breeding and genetic animal science

two population regressions

18.1 given:

For sample 1 : n = 28, ∑x² = 142.35, ∑xy = 69.47, ∑y² = 108.77, = 14.7 = 32.0.

For sample 2 : n = 30, ∑x² = 181.32, ∑xy = 97.40, ∑y² = 153.59, = 15.8 = 27.4.

a) Test H₀: β₁ = β₂ vs. H₀: β₁ ≠ β_2.

b) If H₀ in part (a ) is not rejected, test H₀ the elevations the two population regressions are the same, vs. H_A: the two elevation are not the same_.

c)

sample 1 sample 2

n = 28 n = 30

∑x² = 142.35 ∑x² = 181.32

∑xy = 69.47 ∑xy = 97.40

∑y² = 108.77 ∑y² = 153.59

= 14.7 = 15.8

= 32.0 = 27.4

b= = 0.4880 b= =0.5371

residual SS =108.77 – residual SS =153.59 –

=74.8670 = 101.2694

residual DF = 28-2 = 26 residual DF = 30-2 = 28

(s²_y.x)_p= = 3.2617

S_b1-b2 = = 0.20224

T= =

rejectH₀if ≥ t_αbut not reject H₀because < = 2.015

I. Sum of squares of x for common regression A_c = (∑x²)₁ + (∑x²)₂

142.35 +181.32 = 323.67

II. Sum of crossproducts for common regression: B_c = (∑xy)₁+(∑xy)₂

69.47 +97.40 = 166.87

III. Sum of squares of y for common regression: C_c = (∑y²)₁+(∑y²)₂

108.77 +153.59 =262.36

IV. Residual ss for common regression: SS_c = C_c-

=176.32917

V. Residual DF for common regression: DF =n₁+n₂ -3

DF =28+30 -3 =55

VI. Residual MS for common regression:( s²_y.x) =

=3.20598

VII. b_cfor common regression: = = 0.515556

VIII. then, the appropriate test statistic is: t =

t = = =10.6971

reject H₀ because t =10.6971>

comparing two elevation for the tow lines in example 18.8 and fig 18.8 in 2 sample:

Sample1: sample 2:

SS_x = 1470.8712 SS_x =2272.4750

S_cp_xy = 4363.1624 S_cp_{xy =}4928.8100

SS_y= 13299.5296 SS_y=10964.0947

n = 26 n = 30

b = 2.97 b = 2.17

SS_r = 356.7317 SS_r = 273.9142

a ₁= 10.57 a ₂ = 24.91

= 10.57 + 2.97 x_i = 24.91 + 2.17 x_i

= 22.93=18.95

= + & = +

=10.57+ =66.406 & =24.91+ = 65.706

= =1.59264 =

= = 1.0114

t = =0.6921

do not reject H₀because t_c= 0.6921 < = 2.007 & = 2.68

+ نوشته شده در شنبه هفدهم اردیبهشت ۱۳۹۰ ساعت 23:37 توسط مهندس ابراهيم ميرزامحمدي |

near regression equation

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

∑∑ x_ij = 525 ∑∑ y_ij =6037 ∑∑x_ijy_ij=153143

∑∑x_i²_j=13353∑∑y_i²_j=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 x_ij

H₀: the population regression is linear.

H_A: the population regression is not linear.

Ss total: ∑∑y_i²_j- = 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

s_y.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

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ماتریس

تمرين 1: رابطه (AB)^'=B'A' را اثبات كنيد؟

A B

A×B × = =

(AB)= (AB)^'=

A ⇒ A′ B ⇒ B′

B′× A′= = =

⇒ B′× A′ = (AB)^'

تمرين 2:يك ماتريس 3×3 را inverse كنيد؟

A

سطر اول:

6(+1) +3(-1) +5(+1)

6(+1)]40-16] +3(-1)[70-18] +5(+1)[56-36]

6(+1)]24] +3(-1)[52] +5(+1)[+20]

6+]24] +3[-52] +5[+20]

144+(-156)+100 = 88 همسازه سطر اول:

سطر دوم:

7(-1) +4(+1) +2(-1)

7(-1)]30-40] +4(+1)[60-45] +2(-1)[48-27]

