
               student would not have graduated.

          e.   All factors other than the grade point average. For example, extra-curricular activities
               and the type of minor (if any) the graduate has.

   11.7   The straight line appearance on this data plot suggest that the simple linear regression model
          with a positive slope might be appropriate.

   11.8   a.   It is the mean of the service times required when the number of copiers is 4.

          b.   It is the mean of the service times required when the number of copiers is 6.

          c.   The slope parameter equals the change in the mean service time that is associated with
               each additional copier serviced.

          d.   The intercept is the mean service time when there are no copiers. It fails to make
               practical sense because it requires service time when no copiers exist.

          e.   All factors other than the number of copiers serviced.



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                                                                              Chapter 11: Simple Linear Regression Analysis


 11.9   The plot looks reasonably linear.

11.10   a.    Mean demand when price difference is .10.

        b.    Mean demand when price difference is –.05.

        c.    Change in mean demand per dollar increase in price difference

        d.    Mean demand when price difference = 0; yes

        e.    Factors other than price difference; answers will vary.

11.11   a.
                     1000
             y




                      500




                                    0   10   20   30   40   50     60    70    80   90 100
                           0
                                                               x



        b.    Yes, the plot looks linear, positive slope

11.12   a.    Mean labor cost when batch size = 60

        b.    Mean labor cost when batch size = 30

        c.    Change in mean labor cost per unit increase in batch size

        d.    Mean labor cost when batch size = 0; questionable

        e.    Factors other than batch size; answers will vary.

11.13   a.
                     200
             Price




                     150




                     100       10                 15                20                25
                                                        size


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Chapter 11: Simple Linear Regression Analysis


          b.   Yes, the relationship looks to be linear with a positive slope.

  11.14   a.   Mean sales price when home size = 20 (2000 square feet)

          b.   Mean sales price when home size = 18 (1800 square feet)

          c.   Change in mean sales price per one unit (100 square foot) increase in home size

          d.   Mean sales price when home size = 0; no

          e.   Factors other than home size; answers will vary.

  11.15   The quality or “goodness” of the fit of the least squares line to the observed data.

  11.16   The “best” line that can be fitted to the observed data. The slope and the intercept of the least
          squares line.

  11.17   Evaluate y  b0  b1x for the given value of x.
                   ˆ

  11.18   Because we do not know how y and x are related outside the experimental region.

  11.19   a.   b0 = 14.8156 b1 = 5.70657

               No. The interpretation of b0 does not make practical sense since it indicates that someone
               with a GPA = 0 would have a starting salary of $14,816, when in fact they would not
               have graduated with a GPA = 0.

          b.    ˆ
                y = 14.8156 + 5.70657(3.25) = 33.362

               That is, $33,362

  11.20   a.   b0 = 11.4641 b1 = 24.6022

               No. The interpretation of b0 does not make practical sense since it indicates that 11.46
               minutes of service would be required for a customer with no copiers.

          b.    ˆ
                y = 11.4641 + 24.6022(4) = 109.873, or 109.9 minutes

  11.21   a.   b0 = 7.81409 b1 = 2.66522

                Yes. The interpretation of b0 does make practical sense since it indicates that 781,409
                bottles of detergent would be demanded when the price difference with other products
                is zero.

          b.   ˆ
               y = 7.814088 + 2.665214(.10) = 8.081
          c.
                y  b0  b1 x
                ˆ

               8.5  7.81409  2.6652 x

                    .68591
               x           .257 , or about 26 cents
                    2.6652

  11.22   a.


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                                                   Chapter 11: Simple Linear Regression Analysis


            xi                      yi                        xi2                xi yi

             5                      71                       25                  355
            62                     663                      3844                41106
            35                     381                      1225                13335
            12                     138                       144                1656
            83                     861                      6889                71463
            14                     145                       196                2030
            46                     493                      2116                22678
            52                     548                      2704                28496
            23                     251                       529                5773
           100                    1024                     10000               102400
            41                     435                      1681                17835
            75                     772                      5625                57900

        xi  548              yi  5782              xi2  34978        xi yi  365027
                      ( xi )(  yi )
     SS xy   xi yi 
                               n
                         (548)(5,782 )
            365,027                       100,982 .32
                                 12
                     ( xi ) 2
     SS xx   xi2 
                        n
                       (548) 2
            34,978               9,952 .667
                          12
             SS xy 100,982 .32
       b1                            10.1463
             SS xx 9,952 .667
                        5782                 548 
       b0  y  b1 x              10.1463        18.4880
                        12                   12 

b.   b1 is the estimated increase in mean labor cost (10.1463) for every 1 unit increase in the
     batch size.
     b0 is the estimated mean labor cost (18.4880) when batch size = 0; no.

