Question

QUESTION 1 [38] The data in the table below come from the comparison of the growth rates for bacteria types A and B. The growth y recorded at nine equally spaced (and coded) points of time is show in the table. Bacteria type A B -4 6.8 8.8 Time -3 -2 -1 0 7.5 8.0 9.0 9.1 10.2 9.2 10.0 10.3 12.2 12.6 2 3 4 10.4 10.8 11.4 13.9 14.4 15.2 (1.1) Fit the linear model y = Bo + B1x1 + B2X2 + B3X1X2 +€ where xı = 1 refers to bacteria type B and x1 = 0 refers to bacteria type A, and x2 refers to the coded time. Use Excel. (6) (1.2) Predict the growth rate of type A bacteria at time x2 = 0 and compare the answer with the observed value. (6) (1.3) Calculate the residuals and plots them against the fitted values. Give a table using Excel. (8) (1.4) Calculate a 90% confidence interval for the expected growth for type B at time x2 = 1. (8) (1.5) Calculate a 90% prediction interval for the expected growth for type B at time x2 = 1. (8) (1.6) Compare the results in parts (1.4) and (1.5). (2)

Answer 1

SUMMARY OUTPUT
Regression Statistics
Multiple R	0.9934
R Square	0.9868
Adjusted R Square	0.9840
Standard Error	0.2994
Observations	18
ANOVA
	df	SS	MS	F	Significance F
Regression	3	93.8897	31.2966	349.1868	0.0000
Residual	14	1.2548	0.0896
Total	17	95.1444
	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%
Intercept	9.2444	0.0998	92.6366	0.0000	9.0304	9.4585
x1	2.6000	0.1411	18.4230	0.0000	2.2973	2.9027
x2	0.5717	0.0386	14.7911	0.0000	0.4888	0.6546
x1x2	0.2833	0.0547	5.1837	0.0001	0.1661	0.4006

1.3)

RESIDUAL OUTPUT
Observation	Predicted Y	Residuals
1	6.957777778	-0.15778
2	7.529444444	-0.02944
3	8.101111111	-0.10111
4	8.672777778	0.327222
5	9.244444444	-0.14444
6	9.816111111	0.383889
7	10.38777778	0.012222
8	10.95944444	-0.15944
9	11.53111111	-0.13111
10	8.424444444	0.375556
11	9.279444444	-0.07944
12	10.13444444	-0.13444
13	10.98944444	-0.68944
14	11.84444444	0.355556
15	12.69944444	-0.09944
16	13.55444444	0.345556
17	14.40944444	-0.00944
18	15.26444444	-0.06444

Residuals VS Predicted Plot 0.6 0.4 0.2 0 Residuals 0 2 4 6 16 14 16 18 -0.2 -0.4 -0.6 -0.8 Predicted

The above plot does not have any pattern. Hence, the independence assumption between the residulas against predicted is satisfied but may include an outlier.

The regression equation is ŷ - 9. 2444 +2.60oon, +05717 12 +0.28 332,22 1:2 growth rate The predicted of type A bateria at time, n2=0 and compare 989.2944 +0.6000X0705717 x 0 +0.28 33 x0x0 59.2444 obsenned value when of growth rate when type A bateria at timi 9.1. So, the observed valen is smaller than predicted value. The corresponding n2=0 is 1.49 9 120 MORE 60 0.1111 0.0167 -0,0167 -0.0167 0.0333 i) et represent the corresponding raku on a for type B at time x2=1] emor, s = 0.2994 The 90% confidence interval for the expected Standard growth ý tt 8. 0.7, 18-4 CT 9,2444 + 2.6000X1+0,541FX1+ 0.2833*1*1=12,6994 0,40,14 = 1.7613 314

Now 12.6993 +1.7613 x 0.2994 [a'kila = 12.6993 I 1.7613 x 0.2994 $0.1278 (12.5108, 12,8278) 1.5) The 98% prediction internal for the expected growth for type B = 12.6993 ± 0.1885 at time is ŷ it 0.10 27 18-4 It a' (xxja = 12.69937 1.7613x 0.2994 Vitol +0.1278 =12.6993 +0,5600 = (12.139 2.1393, 13.2 .2593) 1.6) The confidence interval in (1.4) has shortes width than the predicted interval in (1.5).