Problem 8-25 Air pollution control specialists in southern California monitor th
ID: 3074612 • Letter: P
Question
Problem 8-25
Air pollution control specialists in southern California monitor the amount of ozone, carbon dioxide, and nitrogen dioxide in the air on an hourly basis. The hourly time series data exhibit seasonality, with the levels of pollutants showing patterns that vary over the hours in the day. On July 15, 16, and 17, the following levels of nitrogen dioxide were observed for the 12 hours from 6:00 A.M. to 6:00 P.M.
Click on the datafile logo to reference the data.
Hour
Level
1
25
2
28
3
35
4
50
5
60
6
60
7
40
8
35
9
30
10
25
11
25
12
20
1
28
2
30
3
35
4
48
5
60
6
65
7
50
8
40
9
35
10
25
11
20
12
20
1
35
2
42
3
45
4
70
5
72
6
75
7
60
8
45
9
40
10
25
11
25
12
25
July 15:
25
28
35
50
60
60
40
35
30
25
25
20
July 16:
28
30
35
48
60
65
50
40
35
25
20
20
July 17:
35
42
45
70
72
75
60
45
40
25
25
25
Use a multiple linear regression model with dummy variables as follows to develop an equation to account for seasonal effects in the data:
Hour1 = 1 if the reading was made between 6:00 A.M. and 7:00A.M.; 0 otherwise
Hour2 = 1 if the reading was made between 7:00 A.M. and 8:00 A.M.; 0 otherwise
.
.
.
Hour11 = 1 if the reading was made between 4:00 P.M. and 5:00 P.M., 0 otherwise
Note that when the values of the 11 dummy variables are equal to 0, the observation corresponds to the 5:00 P.M. to 6:00 P.M. hour.
If required, round your answers to three decimal places. For subtractive or negative numbers use a minus sign even if there is a + sign before the blank. (Example: -300)
Value = + Hour1 + Hour2 + Hour3 + Hour4 + Hour5 + Hour6 + Hour7 + Hour8 + Hour9 + Hour10 + Hour11
Using the equation developed in part (b), compute estimates of the levels of nitrogen dioxide for July 18.
If required, round your answers to three decimal places.
6:00 a.m. - 7:00 a.m. forecast
7:00 a.m. - 8:00 a.m. forecast
8:00 a.m. - 9:00 a.m. forecast
9:00 a.m. - 10:00 a.m. forecast
10:00 a.m. - 11:00 a.m. forecast
11:00 a.m. - noon forecast
noon - 1:00 p.m. forecast
1:00 p.m. - 2:00 p.m. forecast
2:00 p.m. - 3:00 p.m. forecast
3:00 p.m. - 4:00 p.m. forecast
4:00 p.m. - 5:00 p.m. forecast
5:00 p.m. - 6:00 p.m. forecast
Let t = 1 to refer to the observation in hour 1 on July 15; t = 2 to refer to the observation in hour 2 of July 15; ...; and t = 36 to refer to the observation in hour 12 of July 17. Using the dummy variables defined in part (b) and ts, develop an equation to account for seasonal effects and any linear trend in the time series.
If required, round your answers to three decimal places. For subtractive or negative numbers use a minus sign even if there is a + sign before the blank. (Example: -300)
Value = + Hour1 + Hour2 + Hour3 + Hour4 + Hour5 + Hour6 + Hour7 + Hour8 + Hour9 + Hour10 + Hour11 + t
Based on the seasonal effects in the data and linear trend estimated in part (d), compute estimates of the levels of nitrogen dioxide for July 18.
If required, round your answers to three decimal places.
6:00 a.m. - 7:00 a.m. forecast
7:00 a.m. - 8:00 a.m. forecast
8:00 a.m. - 9:00 a.m. forecast
9:00 a.m. - 10:00 a.m. forecast
10:00 a.m. - 11:00 a.m. forecast
11:00 a.m. - noon forecast
noon - 1:00 p.m. forecast
1:00 p.m. - 2:00 p.m. forecast
2:00 p.m. - 3:00 p.m. forecast
3:00 p.m. - 4:00 p.m. forecast
4:00 p.m. - 5:00 p.m. forecast
5:00 p.m. - 6:00 p.m. forecast
Is the model you developed in part (b) or the model you developed in part (d) more effective?
If required, round your answers to three decimal places.
Model developed in part (b)
Model developed in part (d)
MSE
can you help with part F
can you do part (F)
Hour
Level
1
25
2
28
3
35
4
50
5
60
6
60
7
40
8
35
9
30
10
25
11
25
12
20
1
28
2
30
3
35
4
48
5
60
6
65
7
50
8
40
9
35
10
25
11
20
12
20
1
35
2
42
3
45
4
70
5
72
6
75
7
60
8
45
9
40
10
25
11
25
12
25
Explanation / Answer
ANSWER:
1.
Therefore adjacent r2=0.8272
2.
3.Here we have to findout the value=11.167+12.479*H1+16.042*H2+20.604*H3+37.837*H4+45.396*H5+47.625*H6+30.521*H7+20.083*H8+14.646*H9+4.208*H10+2.104*H11+0.437*T.
4.
5.Here RMSE value of mod partb=5.467073
RMSE value of mod partd=3.393212
Estimate Intercept 21.667 H1 7.667 H2 11.667 H3 16.667 H4 34.333 H5 42.333 H6 45 H7 28.333 H8 18.33 H9 13.333 H10 3.333 H11 1.667