c.   y  18.488010.1463x
     ˆ

d.   ˆ
     y = 18.4880 + 10.1463(60) = 627.266




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Chapter 11: Simple Linear Regression Analysis


  11.23   a.     MINITAB output

                 Regression Analysis: Sale Price versus Size
                 The regression equation is
                 Sale Price = 48.0 + 5.70 Size
                 Predictor         Coef           SE Coef             T           P
                 Constant         48.02             14.41          3.33       0.010
                 Size            5.7003            0.7457          7.64       0.000
                 S = 10.59               R-Sq = 88.0%            R-Sq(adj) = 86.5%

          b.     b1 is the estimated increase in mean sales price (5.7003) for every hundred square foot
                 increase in home size.

                 b0 is the estimated mean sales price when square footage = 0. No, the interpretation of
                 b0 makes no practical sense.

          c.     y  48.02  5.7003 x.
                 ˆ

          d.     ˆ
                 y = 48.02 + 5.7003 (20) = 162.026

                 That is, $162,026.


  11.24   (1) Mean of error terms = 0
          (2) Constant variance
          (3) Normality
          (4) Independence
          See page 466 in the text.

  11.25    2 ;  That is, the constant variance and standard deviation of the error term populations.
                 SSE 1.438
  11.26   s2              .2876
                 n2 72

          s  s 2  .2876  .5363

                 SSE 191.70166
  11.27   s2                  21.30018
                 n2   11  2


          s  s 2  21.30018  4.61521
  11.28   s 2  2. 8059  .1002, s 
                   28                    s2  .3166

  11.29   s 2  747  74.7, s  8.643
                10

  11.30          SSE 896 .8
          s2                 112 .1
                 n  2 10  2

          s  s 2  112 .1  10.588

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                                                                  Chapter 11: Simple Linear Regression Analysis


11.31          SSE 222.8242
        s2                 27.8530
               n2   10  2

        s  s 2  27.8530  5.2776

11.32   Strong ( = .05) or very strong ( = .01) evidence that the regression relationship is
        significant.

11.33   Explanations will vary.

11.34   a.     b0 = 14.816            b1 = 5.7066

        b.     SSE = 1.4381 s2 = .2876 s = .5363

        c.     sb1 = .3953            t = 14.44

               t = b1 / sb1 = 5.7066 /.3953 = 14.44

        d.     df = 5 t.025 = 2.571            Reject Ho, Strong evidence of a significant relationship between
               x and y.

        e.     t.005 = 4.032          Reject Ho, Very strong evidence of a significant relationship between x
               and y.

        f.     p-value=.000            Reject at all , Extremely strong evidence of a significant
               relationship between x and y.

        g.     95% Cl: [ b1 ± t.025 s b1 ] = 5.7066 ± (2.571)(.3953) = [4.690, 6.723]

               We are 95% confident that the mean starting salary increases by between $4690 and
               $6723 for each 1.0 increase in GPA.

        h.     99% Cl: [ b1 ± t.005 sb1 ] = 5.7066 ± (4.032)(.3953) = [4.113, 7.300]

               We are 99% confident that the mean starting salary increases by between $4113 and
               $7300 for each 1.0 increase in GPA.

        i.     sb0 = 1.235            t = 12.00

               t = b0 / sb0 = 14.816 / 1.235 = 12.00

        j.     p-value=.000           Reject at all , Extremely strong evidence that the y-intercept is
               significant.
                        s              .5363
               s b1                              .3953
                          SS xx        1.8407
        k.
                           1   x2         1 (3.0814) 2
               s b0  s            .5363             1.235
                           n SS xx        7  1.8407




