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chap13_part2

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									                                                            Solutions to End-of-Section and Chapter Review Problems     325


13.73   (h)
                                                                             E.R.A. Residual Plot

                                                  15

                                                  10

                                                   5
                                      Residuals
                                                   0

                                                   -5

                                                  -10

                                                  -15

                                                  -20
                                                        0         1               2           3           4         5         6
                                                                                            E.R.A.

                                  Based on a visual inspection of the graphs of the distribution of studentized residuals
                                  and the residuals versus E.R.A., there is no pattern. The model appears to be
                                  adequate.
        (i)                       p-value = 7.28222E-05 < 0.05. Reject H0. There is evidence that the fitted linear
                                  regression model is useful.
        (j)                       81.5457  Y | X  888.2396
        (k)                       68.8964  YI  100.8889
        (l)                       21.7459  1  8.4395
        (m)                       The “population” might be considered to be all the recent years in which baseball has
                                  been played.
        (n)                       Other independent variables might be considered for inclusion in the models are (i)
                                  runs scored, (ii) hits allowed, (iii) walks allowed, (iv) number of errors, etc.

13.74   (a)
                                                                      Scatter Diagram

                             80
                             70
                             60
          Price per Person




                             50
                             40
                             30
                             20
                             10
                             0
                                  0                 10       20       30     40       50      60     70       80   90
                                                                           Sum of Ratings
326     Chapter 13: Simple Linear Regression


13.74    (b)                 ˆ
                             Y  24.2468  1.0464X
cont.    (c)                 Since no restaurant will receive a summated rating of 0, it is not meaningful to
                             interpret b0. For each additional unit of increase in summated rating, the estimated
                             average price per person will increase by $1.0464.
         (d)                 Y  24.2468  1.0464 50  $28.07
                             ˆ
         (e)                 SYX = 6.0352.
         (f)                 r2 = 0.6584. 65.84% of the variation in price per person can be explained by the
                             variation in summated rating.
         (g)                 r  r 2  0.6584  0.8114 .
         (h)
                                           Summated Rating Residual Plot

                       20

                       15

                       10
           Residuals




                        5

                        0

                        -5

                       -10

                       -15

                       -20
                             0     10      20     30      40     50      60     70     80      90
                                                       Summated Rating

                             Based on a visual inspection of the residual plot of summated rating, there may be a
                             nonlinear relationship between price per person and summated rating. A quadratic
                             model may be more appropriate for the data.
         (i)                 p-value is virtually 0. Reject H0. There is very strong evidence to conclude that there
                             is a linear relationship between price per persona and summated rating.
         (j)                 $26.55  Y | X  $29.59
         (k)                 $16.00  YI  $40.15
         (l)                 0.8953  1  1.1975
         (m)                 The linear regression model appears to have provided an adequate fit and shown a
                             significant linear relationship between price per person and summated rating. Since
                             65.84% of the variation in price per person can be explained by the variation in
                             summated rating, price per person is moderately useful in predicting the price. Given
                             the parabolic pattern in the summated rating residual plot, a quadratic regression
                             model may perform better in predicting price per person using summated rating.
                                Solutions to End-of-Section and Chapter Review Problems          327


13.75   (a)
                                        Scatter Plot

                  4500000
                  4000000
                  3500000
                  3000000
          Sales




                  2500000
                  2000000
                  1500000
                  1000000
                  500000
                       0
                            0   10000   20000    30000    40000     50000    60000
                                                Income

        (b)
                    b0  299876.8059             b1  39.1698
                    Y  299876.8059+39.1698X
        (c) Since median family income of customer base cannot be 0, b0 just captures the portion of
                the latest one-month sales total that varies with factors other than income.
                 b1  39.1698 means that as the median family income of customer base increases by
                one dollar, the estimated average latest one-month sales total will increase by $39.17.
        (d) SYX  849860.17
        (e) r 2  0.1472 . 14.72% of the total variation in the franchise's latest one-month sales total
                 can be explained by using the median family income of customer base.
        (f) r  r  0.3837 . There is not a very strong positive linear relationship between latest
                  2

                one-month sales total and median family income of customer base.
328     Chapter 13: Simple Linear Regression


13.75    (g)
cont.