7(-1)]-10] +4(+1)[15] +2(-1)[ 21]

7[+10] +4[+15] +2[-21]

همسازه سطردوم:70+60+(-42) = 88

سطر سوم:

9(+1) +8(-1) +10(+1)

9(+1)]6-20] +8(-1)[12-35] +10(+1)[24-21]

9(+1)]-14] +8(-1)[-23] +10(+1)[+3]

9-]14] +8[+23] +10[+3]

همسازه سطرسوم:(-126) +184+30 = 88

C= C′=

A^-1 = × C′ A^-1= × =

تمرين 3: ثابت كنيدA= I ×A^-1

A │A│ = 8(9) – 5(6) = 72 – 30 = 42

A^-1 = = ⇒ A , A^-1=

A^-1× A =

× = =

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باند اعتماد

تمرين 1: براي داده هاي زير، باند اعتماد را براي ŷ و رسم كنيد.

طول بال	1.4	1.5	2.2	2.4	3.1	3.2	3.2	3.9	4.1	4.7	4.5	5.2	5
روز	3	4	5	6	8	9	10	11	12	14	15	16	17

ü باند اعتماد براي ŷ:

a= 0.715 cm b = 0.27cm ss_x = 262 day²

n = 13 Ms e = 0.047701 = 2.201

s_ŷ = ^ŷ_i = a +bx_iL =ŷ_i± tsŷ_i

_0.0769

1. X_i=3 ŷ₃ = 0.715+0.27(3)= 1.525 s_ŷ = ₌=0.111

L =1.525± 2.201(0.111 ) ⇒ L₁ = 1.525+0.244311= 1.769311 & L₂ = 1.525+0.244311=1.280

2. X_i= 4 ŷ₃ = 0.715+0.27(4)= 1.795 s_ŷ = ₌0.101

L= 1.795 ± 2.201(0.101 ) ⇒ L₁ = 1.795+0.222301= 2.017301 & L₂ = 1.795-0.222301=1.572699

3. X_i=5 ŷ₃ = 0.715+0.27(5)= 2.065 s_ŷ = ₌0.0906

L = 2.065± 2.201(0.0906) ⇒ L₁ = 2.065+0.1994106=2.2644106 & L₂ = 2.065 +0.1994106=1.8655894

4. X_i= 6 ŷ₃ = 0.715+0.27(6)= 2.335 s_ŷ = ₌0.0811

L = 2.335 ± 2.201(0.0811) ⇒ L₁ = 2.335+0.1785011=2.5135011 & L₂ = 2.335 +0.1785011=2.1564989

5. X_i= 8 ŷ₃ = 0.715+0.27(8)= 2.875 s_ŷ = ₌0.0663

L = 2.875± 2.201(0.0663) ⇒ L₁ = 2.875+0.1459263=3.0209263 & L₂ = 2.875+0.1459263=2.7290737

6. X_i=9 ŷ₃ = 0.715+0.27(9)= 3.145 s_ŷ = ₌0.0620

L = 3.145± 2.201(0.0620 ) ⇒ L₁ = 3.145+0.136462=3.281462 & L₂ = 3.145+0.136462=3.008538

7. X_i= 10 ŷ₃ = 0.715+0.27(10)= 3.415 s_ŷ = ₌0.0605

L = 3.415 ± 2.201(0.0605) ⇒ L₁ = 3.415+0.1331605=3.5481605 & L₂ = 3.415+0.1331605=3.2818395

8. X_i= 11 ŷ₃ = 0.715+0.27(11)= 3.685 s_ŷ = ₌0.0620

L = 3.685± 2.201(0.0620) ⇒ L₁ = 3.685+0.136462=3.821462 & L₂ = 3.685+0.136462=3.548538

9. X_i= 12 ŷ₃ = 0.715+0.27(12)= 3.955 s_ŷ = ₌0.0663

L = 3.955± 2.201(0.0663 ) ⇒ L₁ = 3.955+0.1459263=4.1009263 & L₂ = 3.955+0.1459263=3.8090737

10. X_i=14 ŷ₃ = 0.715+0.27(14)= 4.495 s_ŷ = ₌0.0811