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Chapter 11: Simple Linear Regression Analysis


  11.35   a.   b0 = 11.4641      b1 = 24.6022

          b.   SSE = 191.7017            s2 = 21.3002      s = 4.615

          c.   sb1 = .8045       t = 30.580

               t = b1 / sb1 = 24.602 /.8045 = 30.580

          d.   df = 9 t.025 = 2.262      Reject Ho, strong evidence of a significant relationship between
               x and y.

          e.   t.005 = 3.250     Reject Ho, very strong evidence of a significant relationship between x
               and y.

          f.   p-value=.000 Reject at all , extremely strong evidence of a significant relationship
               between x and y.

          g.   [24.6022  2.262(.8045)] = [22.782, 26.422]

          h.   [24.6022  3.250(.8045)] = [21.987,27.217]

          i.   sb0 = 3.4390      t = 3.334

               t = b0 / sb0 = 11.464 / 3.439 = 3.334

          j.   p-value=.0087          Reject at all  except .001
                       s      4.61521
          k.   sb1                   .8045
                      SS xx    32.909

                         1   x2             1 3.909 2
               sb0  s           4.61521            3.439
                         n SS xx           11 32.909


  11.36   See the solutions to 11.34 for guidance.

          a.   b0  7.814, b1  2.665

          b.   SSE  2.806, s 2  .100, s  .3166

          c.   s b1  .2585, t  10.31

          d.   Reject H 0 .

          e.   Reject H 0 .

          f.   p-value = less than .001; reject H 0 at each value of 

          g.   [2.665 ± 2.048(.2585)] = [2.136, 3.194]

          h.   [2.665 ± 2.763(.2585)] = [1.951, 3.379]



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                                                                Chapter 11: Simple Linear Regression Analysis


        i.   s b0  .0799, t  97.82

        j.   p-value = less than .001; reject H 0 .
                        s          .31656
        k.   sb1                             .2585
                       SS xx       1.49967

                        1   x2            1   .2133 2
             sb0  s            .31656              .079883
                        n SS xx          30 1.49967

11.37   See the solutions to 11.34 for guidance.

        a.   b0 18.488, b1 10.1463

        b.   SSE = 747, s 2  75, s = 8.642

        c.   sb1  .0866, t 117.13

        d.   Reject H 0 .

        e.   Reject H 0 .

        f.   p-value = .000; reject H 0 at each value of 

        g.   [10.1463 ± 2.228(.0866)] = [9.953, 10.339]

        h.   [10.1463 ± 3.169(.0866)] = [9.872, 10.421]

        i.   sb0  4.677, t  3.95

        j.   p-value = .003; fail to reject H 0 at  = .001. Reject H 0 at all other values of 
                     s       8.64154
        k.   sb1                        .086621
                    SS xx    9952 .667

                        1   x2             1   45.667 2
             sb0  s            8.64154               4.67658
                        n SS xx           12 9952 .667


11.38   See the solutions to 11.34 for guidance.

        a.   b0 = 48.02            b1 = 5.7003

        b.   SSE = 896.8           s2 = 112.1       s = 10.588

        c.   sb1 = .7457           t = 7.64

             t = b1 / sb1 = 5.7003 /.7457 = 7.64

        d.   df = 8 t.025 = 2.306             Reject Ho



                                                          159
Chapter 11: Simple Linear Regression Analysis


          e.   t.005 = 3.355          Reject Ho

          f.   p-value=.000           Reject at all 

          g.   [3.9807,7.4199]

          h.   [3.198,8.202]

          i.   sb0 = 14.41            t = 3.33

               t = b0 / sb0 = 48.02 / 14.41 = 3.33

          j.   p-value=.010           Reject at all  except .01 and .001
                          s          10.588
          k.   sb1                            .7457
                         SS xx        201 .6

                           1   x2            1 18.8 2
               sb0  s             10.588           14.41
                          10 SS xx          10 201 .6


  11.39   Find sb1 from Minitab
              The regression equation is
       sales = 66.2 + 4.43 ad exp

       Predictor                Coef           SE Coef                 T       P
       Constant               66.212             5.767             11.48   0.000
       Ad exp                 4.4303            0.5810              7.62   0.000



          95% C.I. for β1 [ 4.4303  2.306(.5810) ] = [3.091,5.770]

  11.40   a.   b0 = -.1602            b1 = 1.2731

          b.   SSE = .1343

          c.   s2= .0336 s = .1833

          d.   sb1 = .1025

          e.   t = 12.4209

          f.   p-value=.00024                    Reject at all 

          g.   [.9885, 1.5577], 95% confident the mean preference would increase.