                                                                Income Residual Plot

                                   2000000
                      Residuals
                                                    0
                                                            0            10000 20000 30000 40000 50000 60000
                                  -2000000
                                                                                        Income

                There is a slight increase in the variance of the residuals at the higher end of the
                median family income. In general, however, the assumption of homoscedasticity
                seems to be intact.
                                  10                                               100.00%
                                   9                                               90.00%
                                   8                                               80.00%
                    Frequency




                                   7                                               70.00%
                                   6                                               60.00%        Frequency
                                   5                                               50.00%
                                   4                                               40.00%        Cumulative %
                                   3                                               30.00%
                                   2                                               20.00%
                                   1                                               10.00%
                                   0                                               .00%
                                       -1500000
                                                  -500000


                                                                         1500000
                                                                500000




                                                  Residual


                The histrogram does not suggest severe asymmetry nor abnormal extreme
                observations. So the normality assumption also seems to be intact.
         (h)     H0 :   0                                 H1 :   0
                                                                  r
                Test statistic: t         2.4926
                                    1 r2
                                    n2
                Decision rule: Reject H 0 when |t|>2.0289.
                Decision: Since t = 2.4926 is above the upper critical bound 2.4926, reject H 0 .
                There is enough evidence to conclude that there is a linear relationship between one-
                month sales total and median income of customer base.
         (i)     b1  tn2 Sb1  39.1697  2.028115.7143                                               7.2995< <71.0400
                                      Solutions to End-of-Section and Chapter Review Problems            329


13.75   (j)   (a)
cont.

                                                              Scatter Plot

                                 5000000

                                 4000000

                                 3000000
                    Sales


                                 2000000

                                 1000000

                                       0
                                           0      5      10      15         20   25       30        35    40
                                                                           Age


              (b)                 Y  931626.16+21782.76X
              (c)                Since median age of customer base cannot be 0, b0 just captures the portion of
                                 the latest one-month sales total that varies with factors other than age.
                                 b1  21782.76 means that as the median age of customer base increases by
                                 one year, the estimated average latest one-month sales total will increase by
                                 $21782.76.
              (d)                 SYX  919492.84
              (e)                 r 2  0.0017 . Only 0.17% of the total variation in the franchise's latest one-
                                  month sales total can be explained by using the median age of customer base.
              (f)                 r  r 2  0.0413 . There is essentially no linear relationship between latest
                                  one-month sales total and median age of customer base.
              (g)

                                                 Age Residual Plot

                                  4000000
                     Residuals




                                  2000000
                                           0
                                 -2000000 0             10            20          30           40
                                                                      Age

                                  The residuals are very eventually spread out across different range of median
                                  age.
330     Chapter 13: Simple Linear Regression


13.75    (j)    (h)         H0 :   0             H1 :   0
                                                    r
                            Test statistic: t         0.2482
                                                1 r2
                                                n2
                            Decision rule: Reject H 0 when |t|>2.0289.
                            Decision: Since t = 0.2482 is below the upper critical bound 2.4926, do not
                            reject H 0 . There is not enough evidence to conclude that there is a linear
                            relationship between one-month sales total and median age of customer base.
                (i)         b1  tn2 Sb1  21782.76354  2.028187749.63
                            -156181.50< <199747.02
         (k)    (a)
                                                   Scatter Diagram

                          4500000
                          4000000
                          3500000
                          3000000
                  Sales




                          2500000
                          2000000
                          1500000
                          1000000
                          500000
                                0
                                    0         20          40         60        80         100
                                                                HS

