Question

The data in Table B.5 present the performance of a chemical process as a

function of sever controllable process variables.

a. Fit a multiple regression model relating CO2 product (y) to total solvent (x6) and hydrogen consumption (x7).

Problem 2.4 you were asked to compute a 95% CI on mean gasoline pre-

b. Test for significance of regression. Calculate R2 and R2 . Adj

c. Using t tests determine the contribution of x6 and x7 to the model.

d. Construct 95% CIs on β6 and β7.

e. Refit the model using only x6 as the regressor. Test for significance of

regression and calculate R2 and R2 . Discuss your findings. Based on these Adj

statistics, are you satisfied with this model?

f. Construct a 95% CI on β6 using the model you fit in part e. Compare the length of this CI to the length of the CI in part d. Does this tell you any- thing important about the contribution of x7 to the model?

g. Compare the values of MSRes obtained for the two models you have fit (parts a and e). How did the MSRes change when you removed x7 from the model? Does this tell you anything importaut about the contributiou of x7 to the model?

y	x1	x2	x3	x4	x5	x6	x7
36.98	5.1	400	51.37	4.24	1484.83	2227.25	2.06
13.74	26.4	400	72.33	30.87	289.94	434.9	1.33
10.08	23.8	400	71.44	33.01	320.79	481.19	0.97
8.53	46.4	400	79.15	44.61	164.76	247.14	0.62
36.42	7	450	80.47	33.84	1097.26	1645.89	0.22
26.59	12.6	450	89.9	41.26	605.06	907.59	0.76
19.07	18.9	450	91.48	41.88	405.37	608.05	1.71
5.96	30.2	450	98.6	70.79	253.7	380.55	3.93
15.52	53.8	450	98.05	66.82	142.27	213.4	1.97
56.61	5.6	400	55.69	8.92	1362.24	2043.36	5.08
26.72	15.1	400	66.29	17.98	507.65	761.48	0.6
20.8	20.3	400	58.94	17.79	377.6	566.4	0.9
6.99	48.4	400	74.74	33.94	158.05	237.08	0.63
45.93	5.8	425	63.71	11.95	130.66	1961.49	2.04
43.09	11.2	425	67.14	14.73	682.59	1023.89	1.57
15.79	27.9	425	77.65	34.49	274.2	411.3	2.38
21.6	5.1	450	67.22	14.48	1496.51	2244.77	0.32
35.19	11.7	450	81.48	29.69	652.43	978.64	0.44
26.14	16.7	450	83.88	26.33	458.42	687.62	8.82
8.6	24.8	450	89.38	37.98	312.25	468.38	0.02
11.63	24.9	450	79.77	25.66	307.08	460.62	1.72
9.59	39.5	450	87.93	22.36	193.61	290.42	1.88
4.42	29	450	79.5	31.52	155.96	233.95	1.43
38.89	5.5	460	72.73	17.86	1392.08	2088.12	1.35
11.19	11.5	450	77.88	25.2	663.09	994.63	1.61
75.62	5.2	470	75.5	8.66	1464.11	2196.17	4.78
36.03	10.6	470	83.15	22.39	720.07	1080.11	5.88

Answer 1

Using Excel we get following output for 1^st model of X₆ and X₇

SUMMARY OUTPUT Regression Statistics Multiple R 0.83644727 R Square 0.69964403 Adjusted R Square 0.67461437 Standard Error 9.92437059 Observations 27 ANOVA df SS MS F Significance F Regression 2 5506.27694 2753.138 27.95259 5.39058E-07 Residual 24 2363.83516 98.49313 Total 26 7870.1121

	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%
Intercept	2.52646008	3.61005487	0.69984	0.490756	-4.924326923	9.9772471
x6	0.018522	0.00274725	6.742016	5.66E-07	0.012851956	0.024192
x7	2.18575295	0.97269576	2.247109	0.034098	0.178207574	4.1932983

a.Fit a multiple regression model relating CO2 product (y) to total solvent (x6) and hydrogen consumption (x7).

     Ans.

Y=2.52646008+0.018522X₆+2.18575295X₇

Here,

B₀=2.52646008     B6=0.018522     B7=2.18575295

(We can also take Rounded value)

b.Test for significance of regression. Calculate R2 and R2 . Adj

Ans.

R Square=0.69964403 =69.96%

Adj R square=0.67461437 = 67.46% (in %)

Significance of regression,

F test statistics is 27.95259,

Overall significance is 5.39058E-07 which very less than significance level hence we can say that model provide better fit and model is significant.

c.Using t tests determine the contribution of x6 and x7 to the model.

Ans.

H₀: β6=0VsH1: β6‡0

Test statistics is 6.742016 and pvalue of the test statistics is 5.66E-07 which is less than 0.05 hence we reject null hypothesis and conclude that β6 or X6 contribute significantly in the model.

Similarly for β7

H₀: β7=0VsH1: β7‡0

Test statistics is 2.247109 and pvalue of the test statistics is 0.034098 which is less than 0.05 hence we reject null hypothesis and conclude that β7 or X7 contribute significantly in the model.

d.Construct 95% CIs on β6 and β7.

Ans.

Confidence Interval on β6 is (0.012851956, 0.024192)

Lower limit is 0.012851956 and Upper limit is 0.024192

Similarly

Confidence Interval on β7 is (0.178207574, 4.1932983)

Lower limit is 0.178207574 and Upper limit is 4.1932983

We refit model using x6 in Excel and we get following Output

SUMMARY OUTPUT
Regression Statistics
Multiple R		0.79777843
R Square		0.63645043
Adjusted R Square		0.62190844
Standard Error		10.6979922
Observations		27
ANOVA
	df		SS		MS		F		Significance F
Regression	1		5008.93619		5008.936		43.76641		6.23754E-07
Residual	25		2861.1759		114.447
Total	26		7870.1121
	Coefficients		Standard Error		t Stat		P-value		Lower 95%		Upper 95%
Intercept	6.14418064		3.48306431		1.764016		0.089948		-1.029324521		13.317686
x6	0.01939474		0.00293166		6.615619		6.24E-07		0.013356876		0.0254326

e. Refit the model using only x6 as the regressor. Test for significance of regression and calculate R2 and R2 . Discuss your findings. Based on these Adj statistics, are you satisfied with this model?

Ans.

Y=6.14418064+0.01939474X6

Here,

Β0=6.14418064 and B6= 0.01939474

Significance Regression Test is

F statistics is 43.76641 and Significance F value or P value is 6.23754E-07 which is very less hence we can say that model is significant.

R Square = 0.63645043 = 63.65%

Adjusted R Sq = 0.62190844 = 62.19% (in %)

f. Construct a 95% CI on β6 using the model you fit in part e. Compare the length of this CI to the length of the CI in part d. Does this tell you any- thing important about the contribution of x7 to the model?

Ans.

Confidence interval for β6 is (0.013356876, 0.0254326)

Lower Limit is 0.013356876 and Upper Limit is 0.0254326

Confidence interval in part e and part d is quite similar.

Step involved to get output in Excel

1 copy data on excel sheet

2 go to option Data > Data Analysis > Regression > Input Y Range: select Y coloumn and Input X Range: select X6 and X7 coloumn , select label and confidence level at 95% and give output range (select empty region or were we want to print the output) > ok.