Chapter 13, Problem 11RE

Air pollution: Following are measurements of particulate matter (PM) concentration (in micrograms per cubic meter), temperature, in degrees Fahrenheit wind speed in miles per hour, and humidity in percent for 38 days in Denver, Colorado.

1. Let y represent PM, $x_1$ represent temperature, $x_2$ represent wind speed, and $x_3$ represent humidity. Construct the multiple regression equation $\hat{y}=b_0+b_1{x}_1+b_2{x}_2+b_3{x}_3$.
2. Predict the particulate concentration on a day where the temperature is 40 degrees, the wind speed is 5 miles per hour, and the humidity is 30 percent.
3. Refer to part (b). Construct a 95% confidence interval for the particulate concentration.
4. Refer to part (b). Construct a 95% prediction interval for the particulate concentration.
5. Are all the values in the prediction interval reasonable? Explain.
6. What percentage of the variation in particulate concentration is explained by the model?
7. Is the model useful for prediction? Why or why not? Use the $\alpha =0.05$ level.
8. Test $H_0:{\beta }_1=0$ versus $H_1:{\beta }_1\ne 0$ at the $\alpha =0.05$ level. Can you reject $H_0$ ? Repeat for ${\beta }_2$ and ${\beta }_3$.

To determine

The multiple regression equation $\hat{y}=b_0+b_1{x}_1+b_2{x}_2+b_3{x}_3$

Answer to Problem 11RE

  $\hat{y}=-42.5163427+(0.6557601)x_1+(3.6607217)x_2+(0.5806123)x_3$

Explanation of Solution

It is known that for estimating coefficient below given formula is used

  $b = {\left(X\text{'}X\right)}^{-1}X\text{'}Y$

Where,

  $b=\left[\begin{array}{l}{b}_0\\ b_1\\ b_2\\ b_3\end{array}\right]$ And $X$ is the matrix of observations.

Now, using ‘R’ software coefficients are calculated.

  $\begin{array}{l}\hat{y}=b_0+b_1{x}_1+b_2{x}_2+b_3{x}_3\\ \hat{y}=-42.5163427+(0.6557601)x_1+(3.6607217)x_2+(0.5806123)x_3\end{array}$

To determine

The particulate concentration on a day where temperature, wind speed and humidity is $(40,5,30)$

Answer to Problem 11RE

  $\hat{y}=19.43604$

Explanation of Solution

From part (a)

  $\hat{y}=-42.5163427+(0.6557601)x_1+(3.6607217)x_2+(0.5806123)x_3$

Now putting the values of $(x_1,x_2,x_3)$ as $(40,5,30)$ in the above equation.

Using calculator,

  $\hat{y}=19.43604$

To determine

The 95% confidence interval for the particulate concentration

Answer to Problem 11RE

The 95% confidence interval for the particulate concentration is given below:

fitted	lower	upper
20.47848	15.11886	25.8381
52.31326	41.71678	62.90975
40.00498	32.93215	47.07781
24.39183	18.80337	29.98028
48.76496	38.10801	59.42191
12.63185	4.863304	20.40039
17.92577	12.6039	23.24764
23.18885	16.01585	30.36184
17.51326	10.06084	24.96568
20.19053	14.54028	25.84077
53.03338	43.31002	62.75673
19.38092	11.9622	26.79964
11.56664	3.850621	19.28265
9.147208	1.815757	16.47866
25.77363	19.68499	31.86227
37.31334	29.49157	45.13512
41.46232	32.1341	50.79054
40.68683	31.59058	49.78308
25.38821	19.62186	31.15456
20.29095	14.17048	26.41141
9.484433	0.412967	18.5559
21.81731	15.19766	28.43697
22.40822	13.90819	30.90825
15.21627	7.948146	22.48438
17.56946	11.07772	24.0612
12.52526	6.282131	18.76839
14.68255	7.538868	21.82623
16.894	8.889021	24.89897
18.27573	10.99998	25.55148
17.17309	9.305811	25.04037
27.14994	21.18848	33.11141
25.14871	16.67252	33.62489
24.66506	16.59167	32.73845
27.16704	18.07744	36.25665
33.64924	25.76426	41.53423
43.80541	35.69192	51.9189
21.51525	17.32562	25.70487
35.60585	29.60614	41.60555