                            There appear to be some positive linear relationship between total sales and
                            percentage of customer base with high school diploma.
                (b)          Y  -2969741.23+59660.09X
                (c)         Since the percent of customer base with high school diploma cannot be 0, b0
                            just captures the portion of the latest one-month sales total that varies with
                            factors other than HS. b1  59660.09 means that as one more percent of the
                            customer base have received a high school diploma, the estimated average
                            latest one-month sales total will increase by $59660.09.
                (d)         SYX  802003.81
                (e)         r 2  0.2405 . 24.05% of the total variation in the franchise's latest one-
                            month sales total can be explained by the percentage of customer base with a
                            high school diploma.
                (f)         r  r 2  0.4904 . There is some positive linear relationship between latest
                            one-month sales total and percentage of customer base with a high school
                            diploma.
                                         Solutions to End-of-Section and Chapter Review Problems         331


13.75   (k)   (g)
cont.

                                                  HS Residual Plot

                                 2000000

                    Residuals             0
                                              0     20        40         60        80        100
                                -2000000
                                                                    HS

                                 The residual plot suggests there might be a violation of the homoscedasticity
                                 assumption since the variance of the residuals increases as the percentage of
                                 customer base with a high school diploma increases.
              (h)                 H0 :   0            H1 :   0
                                                          r
                                 Test statistic: t         3.3766
                                                    1 r2
                                                     n2
                                 Decision rule: Reject H 0 when |t|>2.0289.
                                 Decision: Since t = 3.3766 is above the upper critical bound 2.4926, reject
                                  H 0 . There is enough evidence to conclude that there is a linear relationship
                                 between one-month sales total and percentage of customer base with a high
                                 school diploma.
              (i)                 b1  tn2 Sb1  59660.09  2.028117668.885 23825.98< <95494.21
        (l)   (a)
                                                        Scatter Diagram

                         4500000
                         4000000
                         3500000
                         3000000
                Sales




                         2500000
                         2000000
                         1500000
                         1000000
                                500000
                                     0
                                         0         10          20             30        40         50
                                                                    Collge

                                 There is a positive linear relationship between total sales and percentage of
                                 customer base with a college diploma.
              (b)                Y  789847.38+35854.15X
332     Chapter 13: Simple Linear Regression


13.75    (l)    (c)     Since the percent of customer base with college diploma cannot be 0, b0 just
cont.                   captures the portion of the latest one-month sales total that varies with factors
                        other than College. b1  35854.15 means that as one more percent of the
                        customer base have received a college diploma, the estimated average latest
                        one-month sales total will increase by $35854.15.
                (d)     SYX  871329.93
                (e)     r 2  0.1036 . 10.36% of the total variation in the franchise's latest one-
                        month sales total can be explained by the percentage of customer base with a
                        college diploma.
                (f)     r  r 2  0.3218 . There is some positive linear relationship between latest
                        one-month sales total and percentage of customer base with a college
                        diploma.
                (g)

                                                      College Residual Plot

                                         4000000
                            Residu als




                                         2000000
                                               0
                                         -2000000 0         10     20     30   40      50
                                                                    College

                        The residuals are quite evenly spread out around zero even though there
                        might be a slight tendency for the variance to increase as the percentage of
                        customer base with a college diploma increases.
                (h)     H0 :   0                    H1 :   0
                                                        r
                        Test statistic: t         2.0392
                                            1 r2
                                            n2
                        Decision rule: Reject H 0 when |t|>2.0289.
                        Decision: Since t = 2.0392 is above the upper critical bound 2.4926, reject
                        H 0 . There is enough evidence to conclude that there is a linear relationship
                        between one-month sales total and percentage of customer base with a
                        college diploma.
                (i)     b1  tn2 Sb1  35854.15  2.028117582.269  195.75< <71512.60
                                  Solutions to End-of-Section and Chapter Review Problems             333


13.75   (m)   (a)
cont.
                                                   Scatter Diagram

                         4500000
                         4000000
                         3500000
                         3000000
                Sales

                         2500000
                         2000000
                         1500000
                         1000000
                             500000
                                  0
                        -5            0        5           10        15        20         25
                                                        Growth