  11.41   The distance between xo and x , the average of the previously observed values of x.

  11.42   A confidence interval is for the mean value of y. A prediction interval is for an individual
          value of y.

  11.43   The smaller the distance value, the shorter the lengths of the intervals.

  11.44   a.   33.362, [32.813, 33.911]


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                                                             Chapter 11: Simple Linear Regression Analysis


        b.    33.362, [31.878, 34.846]

                                 1 (3.25  3.0814 ) 2
        c.    Distance Value                         .1583
                                 7      1.8407

              [33.362 ± 2.571(.5363) .1583 ] = [32.813, 33.911]

              [33.362 ± 2.571(.5363) 1  .1583 ] = [31.878, 34.846]

11.45   a.    109.873, [106.721, 113.025]

        b.    109.873, [98.967, 120.779]

        c.    113 minutes

11.46   a.    8.0806; [7.948, 8.213]

        b.    8.0806; [7.419, 8.743]

        c.    See graph in text, page 483
                                                         2
                                                 .065 
        d.    s dist  .065, s  .3166 , dist           .04215
                                                 .3166 
              99% C.I.: [8.0806 ± 2.763(.065)] = [7.9010, 8.2602]

                                                                
              99% P.I.: 8.0806  2.763(.3166 ) 1.04215  7.1877 ,8.9735 


        e. (1) 8.4804; [8.360, 8.600]
           (2) 8.4804; [7.821, 9.140]
                                                             2
                                                    .059 
             (3) s dist  .059, s  .3166 , dist           .03473
                                                    .3166 
              99% C.I.: [8.4804 ± 2.763(.059)] = [8.3174, 8.6434]

                                                            
             99% P.I.: 8.4804  2.763(.3166 ) 1.03473  7.5907 ,9.3701


11.47   a.    627.26, [621.05, 633.47]

        b.    627.26, [607.03, 647.49]
                                                         2
                                                2.79 
        c.    s dist  2.79, s  8.642, dist           .104227
                                                8.642 
              99% C.I.: [627.26 ± 3.169(2.79)] = [(618.42, 636.10)]
                                                                    
              99% P.I.: 627.26  3.169(8.642) 1.104227  598.48,656.04




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Chapter 11: Simple Linear Regression Analysis


  11.48   a.   162.03, [154.04, 170.02]

          b.   162.03, [136.33, 187.73]

  11.49   2.3429, [1.7367, 2.9491]

  11.50   Use a computer program to find the prediction equation y  109 1.075x .
                                                                 ˆ
          Point predictions and prediction intervals are:

          a.   87.5, [57.28, 117.72]

          b.   76.75, [48.04, 105.46]

          c.   66.00, [37.82, 94.18]

          d.   55.25, [26.55, 83.96]

          e.   44.5, [14.28, 74.72]

  11.51 From Minitab
               The regression equation is
               Market Rate = 0.85 + 0.610 Accounting Rate

               Predictor            Coef         SE Coef              T         P
               Constant            0.847           1.975           0.43     0.670
               Accounti           0.6105          0.1431           4.27     0.000

               Predicted Values for New Observations

               New Obs       Fit         SE Fit             95.0% CI                   95.0% PI
               1          10.004          0.753     (     8.494, 11.514)      (     -0.310, 20.318)

               Values of Predictors for New Observations

               New Obs    Accounti
               1              15.0
          a.    y  b0  b1 (15 .00 )  .847  .6105 (15 .00 )  10 .0045
                ˆ
               95% C.I.: [8.494, 11.514]

          b.   10.0045
               95% P.I.: [–.310, 20.318]

  11.52   Total variation: measures the total amount of variation exhibited by the observed values of y.
          Unexplained variation: measures the amount of variation in the values of y that is not
          explained by the model (predictor variable).
          Explained variation: measures the amount of variation in the values of y that is explained by
          the predictor variable.

