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 t_s, 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)

Answer 1

1.

	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

Therefore adjacent r2=0.8272

2.

1	29.33
2	33.333
3	38.333
4	56.000
5	64.000
6	66.667
7	50.000
8	40.000
9	35.000
10	25.000
11	23.333
12	21.667

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.

1	39.833
2	43.833
3	48.833
4	66.500
5	74.500
6	77.167
7	60.500
8	50.500
9	45.500
10	35.500
11	33.833
12	32.167

5.Here RMSE value of mod partb=5.467073

RMSE value of mod partd=3.393212