                             It is not obvious that there is any linear relationship between total sales and
                             annual population growth rate of customer base over the past 10 years.
              (b)            Y  1595571.48+26833.54X
              (c)            b0 =1595571 means the estimated average latest one-month sales total is
                             $1595571 when the annual population growth rate of customer base over the
                             past 10 years is zero. b1  26833.54 means that as the annual population
                             growth rate increases by 1%, the estimated average latest one-month sales
                             total will increase by $26833.54.
              (d)            SYX  914466.58
              (e)            r 2  0.0126 . Only 1.26% of the total variation in the franchise's latest one-
                             month sales total can be explained by the annual population growth rate of
                             customer base over the past 10 years.
              (f)            r  r 2  0.1122 . If there is any linear relationship between latest one-
                             month sales total and annual population growth rate of customer base over
                             the past 10 years, it will be a very weak positive relationship.
334     Chapter 13: Simple Linear Regression


13.75    (m)    (g)

                                                      Growth Residual Plot

                                         4000000



                            Residu als
                                         2000000
                                               0
                                          -5
                                         -2000000 0        5         10     15   20   25
                                                                   Growth

                        There seems to be a diamond shape pattern of the residual distribution and,
                        hence, a violation of the homoscedasticity assumption. The variance is larger
                        the closer is the growth rate towards zero.
                (h)     H0 :   0                    H1 :   0
                                                       r
                        Test statistic: t         0.6776
                                            1 r2
                                            n2
                        Decision rule: Reject H 0 when |t|>2.0289.
                        Decision: Since t = 0.6776 is below the upper critical bound 2.4926, do not
                        reject H 0 . There is not enough evidence to conclude that there is a linear
                        relationship between one-month sales total and the annual population growth
                        rate of customer base over the past 10 years.
                (i)     b1  tn2 Sb1  26833.54  2.0281 39601.427
                        -53481.77< <107148.86
         (n)    Percentage of customer base with a high school diploma will be the best predictor for
                sales at an individual store location since it provides the highest explanatory power of
                24.05% among the four models.
                                       Solutions to End-of-Section and Chapter Review Problems   335


13.76   (a)
                                                  Scatter Diagram

                            100
                             90
                             80
                             70
                % Passing    60
                             50
                             40
                             30
                             20
                             10
                              0
                                  88        90        92            94   96        98
                                                      % Attendance


              There is a very obvious positive relationship between % of students passing the
              proficiency test and the daily average of the percentage of students attending class.
        (b)   b0  -771.5868                               b1  8.8447
              Y  -771.5869+8.8447X
        (c)   Since it does not make sense for % attendance to be zero, b0  -771.5868 should be
              interpreted as the portion of % passing the proficiency exam that will varies with
              factors other than % attendance. b1  8.8447 implies that as daily average
              percentage of students attending class increases by 1%, the estimated average
              percentage of students passing the ninth-grade proficiency test will increase by
              8.8447%.
        (d)   SYX  10.5787
        (e)   r 2  0.6024 . 60.24% of the total variation in % passing the proficiency test can be
              explained by % attendance.
        (f)   r  r 2  0.7762 . There is a rather strong positive linear relationship between %
              of students passing the proficiency test and daily average of the % of students
              attending class.
336     Chapter 13: Simple Linear Regression


13.76    (g)
cont.

                                           % Attendance Residual Plot

                                50
                    Residuals
                                 0
                                      88       90          92       94   96      98
                                -50
                                                       % Attendance


                The residuals are evenly distributed across difference range of % attendance. There
                is no obvious violation of the homoscedasticity assumption.


                                             Histogram of Residuals

                                16                         100.00%
                                14
                                                           80.00%
                                12
                    Frequency




                                10                         60.00%        Frequency
                                 8
                                                           40.00%        Cumulative %
                                 6
                                 4
                                                           20.00%
                                 2
                                 0                         .00%
                                        -3
                                         4
                                        11
                                        18
                                       -24
                                       -17
                                       -10




                                              Bin



                The distribution of the residuals is left skewed. However, with the exception of 2
                extremely negative residuals, the histrogram is not too badly skewed.