Explanation of Solution

It is known that,

Confidence interval for dependent variable is given by

  ${\hat{y}}_h\pm t_{\frac{\alpha }{2},n-2}\sqrt{MSE\left(\frac{1}{n}+\frac{{\left(x_k-\stackrel{\_}{x}\right)}^2}{{\displaystyle \sum _{i=1}^n\left(X_i-\stackrel{\_}{x^2}\right)}}\right)}$

Where,

  ${\hat{y}}_h$ is the fitted response variable

  $\sqrt{MSE\left(\frac{1}{n}+\frac{{\left(x_k-\stackrel{\_}{x}\right)}^2}{{\displaystyle \sum _{i=1}^n\left(X_i-\stackrel{\_}{x^2}\right)}}\right)}$ is the standard error of the fit

Using ‘R’ programming 95% confidence interval has been calculated.

fitted	lower	upper
20.47848	15.11886	25.8381
52.31326	41.71678	62.90975
40.00498	32.93215	47.07781
24.39183	18.80337	29.98028
48.76496	38.10801	59.42191
12.63185	4.863304	20.40039
17.92577	12.6039	23.24764
23.18885	16.01585	30.36184
17.51326	10.06084	24.96568
20.19053	14.54028	25.84077
53.03338	43.31002	62.75673
19.38092	11.9622	26.79964
11.56664	3.850621	19.28265
9.147208	1.815757	16.47866
25.77363	19.68499	31.86227
37.31334	29.49157	45.13512
41.46232	32.1341	50.79054
40.68683	31.59058	49.78308
25.38821	19.62186	31.15456
20.29095	14.17048	26.41141
9.484433	0.412967	18.5559
21.81731	15.19766	28.43697
22.40822	13.90819	30.90825
15.21627	7.948146	22.48438
17.56946	11.07772	24.0612
12.52526	6.282131	18.76839
14.68255	7.538868	21.82623
16.894	8.889021	24.89897
18.27573	10.99998	25.55148
17.17309	9.305811	25.04037
27.14994	21.18848	33.11141
25.14871	16.67252	33.62489
24.66506	16.59167	32.73845
27.16704	18.07744	36.25665
33.64924	25.76426	41.53423
43.80541	35.69192	51.9189
21.51525	17.32562	25.70487
35.60585	29.60614	41.60555

To determine

The 95% prediction interval for the particulate concentration

Answer to Problem 11RE

The 95% prediction interval for the particulate concentration is given below

fitted	lower	upper
20.47848	-3.39044	44.3474
52.31326	26.75382	77.87271
40.00498	15.69398	64.31598
24.39183	0.470482	48.31317
48.76496	23.18039	74.34954
12.63185	-11.8906	37.15429
17.92577	-5.9347	41.78625
23.18885	-1.15149	47.52918
17.51326	-6.91088	41.93739
20.19053	-3.74533	44.12638
53.03338	27.82339	78.24336
19.38092	-5.03296	43.79479
11.56664	-12.9392	36.07249
9.147208	-15.2403	33.53471
25.77363	1.730515	49.81675
37.31334	12.77398	61.8527
41.46232	16.40208	66.52256
40.68683	15.71201	65.66165
25.38821	1.424684	49.35174
20.29095	-3.76025	44.34214
9.484433	-15.4814	34.45024
21.81731	-2.36573	46.00036
22.40822	-2.35567	47.17211
15.21627	-9.15227	39.5848
17.56946	-6.57888	41.7178
12.52526	-11.5574	36.60796
14.68255	-9.64916	39.01425
16.894	-7.70437	41.49236
18.27573	-6.09508	42.64654
17.17309	-7.38081	41.72699
27.14994	3.13872	51.16117
25.14871	0.392989	49.90442
24.66506	0.044349	49.28577
27.16704	2.194643	52.13945
33.64924	9.089665	58.20882
43.80541	19.17152	68.4393
21.51525	-2.11848	45.14897
35.60585	11.5851	59.62659