  11.53   Proportion of the total variation in the n observed values of y that is explained by the simple
          linear regression model.
  11.54   a.   61.380, 1.438, 59.942, .977, .988; 97.7% of the variation in the observations has been
               explained by the regression model.
          b.   t = 14.44, p-value =.000; Reject Ho at both 
  11.55   a.   20110.54, 191.702, 19918.84, .990, .995; 99.0% of the variation in the observations has
               been explained by the regression model.


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                                                         Chapter 11: Simple Linear Regression Analysis


        b.   t = 30.58, p-value = 000; Reject Ho at both 
11.56   a.   13.459; 2.806; 10.653; r 2  .792 ; r = .890; 79.2% of the variation in the observations
             has been explained by the regression model.
        b.   t = 10.31, p-value = .000; Reject H 0 .
11.57   a.   9952.667; 7.2486; 9945.418; r2  .999; r = .999; 99.9% of the variation in the
             observations has been explained by the regression model.
        b.   t = 117.134, p-value = .000; Reject H 0 .

11.58   a.   7447.5, 896.8, 6550.7, .880, .9381; 88.0% of the variation in the observations has been
             explained by the regression model.

        b.   t=7.64,p-value=.000; Reject Ho at both 

11.59   a.   5.316, .1343, 5.1817, .975, .987; 97.5% of the variation in the observations has been
             explained by the regression model.

        b.   t = 12.421, p-value .000; Reject Ho at both 

11.60   From Minitab
    The regression equation is
    sales = 66.2 + 4.43 ad exp

    Predictor             Coef       SE Coef               T         P
    Constant            66.212         5.767           11.48     0.000
    d exp              4.4303        0.5810            7.62     0.000

    S = 5.278            R-Sq = 87.9%          R-Sq(adj) = 86.4%


    Analysis of Variance

    Source                 DF            SS                MS         F          P
    Regression              1        1619.3            1619.3     58.14      0.000
    Residual Error          8         222.8              27.9
    Total                   9        1842.1



        a.   1842.1, 222.8, 1619.3, r 2  .879 , r  .9375 ; 87.9% of the variation in the observations
             has been explained by the regression model.

        b.   t = 7.62, p-value = .000, Reject Ho.

11.61   From Minitab
    The regression equation is
    Market Rate = 0.85 + 0.610 Accounting Rate

    Predictor             Coef       SE Coef                T        P
    Constant             0.847         1.975             0.43    0.670
    Accounti            0.6105        0.1431             4.27    0.000

    S = 5.085            R-Sq = 25.9%          R-Sq(adj) = 24.5%

    Analysis of Variance

    Source                 DF             SS               MS          F         P



                                                 163
Chapter 11: Simple Linear Regression Analysis

       Regression               1       470.74           470.74    18.21       0.000
       Residual Error          52      1344.33            25.85
       Total                   53      1815.07


          a.   1,815.07; 1,344.933; 470.74

               r2 = .259

               r = .509

          b.   t = 4.27, p-value = .000, Reject H 0 .
  11.62   From Minitab
              The regression equation is
       HeatLoss = 109 - 1.08 Temperature

       Predictor              Coef      SE Coef               T       P
       Constant            109.000        9.969           10.93   0.000
       Temperat            -1.0750       0.2307           -4.66   0.002

       S = 11.30             R-Sq = 75.6%         R-Sq(adj) = 72.1%

       Analysis of Variance

       Source                  DF           SS               MS        F           P
       Regression               1       2773.5           2773.5    21.70       0.002
       Residual Error           7        894.5            127.8
       Total                    8       3668.0


          a.   3,688; 894.5; 2773.5
               r2 = .756
               r = –.870
          b.   t = -4.66, p-value = .002, Reject H 0 .
  11.63   Mildly similar views.
  11.64   H0: 1  0 versus Ha: 1  0 .
  11.65   t-test on 1