         (h)     H0 :   0                   H1 :   0
                                                r
                Test statistic: t         8.2578
                                    1 r2
                                    n2
                Decision rule: Reject H 0 when |t|>2.014.
                Decision: Since t = 8.2578 is above the upper critical bound 2.014, reject H 0 . There
                is enough evidence to conclude that there is a linear relationship between % passing
                and % attendance.

         (i)     b1  tn2 Sb1  8.8447  2.01411.0711                 6.6874< <11.0020
                                      Solutions to End-of-Section and Chapter Review Problems      337


13.76   (j)   (a)
cont.
                                                   Scatter Diagram

                            100
                             90
                             80
                             70
                % Passing
                             60
                             50
                             40
                             30
                             20
                             10
                              0
                                  0        10000     20000      30000       40000      50000
                                                          Salary


                              There seems to be a slightly positive relationship between % of students
                              passing the proficiency test and average teacher salary.
              (b)             Y  23.065+0.0011X
              (c)             Since it does not make sense for teach salary to be zero, b0  23.065 should
                              be interpreted as the portion of % passing the proficiency exam that will
                              varies with factors other than teacher salary. b1  0.0011 implies that as
                              average teacher salary increase by $1, the estimated average percentage of
                              students passing the ninth-grade proficiency test will increase by 0.0011%.
              (d)             SYX  16.3755
              (e)             r 2  0.0474 . Only 4.74% of the total variation in % passing the proficiency
                              test can be explained by average teacher salary.
              (f)             r  r 2  0.2177 . There seems to be a rather weak positive linear
                              relationship between % of students passing the proficiency test and average
                              teacher salary.
338     Chapter 13: Simple Linear Regression


13.76    (j)    (g)
cont.

                                                    Salary Residual Plot

                                  50
                      Residuals

                                   0
                                        0       10000      20000       30000   40000   50000
                                  -50
                                                                   Salary


                The residuals are evenly distributed across difference range of % attendance. There
                is no obvious violation of the homoscedasticity assumption.


                                                Histogram of Residuals

                                  15                               100.00%
                                                                   80.00%
                      Frequency




                                  10                               60.00%        Frequency
                                                                   40.00%        Cumulative %
                                   5
                                                                   20.00%
                                   0                               .00%
                                                     5
                                                          25
                                        -35
                                              -15




                                                    Bin


                The distribution of the residuals is slightly left skewed but not too far from a normal
                distribution.

                (h)                H0 :   0              H1 :   0
                                                               r
                                   Test statistic: t         1.496
                                                      1 r2
                                                       n2
                                   Decision rule: Reject H 0 when |t|>2.014.
                                   Decision: Since t = 1.496 is below the upper critical bound 2.014, do not
                                   reject H 0 . There is not enough evidence to conclude that there is a linear
                                   relationship between % passing and average teacher salary
                (i)                b1  tn2 Sb1  0.0011  2.0141 0.00073           -0.000375< <0.002542
                                                  Solutions to End-of-Section and Chapter Review Problems      339


13.76   (k)   (a)
                                                               Scatter Diagram

                               100
                                90
                                80
                                70
                % Passing       60
                                50
                                40
                                30
                                20
                                10
                                 0
                                         0              1000           2000      3000       4000
                                                                    Spending

                                         There seems to be a slightly positive relationship between % of students
                                         passing the proficiency test and spending per pupil.
              (b)                        Y  35.7843+0.0109X
              (c)                        Since it does not make sense for spending per pupil to be zero, b0  35.7843
                                         should be interpreted as the portion of % passing the proficiency exam that
                                         will varies with factors other than spending. b1  0.0109 implies that as
                                         spending per pupil increase by $1, the estimated average percentage of
                                         students passing the ninth-grade proficiency test will increase by 0.019%.
              (d)                        SYX  15.9984
              (e)                        r 2  0.0907 . Only 9.07% of the total variation in % passing the proficiency
                                         test can be explained by spending per pupil.
              (f)                        r  r 2  0.3012 . There seems to be a rather weak positive linear
                                         relationship between % of students passing the proficiency test and spending
                                         per pupil.


