The data in the accompanying table come from the comparison of the growth rates
ID: 3180424 • Letter: T
Question
The data in the accompanying table come from the comparison of the growth rates for bacteria types A and B. The growth Y recorded at five equally spaced (and coded) points of time is shown in the table. a Fit the linear model Y = beta_0 +beta_1x_1 + beta_2x_2 + beta_3x_1x_2 + epsilon to the n = 10 data points. Let x1 = 1 if the point refers to bacteria type B and let x1 =0 if the point refers to type A. Let x2 = coded time. b Plot the data points and graph the two growth lines. Notice that beta 3 is the difference between the slopes of the two lines and represents time-bacteria interaction. c Predict the growth of type A at time x2 = 0 and compare the answer with the graph. Repeat the process for type B. d Do the data present sufficient evidence to indicate a difference in the rates of growth for the two types of bacteria? e Find a 90% confidence interval for the expected growth for type B at time x2 = 1. f Find a 90% prediction interval for the growth Y of type B at tiExplanation / Answer
Answer:
x1
x2
x1*x2
y
0
-2
0
8
0
-1
0
9
0
0
0
9.1
0
1
0
10.2
0
2
0
10.4
1
-2
-2
10
1
-1
-1
10.3
1
0
0
12.2
1
1
1
12.6
1
2
2
13.9
Regression Analysis
R²
0.975
Adjusted R²
0.963
n
10
R
0.988
k
3
Std. Error
0.349
Dep. Var.
y
ANOVA table
Source
SS
df
MS
F
p-value
Regression
28.9300
3
9.6433
79.15
3.24E-05
Residual
0.7310
6
0.1218
Total
29.6610
9
Regression output
confidence interval
variables
coefficients
std. error
t (df=6)
p-value
90% lower
90% upper
Intercept
9.3400
0.1561
59.834
1.46E-09
9.0367
9.6433
x1
2.4600
0.2208
11.144
3.11E-05
2.0310
2.8890
x2
0.6000
0.1104
5.436
.0016
0.3855
0.8145
x1*x2
0.4100
0.1561
2.627
.0392
0.1067
0.7133
Predicted values for: y
90% Confidence Interval
90% Prediction Interval
x1
x2
x1*x2
Predicted
lower
upper
lower
upper
Leverage
1
1
1
12.8100
12.4385
13.1815
12.0367
13.5833
0.300
a)
y= 9.34+2.46*x1+0.6*x2+0.41*x1x2
b).
Observation
y
Predicted
Residual
1
8.00
8.14
-0.14
2
9.00
8.74
0.26
3
9.10
9.34
-0.24
4
10.20
9.94
0.26
5
10.40
10.54
-0.14
6
10.00
9.78
0.22
7
10.30
10.79
-0.49
8
12.20
11.80
0.40
9
12.60
12.81
-0.21
10
13.90
13.82
0.08
c).
For A,
when x1=0, x2=0, estimated y= 9.34+2.46*0+0.6*0+0.41*0 =9.34
For B,
when x1=1, x2=0, estimated y= 9.34+2.46*1+0.6*0+0.41*0 =11.80
y= 9.34+2.46x1+0.6x2+0.41*x1x2
d). calculated F=2.627, P=0.0392 which is < 0.05 level. The interaction is significant.
e). 90% CI = (12.4385, 13.1815)
f).90% PI= 12.0367, 13.5833)
x1
x2
x1*x2
y
0
-2
0
8
0
-1
0
9
0
0
0
9.1
0
1
0
10.2
0
2
0
10.4
1
-2
-2
10
1
-1
-1
10.3
1
0
0
12.2
1
1
1
12.6
1
2
2
13.9
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