Explanation of Solution

It is known that,

Prediction interval for dependent variable is given by

  ${\hat{y}}_h\pm t_{\frac{\alpha }{2},n-2}\sqrt{MSE\left(1+\frac{1}{n}+\frac{{\left(x_k-\stackrel{\_}{x}\right)}^2}{{\displaystyle \sum _{i=1}^n{\left(x_i-\stackrel{\_}{x}\right)}^2}}\right)}$

Where,

  ${\hat{y}}_h$ is the fitted response variable

  $\sqrt{MSE\left(1+\frac{1}{n}+\frac{{\left(x_k-\stackrel{\_}{x}\right)}^2}{{\displaystyle \sum _{i=1}^n{\left(x_i-\stackrel{\_}{x}\right)}^2}}\right)}$ is the standard error of the prediction

Using ‘R’ programming 95% prediction interval has been calculated.

fitted	lower	upper
20.47848	-3.39044	44.3474
52.31326	26.75382	77.87271
40.00498	15.69398	64.31598
24.39183	0.470482	48.31317
48.76496	23.18039	74.34954
12.63185	-11.8906	37.15429
17.92577	-5.9347	41.78625
23.18885	-1.15149	47.52918
17.51326	-6.91088	41.93739
20.19053	-3.74533	44.12638
53.03338	27.82339	78.24336
19.38092	-5.03296	43.79479
11.56664	-12.9392	36.07249
9.147208	-15.2403	33.53471
25.77363	1.730515	49.81675
37.31334	12.77398	61.8527
41.46232	16.40208	66.52256
40.68683	15.71201	65.66165
25.38821	1.424684	49.35174
20.29095	-3.76025	44.34214
9.484433	-15.4814	34.45024
21.81731	-2.36573	46.00036
22.40822	-2.35567	47.17211
15.21627	-9.15227	39.5848
17.56946	-6.57888	41.7178
12.52526	-11.5574	36.60796
14.68255	-9.64916	39.01425
16.894	-7.70437	41.49236
18.27573	-6.09508	42.64654
17.17309	-7.38081	41.72699
27.14994	3.13872	51.16117
25.14871	0.392989	49.90442
24.66506	0.044349	49.28577
27.16704	2.194643	52.13945
33.64924	9.089665	58.20882
43.80541	19.17152	68.4393
21.51525	-2.11848	45.14897
35.60585	11.5851	59.62659

To determine

The percentage of variation in particulate concentrationas explained by the model.

Answer to Problem 11RE

The percentage of variation in particulate concentrationas explained by the model is $0.497$

Explanation of Solution

It is known that,

R-squared (R²) is a statistical measure that represents the proportion of the variance for a dependent variable that's explained by an independent variable or variables in a regression model.

In case of multiple regression, adjusted R-squared is used.

Using ‘R’ programming adjusted R-squared is calculated.

  $adj{\text{ R}}^2=0.497$

The percentage of variation in particulate concentration as explained by the model is $0.497$

To determine

Whether the model is useful for prediction or not.

Explanation of Solution

While fitting the model many regression assumptions like autocorrelation, multi-collinearity, heteroscedasticity etc. have not been checked.

So, one cannot say anything about the efficiency and accuracy. One has to consider some other measure like AIC, BIC etc. apart from $adj{\text{ R}}^2=0.497$ to tell whether model is useful for prediction or not.

To determine

Whether one can reject $H_0$ or not for the test $H_0:{\beta }_1=0$ v/s $H_1:{\beta }_1\ne 0$

Answer to Problem 11RE

Coefficients	P-value.
${\beta }_1$	$0.0248$
${\beta }_2$	$7.48e^{-07}$
${\beta }_3$	$0.0322$

At $\alpha =0.05$ , it can be clearly seen that all p-values are less than $\alpha =0.05$ . Hence, null hypothesis will get rejected.