                                                     Spending Residual Plot

                                        50
                            Residuals




                                         0
                                              0          1000          2000      3000     4000
                                        -50
                                                                    Spending

                                         The residuals are evenly distributed across difference range of % attendance.
                                         There is no obvious violation of the homoscedasticity assumption.
340     Chapter 13: Simple Linear Regression


13.76    (k)    (g)
cont.
                                            Histogram of Residuals

                                       15                 100.00%




                           Frequency
                                                          80.00%
                                       10                 60.00%     Frequency
                                       5                  40.00%     Cumulative %
                                                          20.00%
                                       0                  .00%




                                             -5
                                            -35
                                            -25
                                            -15

                                              5
                                             15
                                             25
                                             Bin


                        The distribution of the residuals is slightly left skewed. Excluding the largest
                        negative residual, the histrogram is quite symmetric.
                (h)     H0 :   0                 H1 :   0
                                                    r
                        Test statistic: t         2.1192
                                            1 r2
                                            n2
                        Decision rule: Reject H 0 when |t|>2.014.
                        Decision: Since t = 2.1192 is above the upper critical bound 2.014, reject
                        H 0 . There is enough evidence to conclude that there is a linear relationship
                        between % passing and spending.
                (i)     b1  tn2 Sb1  0.0109  2.0141 0.0052                 0.00054< <0.02129
         (l)    The model with % attendance is the best model to use for predicting % passing since
                it has the highest R-square of 60.24%.

13.77    (a)     Y  0.2141+0.4738X
         (b)    For Exxon, the estimated value of its stock will increase by 0.47% on average when the
                S&P 500 index increases by 1%.
         (c)    (a)     Y  0.6064+0.3607X
                (b)     For Mobil Oil, the estimated value of its stock will increase by 0.36% on
                        average when the S&P 500 index increases by 1%.
         (d)    (a)     Y  -0.2101+0.2068X
                (b)     For International Aluminum, the estimated value of its stock will increase by
                        0.21% on average when the S&P 500 index increases by 1%.
         (e)    (a)     Y  -0.4182+0.5131X
                (b)     For Sears, the estimated value of its stock will increase by 0.51% on average
                        when the S&P 500 index increases by 1%.
         (f)    (a)     Y  -0.4471+1.4286X
                (b)     For BancOne Corporation, the estimated value of its stock will increase by
                        1.43% on average when the S&P 500 index increases by 1%.
         (g)    (a)     Y  0.0361+0.9430X
                (b)     For General Motors, the estimated value of its stock will increase by 0.94%
                        on average when the S&P 500 index increases by 1%.
                            Solutions to End-of-Section and Chapter Review Problems              341


13.78   (a)

                      GM       Ford      IAL             HCR
        GM                  1
        Ford        0.856954         1
        IAL         -0.06409 0.194917         1
        HCR           -0.4403 -0.08868 0.598569                 1
        (b)    There is a strong positive correlation of r = 0.8569 between the stock prices of GM
               and Ford, an also relatively strong positive correlation of r = 0.5986 between the
               prices of IAL and HCR, a moderately negative correlation of r = -0.4403 between the
               prices of GM and HCR, a weak positive correlation of r = 0.1949 between the prices
               of Ford and IAL, a very weak negative correlation of r = -0.08868 between the prices
               of Ford and HCR, and also a very weak negative correlation of r = -0.06408 between
               GM and IAL.
        (c)    It is not a good idea to have all the stocks in an individual's portfolio be strongly
               positively correlated for that will increase the variance, a measure of risk, of the
               portfolio. Some negatively correlated stock prices in a portfolio can reduce the
               combined variance, though at the price of reduced combined expected return.

