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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 ti

Explanation / 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

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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