L = 4.495 ± 2.201(0.0811 ) ⇒ L₁ = 4.495+0.1785011=4.6735011 & L₂ = 4.495+0.1785011=4.3164989

11. X_i=15 ŷ₃ = 0.715+0.27(15)= 4.765 s_ŷ = ₌0.0906

L = 4.765± 2.201(0.0906 ) ⇒ L₁ = 4.765+0.1994106=4.9644106 & L₂ = 4.765+0.1994106=4.5655894

12. X_i=16 ŷ₃ = 0.715+0.27(16)= 5.035 s_ŷ = ₌0.101

L = 5.035± 2.201(0.101) ⇒ L₁ = 5.035+0.222301=5.257301 & L₂ = 5.035+0.222301=4.812699

13. X_i= 17 ŷ₃ = 0.715+0.27(17)= 5.305 s_ŷ = ₌0.111

L = 5.305± 2.201( 0.111 ) ⇒ L₁ = 5.305+0.244311= 5.549311 & L₂ = 5.305+0.244311=5.060689

ü باند اعتماد براي :

a = 0.715 b = 0.270 = 2.201 s²b = 0.0135

= k = b²-t²×s²b + ±

K= 0.27²-2.201² ×(0.0135)²= 0.0720

1. Y_i= 1.4 _1.4= = 2.53

+ ±

10+(-7.556)±30.569 2.431±2.034 ⇒L₁=0.397 L₂= 4.465

2. Y_i= 1.5 _1.4= = 2.90

+ ±

10+(-7.18125)±30.569 2.81875±2.018 ⇒L₁=0.80075 L₂=4.83675

3. Y_i= 2.2 _1.4= = 5.5

+ ±

10+(-4.55625)±30.569 5.44375 ± 1.1923 ⇒L₁=4.25145 L₂=6.63605

4. Y_i= 2.4 _1.4= = 6.24

+ ±

10+(-3.80625)±30.569 6.19375± 1.761 ⇒L₁=4.43275 L₂=7.95475

5. Y_i= 3.1 _1.4= = 8.83

+ ±

10+(-1.18125)±30.569 8.81875± 1.83 ⇒L₁=6.98875 L₂=10.64875

6. Y_i= 3.2 _1.4= = 9.20

+ ±

10+(-0.80625)±30.569 9.19375±1.84 ⇒L₁=7.35375 L₂=11.03375

7. Y_i= 3.2 _1.4= = 9.20

+ ±

10+(-0.80625)±30.569 9.19375 ± 1.84 ⇒L₁=7.35375 L₂=11.03375

8. Y_i= 3.9 _1.4= = 11.79

+ ±

10+(1.81875)±30.569 11.81875 ±1.869 ⇒L₁=9.94975 L₂=13.68775

9. Y_i= 4.1 _1.4= = 12.53

+ ±

10+(2.56875)±30.569 12.56875±1.866 ⇒L₁=10.70275 L₂=14.43475

10. Y_i= 4.7 _1.4= = 14.75

+ ±

10+(4.81875)±30.569 14.81875±1.930 ⇒L₁= 12.88875 L₂=16.74875

11. Y_i= 4.5 _1.4= = 14.01

+ ±

10+(4.06875)±30.569 14.06875± 1.911 ⇒L₁=12.15775 L₂=15.97975

12. Y_i= 5.2 _1.4= = 16.61

+ ±

10+(6.69375)±30.569 16.69375±1.997 ⇒L₁=14.69675 L₂=18.69075

13. Y_i= 5 _1.4= = 15.87

+ ±

10+(5.94375)±30.569 15.94375± 1.969 ⇒L₁= 13.97475 L₂=17.91275

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تابعیت

17.1. the following data are the rates of oxygen consumption of birds, measures at different environmental tem-

peratures:

Temperature

( ̊C)

oxygen consumption (ml/g/hr)