                        SSXY
13.79   (a)    r                   0.2196
                      SSX SSY
        (b)    H0 :                    H1 :   0
                                     r
               Test statistic: t          = -1.7146
                                    1 r2
                                    n2
               Decision rule: Reject H 0 if the p-value is less than the level of significance
                 0.05 .
               Decision: Since the p value of 0.0918 is greater than the level of significance, do not
               reject the null hypothesis. There is not enough evidence to conclude that there is a
               linear relationship between the two variables.
        (c)    The 10% level of significance is greater than the p value of 0.0918 and, hence, the
               null hypothesis will be rejected. There is enough evidence at the 10% level of
               significance to conclude that there is a linear relationship between the two variables.
342   Chapter 13: Simple Linear Regression


                                      The Springville Herald Case

SH13.1        The method was placing too much emphasis on the last two time periods instead of
              considering a large amount of data that was available. This overemphasis would tend
              to lead to large fluctuations in the forecast accuracy when the trend differed from
              what had occurred in the last two months.

SH13.2        There are many factors that might be considered other than the number of outlets.
              Among them are:
              1. The amount of newspaper advertising for new subscriptions.
              2. The amount of radio and television advertising for new subscriptions.
              3. Whether or not special promotional campaigns were being used in a particular
              month.
              4. The average experience of the telemarketers used in a given month.
              5. The training programs provided to the telemarketers in a given month.

SH13.3 (a)    The statistical model initially fit to the data is a simple linear model that predicts
              New Subscriptions = -413.82 + 4.40795 Telemarketing Hours

              The Y intercept (b0) equal to -413.82, has no direct interpretation since sales of
              below zero new subscriptions is not feasible. We can say however, that -413.82
              represents a constant portion of the predicted average new subscriptions that varies
              with factors other than the number of telemarketing hours.

               The slope (b1) equal to 4.40795, can be interpreted to mean that for each increase of
              one telemarketing hour, average new subscriptions are predicted to increase by
              4.40795 per month.

              The r2 value of .8617 can be interpreted to mean that 86.17% of the variation in new
              subscriptions can be explained by variation in the number of telemarketing hours
              from month to month.

              Since the data was collected for 24 consecutive months, it is important for us to
              determine whether there is any autocorrelation among the residuals. The Durbin-Watson
              D statistic of 1.7009 is > 1.45, the upper critical value for n = 24 and   0.05 . This
              leads us to believe that there isn't any positive autocorrelation among the residuals. We
              could also examine the plot of the standardized residuals over time to see whether a
              pattern existed. In this case this plot does not seem to indicate any evidence of positive
              association among consecutive residuals.

              Before moving further with any predictions based on the model, we need to
              determine whether the simple linear model was appropriate for these data The
              residual plot of the standardized residuals versus the fitted Y values (or the X values)
              indicates no evidence of any pattern. Thus, we may conclude that the simple linear
              model was appropriate for these data.
                          Solutions to End-of-Section and Chapter Review Problems            343


SH13.3 (a)
cont.        We may examine the validity of the assumption of normality among the residuals by
             studying the normal probability plot. If the points in this plot fall on an approximate
             straight line, there would be no reason to suspect serious departure from normality.
             Since that seems to be the case here, the normality assumption does not appear to
             have been seriously violated.
       (b)   For X = 1,000, the predicted value of
             New Subscriptions = -413.82 + 4.40795 (1000) = 3,994.1


       (c)   We note that the value of 2,000 for is beyond the range of our X values. Thus, we
             would be assuming that the regression model fit for a range of 704 to 1,498
             telemarketing hours would be valid for a month in which the telemarketing hours was
             2,000. This represents a situation which is well beyond the range of the number of
             telemarketing hours that we have used in the past 24 months. Many things could
             change with this increase of telemarketing hours, and the type of increase in new
             subscriptions that we have had with expanding telemarketing hours may not
             continue when we further expand by another 25% above the maximum amount that
             we have used in the past.

								
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