Question

An assistant in the district sales office of a national cosmetics firm obtained data on advertising expenditures and sales last year in the district’s 44 territories. Data is consmetics.csv. Use R. I don't want answers in Excel or SAS :)

X1: expenditures for point-of-sale displays in beauty salons and department stores (X$1000).

X2: expenditures for local media advertising.

X3: expenditures for prorated share of national media advertising.

Y: Sales (X$1000).

6. (4) Are there any influential points?

7. Is there a serious multicollinearity problem?

(3) Include an appropriate scatterplot and correlation values between the explanatory variables.

(3) Judge by VIF, do you think there is a problem with multicollinearity? (Hint: VIP or tolerance)

(3) Compare your answers in parts i and ii. Are your conclusions the same or different? Please explain your answer.

Data:

y	x1	x2	x3
12.85	5.6	5.6	3.8
11.55	4.1	4.8	4.8
12.78	3.7	3.5	3.6
11.19	4.8	4.5	5.2
9	3.4	3.7	2.9
9.34	6.1	5.8	3.4
13.8	7.7	7.2	3.8
8.79	4	4	3.8
8.54	2.8	2.3	2.9
6.23	3.2	3	2.8
11.77	4.2	4.5	5.1
8.04	2.7	2.1	4.3
5.8	1.8	2.5	2.3
11.57	5	4.6	3.6
7.03	2.9	3.2	4
0.27	0	0.2	2.7
5.1	1.4	2.2	3.8
9.91	4.2	4.3	4.3
6.56	2.4	2.2	3.7
14.17	4.7	4.7	3.4
8.32	4.5	4.4	2.7
7.32	3.6	2.9	2.8
3.45	0.6	0.8	3.4
13.73	5.6	4.7	5.3
8.06	3.2	3.3	3.6
9.94	3.7	3.5	4.3
11.54	5.5	4.9	3.2
10.8	3	3.6	4.6
12.33	5.8	5	4.5
2.96	3.5	3.1	3
7.38	2.3	2	2.2
8.68	2	1.8	2.5
11.51	4.9	5.3	3.8
1.6	0.1	0.3	2.7
10.93	3.6	3.8	3.8
11.61	4.9	4.4	2.5
17.99	8.4	8.2	3.9
9.58	2.1	2.3	3.9
7.05	1.9	1.8	3.8
8.85	2.4	2	2.4
7.53	3.6	3.5	2.4
10.47	3.6	3.7	4.4
11.03	3.9	3.6	2.9
12.31	5.5	5	5.5

Answer 1

Plot data:

y=c(12.85,11.55,12.78,11.19,9,9.34,13.8,8.79,8.54,6.23,11.77,8.04,5.8,11.57,7.03,0.27,5.1,9.91,6.56,14.17,8.32,7.32,3.45,13.73,8.06,9.94,11.54,10.8,12.33,2.96,7.38,8.68,11.51,1.6,10.93,11.61,17.99,9.58,7.05,8.85,7.53,10.47,11.03,12.31)

x1=c(5.6,4.1,3.7,4.8,3.4,6.1,7.7,4,2.8,3.2,4.2,2.7,1.8,5,2.9,0,1.4,4.2,2.4,4.7,4.5,3.6,0.6,5.6,3.2,3.7,5.5,3,5.8,3.5,2.3,2,4.9,0.1,3.6,4.9,8.4,2.1,1.9,2.4,3.6,3.6,3.9,5.5)

x2=c(5.6,4.8,3.5,4.5,3.7,5.8,7.2,4,2.3,3,4.5,2.1,2.5,4.6,3.2,0.2,2.2,4.3,2.2,4.7,4.4,2.9,0.8,4.7,3.3,3.5,4.9,3.6,5,3.1,2,1.8,5.3,0.3,3.8,4.4,8.2,2.3,1.8,2,3.5,3.7,3.6,5)

x3=c(3.8,4.8,3.6,5.2,2.9,3.4,3.8,3.8,2.9,2.8,5.1,4.3,2.3,3.6,4,2.7,3.8,4.3,3.7,3.4,2.7,2.8,3.4,5.3,3.6,4.3,3.2,4.6,4.5,3,2.2,2.5,3.8,2.7,3.8,2.5,3.9,3.9,3.8,2.4,2.4,4.4,2.9,5.5)

# fit the model using R

fit=lm(y~x1+x2+x3)

fit

Output:

Call:

lm(formula = y ~ x1 + x2 + x3)

Coefficients:

(Intercept)           x1           x2           x3

     1.0233       0.9657       0.6292       0.6760

6. (4) Are there any influential points?

