Question

Use the data set named Store_Visits located in the folder Data Files for HW Assignment (outside of Minitab folder) in the K-drive. The response variable y is the number of visits of a customer to a particular food store in a large suburban area within the period of a month, and the independent variable x is the distance (in miles) of the customer’s home to the store.

Fit a simple linear regression model to the data, and answer the following questions.

a) Give the proportion of the variation in the number of visits per month of a customer explained by the distance of the customer’s home to the store.

b) Submit the residual plot. ^ It appears from the plot that there is a problem with one of the model assumptions. Which one is it, and what would you suggest to remedy the problem?

2. c) Carry out your suggestion to fix the problem of part (b) and submit a new residual plot. Does your suggested remedy work? ^

3. d) Based on your new model, what is the proportion of the variation in the number of visits per month of a customer explained by the distance of the customer’s home to the store? How does it compare to that of the original model?

4. e) Based on your new model, construct a 95% prediction interval for y, the number of visits to the store for a customer who lives 2.5 miles from the store. Interpret the P.I.

K-Drive data. -Minitab

y   x
12   0.8
5   1.2
6   2.3
8   1.5
3   3.2
2   6.3
1   7.9
2   5.3
6   1.5
3   1.9
10   1.7
5   2.6
3   2.9
6   4.2
2   3.9
4   3.1
3   5.8
6   1.7
7   2.2
2   4.5
1   6.1
1   5.8
1   7.4
3   6.4
2   4.7
2   3.9
3   4
4   4.6

Answer 1

The response variable y is the number of visits of a customer to a particular food store in a large suburban area within the period of a month,

And the independent variable x is the distance (in miles) of the customer’s home to the store.

by using Minitab

Fit a simple linear regression model to the data

Steps

1) Enter given data in Minitab coloums

2) Select following options

     Stats Regression Regression Fit Regression Modal

3) Then in obtained box select

   Responses : Column of variable y

And          Continue Predictors : Column of variable x

4) You can select Graph You need ( this can residual graph ) , and storage if wanted

5) then click ok

Now these is our data ( copied from minitab with residuals output )

y	x	RESI1
12	0.8	4.721378
5	1.2	-1.85132
6	2.3	0.323756
8	1.5	1.469155
3	3.2	-1.71482
2	6.3	0.596763
1	7.9	1.305966
2	5.3	-0.47149
6	1.5	-0.53085
3	1.9	-3.10354
10	1.7	3.682805
5	2.6	-0.35577
3	2.9	-2.03529
6	4.2	2.353435
2	3.9	-1.96704
4	3.1	-0.82164
3	5.8	1.062638
6	1.7	-0.31719
7	2.2	1.216931
2	4.5	-1.32609
1	6.1	-0.61689
1	5.8	-0.93736
1	7.4	0.77184
3	6.4	1.703589
2	4.7	-1.11244
2	3.9	-1.96704
3	4	-0.86022
4	4.6	0.780735

And this is minitab output

Regression Analysis: y versus x

Analysis of Variance

Source        DF   Seq SS   Contribution Adj SS   Adj MS    F-Value P-Value
Regression      1   122.25        58.50%     122.25   122.246    36.65     0.000
          x             1   122.25        58.50%   122.25 122.246    36.65     0.000
     Error          26 86.72        41.50% 86.72      3.335
Lack-of-Fit      22    74.72        35.76%      74.72      3.396        1.13 0.510
Pure Error        4    12.00         5.74%     12.00      3.000
Total            27 208.96       100.00%

Model Summary

      S    R-sq R-sq(adj)    PRESS R-sq(pred)
1.82628 58.50%     56.90% 103.156      50.63%

Coefficients

Term       Coef     SE Coef       95% CI         T-Value P-Value    VIF
Constant 8.133    0.760     ( 6.572, 9.695)    10.71    0.000
     x           -1.068    0.176 (-1.431, -0.706)    -6.05   0.000      1.00

Regression Equation

y = 8.133 - 1.068 x

Fits and Diagnostics for Unusual Observations

                                                     Std    Del
Obs       y   Fit    SE Fit      95% CI      Resid Resid Resid        HI Cook’s D    DFITS
1 12.000 7.279   0.637 (5.969, 8.588) 4.721   2.76   3.22 0.121741      0.53 1.19750
11 10.000 6.317   0.511 (5.267, 7.368) 3.683   2.10   2.26 0.078294      0.19 0.65879

Obs
1 R
11 R

R Large residual

Residual Plots for y

Regression Equation

y = 8.133 - 1.068 x

a)

Proportion of the variation in the number of visits per month of a customer explained by the distance of the customer’s home to the store is 58.50%

From output table

Model Summary

      S       R-sq        R-sq(adj)    PRESS   R-sq(pred)
1.82628 58.50%     56.90%    103.156      50.63%

b) Submit the residual plot

It appears from the plot that there is a problem with one of the model assumptions

The plot shows a some U shape pattern i.e plot patterns are non-random , thus variance is not constant .

Suggesting a better fit for a non-linear model .

c)

Suggestion to fix the problem of part (b) it to use transformation ( nonlinear transformation)

A nonlinear transformation changes (increases or decreases) linear relationships between variables and, thus, changes the correlation between variables .

Steps in minitab

Steps

1) For data in Minitab coloums

2) Select following options

     Stats Regression Regression Fit Regression Modal

3) Then in obtained box select

   Responses : Column of variable y

And          Continue Predictors : Column of variable x

4) You can select Graph You need ( this can residual graph ) , and storage if wanted

5) Select Option , you will see box with " No transformation "

     Change it to " = 0.5 (square root)" For square root transformation .