Explanation of Solution

Let’s declare null hypothesis,

  $H_0:{\beta }_1=0$ V/s

  $H_1:{\beta }_1\ne 0$

It is known that for this test, test-statistic is given by

  $t=\frac{\widehat{{\beta }_1}-{({\beta }_1)}_{H_0}}{\sqrt{\frac{RSS}{(n-2){\displaystyle \sum _{i=1}^n{\left(x_i-\stackrel{\_}{x}\right)}^2}}}}~t_{n-2,\alpha }$

Where,

RSS is residual sum of square

  $t_{n-2,\alpha }$ is t- distribution with $n-2$ degree of freedom at alpha level of significance.

Using ‘R’ software p-values for the test is calculated.

Coefficients	P-value.
${\beta }_1$	$0.0248$
${\beta }_2$	$7.48e^{-07}$
${\beta }_3$	$0.0322$

At $\alpha =0.05$ , it can be clearly seen that all p-values are less than $\alpha =0.05$ . Hence, null hypothesis will get rejected.

‘R’ code:

PM=c(24.6,71.7,13.7,19.5,79.6,6.4,13.9,25.1,21.3,19.0,33.5,37.8,8.0,15.1,27.9,35.4,22.0,45.4,23.7
     ,12.6,7.3,17.4,7.7,14.0,20.8,19.8,13.1,10.5,22.7,24.0,39.1,28.2,41.9,25.7,30.1,49.1,11.5,27.1)
Temperature=c(35.8,41.5,38.1,35.6,35.1,30.8,39.7,44.6,53.8,52.5,48.2,29.6,31.7,41.5,55.1,59.9,62.6,51.3,41.5
              ,36.0,47.9,33.5,25.1,28.5,32.8,36.9,43.7,48.3,54.9,57.8,49.2,51.6,54.2,53.8,60.6,44.2,40.9,39.6)
Wind_Speed=c(4.8,13.3,9.5,6.0,12.7,1.3,2.5,2.6,1.4,3.3,11.9,3.2,3.1,2.0,5.2,8.0,9.3,10.7,6.5
             ,3.0,2.5,3.5,4.5,3.1,4.4,2.7,3.8,4.5,3.6,2.7,4.0,2.5,2.6,2.8,5.7,8.7,3.9,7.3)
Humidity=c(37.8,32.6,39.2,37.2,37.5,52.0,43.5,46.4,33.8,27.9,35.1,53.0,37.8,29.5,22.6,19.4,15.3,17.9,29.1,48.6
           ,19.7,50.9,55.1,47.7,38.7,36.1,25.2,19.4,20.0,20.5,39.2,42.5,38.1,41.6,26.8,43.9,39.5,43.8)