  11.66   a.   F = 59.942 / (1.438 / 5) = 208.39
          b.   F.05 = 6.61    df1 = 1, df2 = 5
               Since 208.39 > 6.61, reject H0 with strong evidence of a significant relationship between
               x and y.
          c.   F.01 = 16.26    df1 = 1, df2 = 5
               Since 208.39 > 16.26, reject H0 with very strong evidence of a significant relationship
               between x and y.
          d.   p-value =.000; Reject H0 at all levels of , extremely strong evidence of a significant
               relationship between x and y.
          e.   t2 = (14.44)2 = 208.51 (approximately equals F = 208.39)


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                                                       Chapter 11: Simple Linear Regression Analysis


             (t.025)2 = (2.571)2 = 6.61 = F.05

11.67   a.   F = 19918.844 / (21.30018 / 9) = 935.149

        b.   F.05 = 5.12    df1 = 1, df2 = 9

             Since 935.149 > 5.12, reject H0 with strong evidence of a significant relationship
             between x and y.
        c.   F.01 = 10.56     df1 = 1, df2 = 9

             Since 935.149 > 10.56, reject H0 with very strong evidence of a significant relationship
             between x and y.
        d.   p-value =less than .001; Reject H0 at all levels of , extremely strong evidence of a
             significant relationship between x and y.

        e.   t2 = (30.58)2 = 935.14 (approximately equals F = 935.149)

             (t.025)2 = (2.262)2 = 5.12 = F.05

11.68   a.   F = 106.303

        b.   F.05 =4.20, reject H 0 (df1 = 1, df2 = 28). Strong evidence of a significant relationship
             between x and y.
        c.   F.01 =7.64, reject H 0 (df1 = 1, df2 = 28). Very strong evidence of a significant
             relationship between x and y.
        d.   p-value = less than .001, reject H 0 . Extremely strong evidence of a significant
             relationship between x and y.

        e.   (10.310)2 106.303 (within rounding error)
             (t.025)2 = 4.19 = F.05

11.69   a.   F = 13,720.47

        b.   Reject H 0 .

        c.   Reject H 0 .

        d.   p-value = .000; reject H 0 .

        e.   (117.13)2  13,720.47 (within rounding error)




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Chapter 11: Simple Linear Regression Analysis


  11.70   a.    F = 6550.7 / (896.8 / 8) = 58.43

          b.    F.05 = 5.32    df1 = 1, df2 = 8

                Since 58.43 > 5.32, reject H0.

          c.    F.01 = 11.26    df1 = 1, df2 = 8

                Since 58.43 > 11.3, reject H0.

          d.    p-value =.000; Reject H0 at all levels of 

          e.    t2 = (7.64)2 = 58.37 (approximately equals F = 58.43)

                (t.025)2 = (2.306)2 = 5.32 = F.05

  11.71   a.    F = 5.1817 / (.13435 / 4) = 154.279

          b.    F.05 = 7.71    df1 = 1, df2 = 4

                Since 154.279 > 7.71, reject H0.

          c.    F.01 = 21.2    df1 = 1, df2 = 4

                Since 154.279 > 21.2, reject H0.

          d.    p-value =.0002; Reject H0 at all levels of 

          e.    t2 = (12.4209)2 = 154.279 (approximately equals F = 154.279)

                (t.025)2 = (2.776)2 = 7.71 = F.05

  11.72                                                                        ˆ
          They should be plotted against each independent variable and against y . Funneling or curved
              patterns indicate violations of the regression assumptions.

  11.73   Create a histogram, stem-and-leaf, and normal plot.

  11.74   Transforming the dependent variable.

  11.75   Approximate horizontal band appearance. No violations indicated

  11.76   Possible violations of the normality and constant variance assumptions.

  11.77   No.

  11.78   a.
                3(i )  1 3(4)  1
                                   .3235
                3n  1     33  1
                .5000  .3235  .1765,  z  .46

                3(i )  1 3(10)  1
                                    .8529
                3n  1     33  1
                .8529  .5000  .3529,  z  1.05



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                                                          Chapter 11: Simple Linear Regression Analysis


        b.   No

11.79   The residual plot has somewhat of a cyclical appearance. Since d=.473 is less than dL, 05=1.27,
        we conclude there is positive autocorrelation and since 4 - .473 = 3.527 and this is greater than
        dU,.05 = 1.45 we conclude that there is not negative autocorrelation.