-18	5.2
-15	4.7
-10	4.5
-5	3.6
0	3.4
5	3.1
10	2.7
19	1.8

x_i	-18	-15	-10	-5	0	5	10	19	N=8	∑=-14	X = -1.75
y_i	5.2	4.7	4.5	3.6	3.4	3.1	2.7	1.8	N=8	∑=29	y=3.625

b = (x׳x)^-1x׳y

( * )^-1* *

=

⇒ a= b =

Ss total= y׳y-ny²

Ss total = * - 8(3.625)² = 8.915 ⇒ss total = 8.915

Ss regression: β׳x׳y - ny²

Ss regression: * * - 8(3.625)² =8.7452

Ss residual = ss total – ss regression

⇒ Ss residual = 8.915 - 8.7452 = 0.1698

F

MS

SS

DF

S.O.V

**	309.01	8.7452	8.7452	1	regression
		0.0283	0.1698	6	residual

8.915

7

Total

r ²= = = 0.980

1- r ² = 1-0.980 = 0.02

b =

sb = = = 0.0049

t = : = 17.918  18

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لیون تست

داده های زیر مربوط به وزن تولد یک گله گوسفند میباشند و دارای مادرانی با سنین متفاوت هستنند که شامل مادران شکم اول (2ساله )تا ششم(7 ساله) میباشد تفاوت واریانس میان دسته ها را با آزمون leven test آزمون کنید.

7	6	5	4	3	2	Age of dam→
3.4	3.6	3.5	3.5	3.4	3.3	1
3.5	3.7	3.6	3.5	3.4	3.4	2
3.7	3.8	3.8	3.7	3.6	3.5	3
3.8	4	3.9	3.8	3.7	3.6	4
4	4.2	4.1	4.1	4	3.8	5
18.4	19.3	18.9	18.6	18.1	17.6	∑
3.68	3.86	3.78	3.72	3.62	3.52
0.057	0.058	0.057	0.062	0.062	0.037	var

H₀: = = =

H₀: ≠

First:

ابتدا داده ها را از طریق فرمول = تبدیل میکنیم و داده های جدید ایجاد میکنیم.بدین شکل که هر رکورد را از میانگین گروه خودش کم کرده و از آن قدر مطلق میگیریم.

=

7	6	5	4	3	2

داده های تبدیل شده یا :Z_ij

7	6	5	4	3	2	Age of dam→
0.28	0.26	0.28	0.22	0.22	0.22	1
0.18	0.16	0.18	0.22	0.22	0.12	2
0.02	0.06	0.02	0.02	0.02	0.02	3
0.12	0.14	0.12	0.08	0.08	0.08	4
0.32	0.34	0.32	0.38	0.38	0.28	5
0.92	0.96	0.92	0.92	0.92	0.72	∑
0.184	0.192	0.184	0.184	0.184	0.144
0.01468	0.01192	0.01468	0.01968	0.01968	0.01108	var

W=

N = 6×5 = 30 k = 6 = 0.179

Second:

در ادامه برای بدست آوردن صورت کسر یا N_i ابتدا میانگین هر گروه را از میانگین کل کم کرده وپس از به توان رساندن در تعداد تکرار گروه خودش ضرب کرده و در نهایت عدد حاصل برای تمام گرو هها را با هم جمع نمایید.

N_i ²

5 = 0.0062 & 5 =0.00085 & 5 =0.00085 & 5 =0.00085 & 5 =0.00085 & 5 =0.00013

Result:

→ 0.00973

Third:

در ادامه کار برای بدست آوردن مخرج کسر یا ابتدا تک تک دادها را از میانگین گروه خودش کم کرده و به توان دو رسانده و با هم جمع کرده و در نهایت عدد بدست آمده برای هر گروه را بااعداد بدست آمده دیگر گروه ها جمع نمایید.

Because

∑Z₂:

(0.22-0.144)² = 0.0058 & (0.12-0.144)² =0.00058 & (0.02-0.144)² =0.016 & 0.08-0.144)² = 0.0041 & (0.28-0.144)² =0.0185 = 0.045

∑Z₃:

(0.22-0.184)² =0.0013 & (0.22-0.184)² =0.0013 & (0.02-0.184)² = 0.027 & (0.08-0.184)² = 0.011 & (0.38-0.184)² =0.039 = 0.08

∑Z₄:

(0.22-0.184)² = 0.0013 & (0.22-0.184)² =0.0013 & (0.02-0.184)² =0.027& (0.08-0.184)² = 0.011 & (0.38-00.184)² = 0.039 = 0.08

∑Z₅:

(0.28-0.184)² =0.0093& (0.18-0.184)² =0.000016& (0.02-0.184)² =0.027& (0.12-0.184)² =0.0041 & (0.32-0.184)² = 0.0185 = 0.059

∑Z₆:

(0.26-0.192)²=0.0047& (0.16-0.192)²= 0.0011 & (0.06-0.192)²=0.018 & (0.14-0.192)²=0.0028 & (0.34-0.192)²= 0.022 = 0.049

∑Z₇:

(0.28- )²=0.0093 & (0.18- )²= 0.000016 & (0.02- )²= 0.027 & (0.12- )²= 0.0041 & (0.32- )²= 0.019 = 0.06

Result:

→ = 0.373

W= = =(4.8)( 0.026) = 0.125

= 0.05 =2.62 →w =0.125< F =0.05 =2.62

Therefore do not rejects the null hypothesis that the variances are equal.

+ نوشته شده در جمعه هفدهم اردیبهشت ۱۳۸۹ ساعت 23:39 توسط مهندس ابراهيم ميرزامحمدي |

رگرسیون

1 : n = 33, ∑x² = 744.32, ∑xy = 2341.37, ∑y² = 7498.91,

For sample 2 : n = 34, ∑x² = 973.14, ∑xy = 3147.68, ∑y² = 10366.97

For sample 3 : n = 29, ∑x² = 664.42, ∑xy = 2047.73, ∑y² = 6503.32

For the total of all samples : n = 96, ∑x² = 3146.72, ∑xy = 7938.25, ∑y² = 20599.33

a) Test H₀: β₁ = β₂= β₃ vs. H_A:all ther β’s are not equal _.

b) If H₀ in part (a ) is not rejected, test H₀ the three population regressions lines have the same elevation, vs. H_A: the lines do not have the same elevation_...

_{روی ادامه مطلب کلیک
فرمایید.}

+ نوشته شده در جمعه هفدهم اردیبهشت ۱۳۸۹ ساعت 23:38 توسط مهندس ابراهيم ميرزامحمدي |

نمودار پراكنش و خط تابعيت

بنام خدا

براي داده هاي زير، نمودار پراكنش و خط تابعيت رسم كنيد.

3

1.4

4

1.5

5

2.2

6

2.4

8

3.1

9

3.2

10

3.2

11

3.9

12

4.1

14

4.7

15

4.5

16

5.2

17

5

17.1. the following data are the rates of oxygen consumption of birds, measures at different environmental tem-

peratures: a) calculate a and b for the regression of oxygen consumption rate on temperature.

b) Test, by analysis of variance, the hypothesis H0:= 0.

c) Calculate the standard error of estimate of the regression.

d) Calculate the coefficient of determination of regression.

Temperature

( ̊C)

oxygen consumption (ml/g/hr)

-18	5.2
-15	4.7
-10	4.5
-5	3.6
0	3.4
5	3.1
10	2.7
19	1.8

x_i	y_i	xy	X²	Y²
-18	5.2	-93.6	324	27.04
-15	4.7	-70.5	225	22.09
-10	4.5	-45	100	20.25
-5	3.6	-18	25	12.96
0	3.4	0	0	11.56
5	3.1	15.5	25	9.61
10	2.7	27	100	7.29
19	1.8	34.2	361	3.24
N=8	N=8	N=8	N=8	N=8
∑=-14	∑=29	-150.4	1160	114.04
X = -1.75	y=3.625	-	-	-

b = = = = - 0.0877

a = Y_i – bX = 3.625 – (-0.0877)(-1.75) = 3.4715

Ŷ = 3.4697+ 0.0674 x_i

H₀: b = 0 , H₁: b 0

Ss total: ∑Y_i²_– ₌114.04_-₌8.915

Ss regression = = = 8.745

Ss residval = ss_T – ss_R = 8.915_- 8.7451 = 0.17

F

MS

SS

DF

S.O.V

**	309.01	8.745	8.745	1	regression
		0.0283	0.17	6	residval

8.915

7

Total

آزمون معني دار است يعني تابعيت از نظر آماري وجود دارد.

+ نوشته شده در جمعه هفدهم اردیبهشت ۱۳۸۹ ساعت 23:33 توسط مهندس ابراهيم ميرزامحمدي |

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