# (a)
linear.model=lm(PM~Temperature+Wind_Speed+Humidity)
summ=summary(linear.model)
summ$coefficients
##                Estimate Std. Error   t value     Pr(>|t|)
## (Intercept) -42.5163427 20.9064691 -2.033645 4.985097e-02
## Temperature   0.6557601  0.2791580  2.349065 2.477066e-02
## Wind_Speed    3.6607217  0.6053612  6.047170 7.480612e-07
## Humidity      0.5806123  0.2598741  2.234206 3.215189e-02
#(b)
y=-42.5163427+(0.6557601)*40+( 3.6607217)*5+(0.5806123)*30
y
## [1] 19.43604
#(c)
predict(linear.model, interval ='confidence')
##          fit        lwrupr
## 1  20.478479 15.1188573 25.83810
## 2  52.313262 41.7167782 62.90975
## 3  40.004976 32.9321466 47.07781
## 4  24.391825 18.8033699 29.98028
## 5  48.764964 38.1080145 59.42191
## 6  12.631847  4.8633035 20.40039
## 7  17.925773 12.6039026 23.24764
## 8  23.188846 16.0158506 30.36184
## 9  17.513258 10.0608350 24.96568
## 10 20.190528 14.5402823 25.84077
## 11 53.033375 43.3100223 62.75673
## 12 19.380918 11.9621970 26.79964
## 13 11.566635  3.8506207 19.28265
## 14  9.147208  1.8157572 16.47866
## 15 25.773630 19.6849908 31.86227
## 16 37.313340 29.4915657 45.13512
## 17 41.462320 32.1341000 50.79054
## 18 40.686834 31.5905824 49.78308
## 19 25.388211 19.6218624 31.15456
## 20 20.290945 14.1704778 26.41141
## 21  9.484433  0.4129668 18.55590
## 22 21.817313 15.1976555 28.43697
## 23 22.408222 13.9081921 30.90825
## 24 15.216265  7.9481455 22.48438
## 25 17.569461 11.0777219 24.06120
## 26 12.525258  6.2821313 18.76839
## 27 14.682547  7.5388682 21.82623
## 28 16.893997  8.8890210 24.89897
## 29 18.275732 10.9999846 25.55148
## 30 17.173093  9.3058109 25.04037
## 31 27.149944 21.1884829 33.11141
## 32 25.148706 16.6725233 33.62489
## 33 24.665061 16.5916728 32.73845
## 34 27.167044 18.0774358 36.25665
## 35 33.649244 25.7642561 41.53423
## 36 43.805413 35.6919222 51.91890
## 37 21.515247 17.3256230 25.70487
## 38 35.605845 29.6061374 41.60555
#(d)
predict(linear.model, interval ='prediction')
## Warning in predict.lm(linear.model, interval = "prediction"): predictions on current data refer to _future_ responses
##          fit          lwrupr
## 1  20.478479  -3.39044242 44.34740
## 2  52.313262  26.75381513 77.87271
## 3  40.004976  15.69397519 64.31598
## 4  24.391825   0.47048157 48.31317
## 5  48.764964  23.18039033 74.34954
## 6  12.631847 -11.89059426 37.15429
## 7  17.925773  -5.93469948 41.78625
## 8  23.188846  -1.15148528 47.52918
## 9  17.513258  -6.91087916 41.93739
## 10 20.190528  -3.74532639 44.12638
## 11 53.033375  27.82338810 78.24336
## 12 19.380918  -5.03295643 43.79479
## 13 11.566635 -12.93921571 36.07249
## 14  9.147208 -15.24028905 33.53471
## 15 25.773630   1.73051529 49.81675
## 16 37.313340  12.77398400 61.85270
## 17 41.462320  16.40208203 66.52256
## 18 40.686834  15.71201353 65.66165
## 19 25.388211   1.42468407 49.35174
## 20 20.290945  -3.76025018 44.34214
## 21  9.484433 -15.48137046 34.45024
## 22 21.817313  -2.36573281 46.00036
## 23 22.408222  -2.35567035 47.17211
## 24 15.216265  -9.15226841 39.58480
## 25 17.569461  -6.57888343 41.71780
## 26 12.525258 -11.55744274 36.60796
## 27 14.682547  -9.64916086 39.01425
## 28 16.893997  -7.70436657 41.49236
## 29 18.275732  -6.09507770 42.64654
## 30 17.173093  -7.38080679 41.72699
## 31 27.149944   3.13872014 51.16117
## 32 25.148706   0.39298914 49.90442
## 33 24.665061   0.04434892 49.28577
## 34 27.167044   2.19464261 52.13945
## 35 33.649244   9.08966541 58.20882
## 36 43.805413  19.17152225 68.43930
## 37 21.515247  -2.11847504 45.14897
## 38 35.605845  11.58509698 59.62659
#(e)
summ$adj.r.squared
## [1] 0.4969786
# (f), (g)
summ$coefficients
##                Estimate Std. Error   t value     Pr(>|t|)
## (Intercept) -42.5163427 20.9064691 -2.033645 4.985097e-02
## Temperature   0.6557601  0.2791580  2.349065 2.477066e-02
## Wind_Speed    3.6607217  0.6053612  6.047170 7.480612e-07
## Humidity      0.5806123  0.2598741  2.234206 3.215189e-02