To detect the influential points there is direct command in R

influence.measures(fit)

Output:

Influence measures of

         lm(formula = y ~ x1 + x2 + x3) :

      dfb.1_   dfb.x1   dfb.x2   dfb.x3    dffit cov.r   cook.d    hat inf

1 -0.004003 -0.01553 0.02367 -0.01018 0.04872 1.181 6.08e-04 0.0662   

2 -0.031724 -0.06110 0.06057 0.02859 0.07981 1.322 1.63e-03 0.1655   *

3   0.070146 0.09176 -0.09474 0.02087 0.32625 0.755 2.46e-02 0.0248   

4   0.118258 -0.04666 0.04956 -0.14036 -0.16740 1.219 7.15e-03 0.1128   

5   0.035752 -0.04028 0.04277 -0.03719 0.06299 1.185 1.02e-03 0.0712   

6 -0.079405 -0.04496 -0.08332 0.27876 -0.62822 0.786 9.09e-02 0.0823   

7   0.054999 -0.08175 -0.01744 0.11740 -0.44643 1.170 4.97e-02 0.1536   

8   0.000966 0.02744 -0.03046 -0.01174 -0.10705 1.088 2.91e-03 0.0263   

9 0.129859 0.14546 -0.15695 -0.04914 0.22686 1.121 1.30e-02 0.0750   

10 -0.160591 -0.03721 0.03594 0.11753 -0.19990 1.060 1.00e-02 0.0441   

11 -0.053408 -0.03567 0.03294 0.05924 0.08495 1.242 1.85e-03 0.1140   

12 -0.007740 0.02877 -0.03375 0.02603 0.04375 1.293 4.91e-04 0.1454   

13 -0.015649 0.01800 -0.01732 0.01390 -0.02408 1.334 1.49e-04 0.1709   *

14 0.003829 0.02078 -0.01477 -0.00666 0.04516 1.149 5.22e-04 0.0419   

15 0.016795 0.11599 -0.09709 -0.06883 -0.19640 1.084 9.71e-03 0.0510   

16 -0.390214 0.08274 0.03610 0.06160 -0.61973 0.978 9.23e-02 0.1279   

17 -0.025477 0.23978 -0.20246 -0.04896 -0.29653 1.226 2.22e-02 0.1440   

18 0.037280 0.03407 -0.03469 -0.04316 -0.09325 1.136 2.22e-03 0.0442   

19 -0.012876 -0.03027 0.04342 -0.03343 -0.08752 1.152 1.96e-03 0.0527   

20 0.107908 -0.13563 0.19282 -0.18189 0.42142 0.801 4.15e-02 0.0449   

21 -0.163736 0.03623 -0.07073 0.20341 -0.26079 1.092 1.71e-02 0.0725   

22 -0.087741 -0.12738 0.12487 0.05224 -0.16931 1.195 7.30e-03 0.0983   

23 -0.060632 0.02725 0.01142 -0.04266 -0.18278 1.193 8.50e-03 0.1006   

24 -0.139466 0.17062 -0.16852 0.16797 0.24624 1.385 1.55e-02 0.2185   *

25 -0.012813 0.02110 -0.01806 -0.00178 -0.05279 1.128 7.13e-04 0.0283   

26 -0.010481 0.00798 -0.00989 0.01950 0.02861 1.158 2.10e-04 0.0462   

27 -0.001794 -0.00372 0.00266 0.00269 -0.00671 1.201 1.15e-05 0.0787   

28 -0.108793 -0.23083 0.20589 0.15425 0.32599 1.165 2.67e-02 0.1214   

29 0.041056 -0.07707 0.06971 -0.04308 -0.10309 1.248 2.72e-03 0.1206   

30 -0.465021 -0.41726 0.40908 0.28135 -0.76987 0.405 1.17e-01 0.0480   *

31 0.231108 0.07521 -0.08395 -0.16011 0.25762 1.146 1.67e-02 0.0947   

32 0.407048 0.11020 -0.14871 -0.22532 0.47779 0.901 5.45e-02 0.0740   

33 0.001028 0.02057 -0.02336 0.00520 -0.02853 1.231 2.09e-04 0.1017   

34 -0.213931 0.04803 0.01519 0.03823 -0.33437 1.163 2.81e-02 0.1225   

35 0.003315 -0.09516 0.09307 0.01485 0.16039 1.074 6.48e-03 0.0371   

36 0.165522 0.09625 -0.05904 -0.19597 0.27011 1.153 1.84e-02 0.1015   

37 -0.054049 -0.05080 0.13001 -0.10605 0.35941 1.376 3.28e-02 0.2304   *

38 0.009109 -0.10633 0.04840 0.14990 0.34029 0.965 2.83e-02 0.0562   