11.80   The data plot in Figure 11.40b indicates that as x increases, y increases and becomes more
        variable. The residual plot in Figure 11.40c fans out as x increases, again indicating that y
        becomes more variables as x increases.

11.81   a.   (i) y*  5.0206
                 ˆ

             95% P.I. for y* = [4.3402, 5.7010]

             (ii) y  e 5.0206  151.5022
                  ˆ

                                             
             95% P.I . for y  e 4.3402 , e 5.7010  76.7229,299.1664 

        c.   The residual plot is curved indicating the straight line model does not fit the data
             appropriately.

11.82   a.   Yes

              y  7(25.0069)  175.048
              ˆ
        b.   95%C.I .7(21.4335),7(28.5803)  150.0345,200.0621
             95% P.I .7(13.3044),7(36.7094)  93.1308,256.9658

             Allow 200 minutes.

11.83   a.   Yes; see the plot in part c.

        b.   b0  306.619, b1  27.714

        c.   y  306.619 27.714x
             ˆ


                   260




                   240
             y




                   220

                               2.0           2.4            2.8            3.2
                                                   x


                                                   167
Chapter 11: Simple Linear Regression Analysis


          d.   p-value = .000, reject H 0 , significant

          e.   x0  $2.10; y  248.420; [244.511, 252.327]
                           ˆ
               x0  $2.75; y  230.405; [226.697, 234.112]
                           ˆ
               x0  $3.10; y  220.705; [216.415, 224.994]
                           ˆ

  11.84   a.   b1 = -6.4424 For every unit increase in width difference, the mean number of accidents
               are reduced by 6.4 per 100 million vehicles.

          b.   p-value = .000 Reject H0 at all levels of 

          c.   r2 = .984 98.4% of the variation in accidents is explained by the width difference.

  11.85   a.   No

          b.   Possibly not; Don’t take up smoking

  11.86   a.   Using Figure 11.48, there does seem to be a negative relationship between temperature
               and o-ring failure.

          b.   The temperature of 31 was outside the experimental region.

  11.87   Explanations will vary

  11.88   a.   Argentina, Turkey, Brazil, and Taiwan

          b.   Pakistan and Jordan

  11.89   a.   There is a relationship since F = 21.13 with a p-value of .0002.

          b.   b1 = 35.2877, [19.2202,51.3553]

  11.90   For aggressive stocks a 95% confidence interval for 1 is
                                                               *

           t
          .0163    .025(.003724)    [.0163 2.365(.003724)]  [.00749, .02512] where t.025   is based
          on 7 degrees of freedom.
          We are 95% confident that the effect of a one-month increase in the return length time for an
          aggressive stock is to increase the mean value of the average estimate of 1 by between
          .00749 and .02512.
          For defensive stocks a 95% confidence interval for 1 is
                                                                 *
          [–0.00462 ± 2.365(.00084164)] = [–.00661, –.00263].
          For neutral stocks a 95% confidence interval for 1 is
                                                              *

          [.0087255 ± 2.365(.001538)] = [.005088, .01236].

  11.91   ˆ
          y = 2.0572 + 6.3545(1/5) = 3.3281




                                                    168
                                                                       Chapter 11: Simple Linear Regression Analysis


Internet Exercise

  11.92
                                                      Scatter Plot
                                                     GMAT vs GPA



                          700
                   GMAT




                          650




                          600
                                3.1         3.2            3.3          3.4     3.5       3.6
                                                                  GPA




The regression equation is
GMAT = 184 + 141 GPA

Predictor     Coef SE Coef              T    P
Constant     184.27  84.63            2.18 0.034
GPA          141.08  25.36            5.56 0.000

S = 21.50    R-Sq = 39.2%       R-Sq(adj) = 37.9%

Analysis of Variance

Source             DF          SS       MS         F          P
Regression          1        14316      14316      30.96     0.000
Residual Error     48        22197       462
Total              49        36513

Predicted Values for New Observations

New Obs Fit         SE Fit   95.0% CI         95.0% PI
   1   678.06        5.16 ( 667.68, 688.45) ( 633.60, 722.53)

Values of Predictors for New Observations

New Obs      GPA
1
3.50




                                                                 169

				
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