6) then click ok

Here RESI2 is residual ( e = - ) column

y	x	RESI2
12	0.8	0.745803
5	1.2	-0.37464
6	2.3	0.134675
8	1.5	0.298421
3	3.2	-0.34067
2	6.3	0.175358
1	7.9	0.191528
2	5.3	-0.09363
6	1.5	-0.08052
3	1.9	-0.69036
10	1.7	0.68607
5	2.6	0.001951
3	2.9	-0.42137
6	4.2	0.645756
2	3.9	-0.47022
4	3.1	-0.09962
3	5.8	0.358701
6	1.7	-0.02672
7	2.2	0.304038
2	4.5	-0.30882
1	6.1	-0.29265
1	5.8	-0.37335
1	7.4	0.057034
3	6.4	0.520095
2	4.7	-0.25503
2	3.9	-0.47022
3	4	-0.12548
4	4.6	0.303862

Minitab Output

Regression Analysis: y versus x

Method

Box-Cox transformation λ = 0.5

Analysis of Variance for Transformed Response

Source         DF   Seq SS Contribution Adj SS Adj MS F-Value P-Value
Regression      1   7.7510        66.04% 7.7510 7.7510    50.56    0.000
   x             1   7.7510        66.04% 7.7510 7.7510    50.56    0.000
    Error          26   3.9856        33.96% 3.9856 0.1533
Lack-of-Fit     22   3.3918        28.90% 3.3918 0.1542     1.04    0.553
Pure Error 4   0.5938         5.06% 0.5938 0.1484
Total            27 11.7366       100.00%

Model Summary for Transformed Response

       S         R-sq      R-sq(adj)    PRESS      R-sq(pred)
0.391525 66.04%     64.74%     4.64139      60.45%

Coefficients for Transformed Response

Term        Coef    SE Coef      95% CI       T-Value P-Value   VIF
Constant     2.933     0.163    ( 2.599,   3.268)      18.01    0.000
     x          -0.2690    0.0378 (-0.3467, -0.1912)    -7.11    0.000     1.00

Regression Equation

y^0.5 = 2.933 - 0.2690 x

Fits and Diagnostics for Unusual Observations

Original Response

Obs         y      Fit      95% CI
1 12.0000 7.3891 (5.9414, 8.9946)

Transformed Response

                                                    Std    Del
Obs     y'      Fit       SE Fit      95% CI     Resid   Resid Resid        HI Cook’s D     DFITS
1    3.464 2.718   0.137 (2.437, 2.999) 0.746   2.03    2.17    0.121741      0.29 0.809136

Obs
1 R

y' = transformed response
R Large residual

Residual Plots for y

New residual plot are

We can see a random pattern in residual plot . And hence variance is constant

Yes suggested remedy work . ( i.e square root transformation )

d) For new model , Proportion of the variation in the number of visits per month of a customer explained by the distance of the customer’s home to the store is 66.04%

Model Summary for Transformed Response

       S         R-sq      R-sq(adj)    PRESS      R-sq(pred)
0.391525 66.04%     64.74%     4.64139      60.45%

Compare to that of the original model variance explained is increased by 8 %

For original model R² = 58.50%

For new model R^{2   =} 66.04%

e) Based on your new model, construct a 95% prediction interval for y, the number of visits to the store for a customer who lives 2.5 miles from the store. Interpret the P.I.

For new model 95% confidence interval is for variable x is (-0.3467, -0.1912) and that of constant or intercept is ( 2.599,   3.268)

{   Coefficients for Transformed Response

   Term        Coef    SE Coef      95% CI       T-Value P-Value   VIF
   Constant     2.933     0.163    ( 2.599,   3.268)      18.01    0.000
        x          -0.2690    0.0378 (-0.3467, -0.1912)    -7.11    0.000     1.00

}

Our New Model is

Regression Equation

y^0.5 = 2.933 - 0.2690 x

a prediction interval ( P.I ) is an estimate of an interval in which a future observation will fall, with a certain probability, (here 0.95 )

Construct a 95% prediction interval for y, the number of visits to the store for a customer who lives x =2.5 miles from the store

y^0.5 = 2.933 - 0.2690 x

   = ( 2.933 - 0.2690 x ) ²

for x = 2.5

= ( 2.933 - 0.2690 * 2.5 ) ²   = 5.10986

thus   = 5.10986        at x = 2.5

95% prediction interval for y is given by

t ^*   

Where t   =

Here n = 28 number of observation .

   and k = 1 number of independent variable

At 5 % level of significance

t   =    = =

You can find it from software like minitab , R or by statistical tables

t = = 2.055

And

^{=   S *}

n = 28

x' = 2.5

= mean ( x ) = 3.835714                ( can be calculated manually )

   = 107.1243  

Here S = 0.391525 ( form output table )

    S           R-sq      R-sq(adj)    PRESS      R-sq(pred)
0.391525   66.04%     64.74%     4.64139      60.45%

Hence S² = ( 0.391525 )² = 0.1532918

And

^{=   S * =} 0.391525 *    = 0.05969364

And = 0.003563331

hence 95% prediction interval for y is given by

t ^*

     ⁼ 5.10986 2.055 * ⁼ 5.10986 2.055 * 0.3960494

Thus 95% prediction interval for y is given by ( at x =2.5 )

P. I . = 5.10986 0.8138815

         = ( 4.295979 ,5.923742 )