17.3 the frequency of electrical impulses emitted from electric fish is measured from fish at different temperatures. The resultant data are as follows:

temperature ©

Impulse frequency (number/sec

ni

20

225,230,239

3

22

251,259,265

3

23

266,273,280

3

25

287,295,302

3

27

301,310,317

3

28

307,313,325

3

30

324,330,338

3

 

(a)    Compute a and b for the linear regression equation relation impulse frequency to temperature. (b) test, by analysis of variance, H0:β = 0. (c) calculate the standard error of estimate of the regression. (d) calculate the coefficient of determination of the regression. (e) test H0: the population regression is linear.

 

N = 21

            ∑∑ xij = 525                      ∑∑ yij =6037              ∑∑xijyij =153143

            ∑∑xi2j = 13353                         ∑∑yi2j =1758389        ∑xy = 2218

∑ x2  = 228                       ∑ y2 =22895.23

 = 25                                  =287.47

b =   =  =9.728

a =   - b  =287.47 - 9.728(25) = 44.27

therefore, the least squares regression lin is  = 44.27 + 9.728 xij

H0: the population regression is linear.

HA: the population regression is not linear.

Ss total: ∑∑yi2j -  = 1758389 -  = 22895.238

 

 

 

among groups ss =  -  =  -  =22089.9

among groups DF k-1 = 6

Within groups ss = total ss - among groups ss = 22895.238 - 22089.9  = 805.338

Within groups DF = total DF - among groups DF = 20 – 6 = 14

Regression ss =  =  = 21576.859

Deviations from linearity ss = among groups ss –regression ss = 22089.9 - 21576.859 =513.041 Deviations from linearity DF = among groups DF– regression DF = 6 – 1 = 5

Regression with replication: test for linearity

 

source of variation

ss

Df

MS

total

22895.238

20

-

among groups

22089.9

6

-

linear Regression

21576.859

1

-

Deviations from linearity

513.041

5

102.6082

Within groups

805.338

14

57.52

 

F =  = 1.78387




Since F >table F , do not reject H0

 

source of variation

Ss

Df

MS

total

22895.238

20


linear Regression

21576.859

1

21576.859

residual

1318.379

19

69.38

 

F = 310.995

Since F  > table F , therefore reject H0

r 2=  = 0.942

sy.x=  = 8.329

 

1.      Adjusment for random data that has a binonmal distribut?


73

45

3


69

50

7


80

42

4


83

46

9


74

48

2

379

231

25

x

75.8

46.2

5

    Var 

    31.7

     9.2

    8.5

 

Arc sin =  58.693

Arc sin =42.130

Arc sin  =9.974

Arc sin

Arc sin  = 45.000

Arc sin  =15.341

Arc sin

Arc sin   40.396

Arc sin  = 11.536

Arc sin

Arc sin  =42.705

Arc sin  =17.457

Arc sin

Arc sin  =42.853

Arc sin  = 8.130

 


58.693

42.13

9.974


56.166

56.166

15.341


63.438

40.396

11.536


65.649

42.705

17.457


59.342

42.853

8.13

303.288

224.25

62.438

60.6576

44.85

12.4876