39 0.006866 0.01507 -0.02848 0.03436 0.07403 1.178 1.40e-03 0.0675   

40 0.389962 0.21424 -0.23514 -0.23780 0.48302 0.962 5.64e-02 0.0901   

41 -0.106345 0.00994 -0.01905 0.10628 -0.12748 1.172 4.15e-03 0.0749   

42 -0.035322 -0.02262 0.01698 0.05469 0.08450 1.148 1.82e-03 0.0493   

43 0.166767 0.07541 -0.05798 -0.14659 0.24637 1.018 1.51e-02 0.0451   

44 0.169796 -0.09530 0.09362 -0.18822 -0.23263 1.285 1.38e-02 0.1630   

> # Yes there is influential points, (*) indicates the influential points . 2,13,24,30,37 these are influential points . we also find influential points using cooks distance, A common rule of thumb is that an observation with a value of Cook's D over 1.0 has too much influence.

7. Is there a serious multicollinearity problem?

Using correlation matrix we find multicollinearity

D=data.frame(x1,x2,x3)

cor(D) # to obtain correlation matrix

Output:

         x1        x2        x3

x1 1.0000000 0.9744313 0.3759509

x2 0.9744313 1.0000000 0.4099208

x3 0.3759509 0.4099208 1.0000000

After looking correlation matrix we say that x1 and x2 are highly collinear but there is no multicollinearity problem.

Include an appropriate scatterplot and correlation values between the explanatory variables.

Command:

D=data.frame(x1,x2,x3)

plot(D)    # to plot scatter plot

cor(D)    # to plot correlation values

Output:

Correlation matrix of all explanatory variables

x1                         x2                       x3

x1      1.0000000        0.9744313         0.3759509

x2       0.9744313      1.0000000          0.4099208

x3       0.3759509       0.4099208         1.0000000

Judge by VIF, do you think there is a problem with multicollinearity? (Hint: VIP or tolerance)

We obtain VIF using R

fit=lm(y~x1+x2+x3)

summary(fit)    # to obtain summary of model

Output:

Call:

lm(formula = y ~ x1 + x2 + x3)

Residuals:

    Min      1Q Median      3Q     Max

-5.4217 -0.9115 0.0703 1.1420 3.5479

Coefficients:

            Estimate Std. Error t value Pr(>|t|)

(Intercept)   1.0233     1.2029   0.851   0.4000

x1            0.9657     0.7092   1.362   0.1809

x2            0.6292     0.7783   0.808   0.4237

x3            0.6760     0.3557   1.900   0.0646 .

---

Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 1.825 on 40 degrees of freedom

Multiple R-squared: 0.7417,    Adjusted R-squared: 0.7223

F-statistic: 38.28 on 3 and 40 DF, p-value: 7.821e-12

From output we get R-square value which is : 0.7417 hence

Cammand:

R2=0.7417

(1-R2)       # to get tolarance

[1] 0.2583   #Output

VIF=(1/(1-R2))

VIF

[1] 3.871467

#VIF<5

sqrt(VIF)

[1] 1.967604

Here VIF<5 and square root of VIF is also less than 2 hence we say that there is no multicollinearity problem.

Compare your answers in parts i and ii. Are your conclusions the same or different? Please explain your answer.

Ans: Yes we are getting same result using correlation and VIF there is no multicollinearity problem.