CHAPTER 91 LINEAR REGRESSION
CHAPTER 91 LINEAR REGRESSION EXERCISE 338 Page 958
1. Determine the equation of the regression line of Y on X, correct to 3 significant figures: X
14
18
23
30
50
Y
900
1200
1600
2100
3800
X
Y
X2
XY
Y2
14
900
196
12 600
810 000
18
1200
324
21 600
1 440 000
23
1600
529
36 800
2 560 000
30
2100
900
63 000
4 410 000
50
3800
2500
190 000
14 440 000
∑ Y = 9600 ∑ X
∑ X = 135
Y ∑=
Substituting into
2
∑ XY = 324 000
= 4449
∑Y
2
= 23 660 000
a0 N + a1 ∑ X
= ∑ XY a0 ∑ X + a1 ∑ X 2
and gives:
9600 = 5 a0 + 135 a1
(1)
and
324 000 = 135 a0 + 4449 a1
(2)
27 × (1) gives:
259 200 = 135 a0 + 3645 a1
(3)
804 a1
from which, a1 =
64 800 = 80.6 804
Substituting in (1) gives: 9600 = 5 a0 + 135(80.6)
from which, a0 =
9600 − 135(80.6) = –256 5
(2) – (3) gives:
64 800 =
Hence, the equation of the regression line of Y on X is: i.e.
Y= a0 + a1 X Y = –256 + 80.6X
2. Determine the equation of the regression line of Y on X, correct to 3 significant figures: X
6
3
9
15
2
14
21
13
Y
1.3 0.7 2.0 3.7 0.5
2.9
4.5 2.7
1399
© 2014, John Bird
X
Y
X2
XY
Y2
6
1.3
36
7.8
1.69
3
0.7
9
2.1
0.49
9
2.0
81
18.0
4.0
15
3.7
225
55.5
13.69
2
0.5
4
1.0
0.25
14
2.9
196
40.6
8.41
21
4.5
441
94.5
20.25
13
2.7
169
35.1
7.29
∑ X = 83 ∑ Y = 18.3 ∑ X Substituting into and gives: and
Y ∑=
2
∑ XY = 254.6 ∑ Y
= 1161
2
= 56.07
a0 N + a1 ∑ X
= ∑ XY a0 ∑ X + a1 ∑ X 2
18.3 = 8 a0 + 83 a1
(1)
254.6 = 83 a0 + 1161 a1
(2)
83 × (1) gives:
1518.9 = 664 a0 + 6889 a1
(3)
8 × (2) gives:
2036.8 = 664 a0 + 9288 a1
(4)
(4) – (3) gives:
517.9 =
from which, a1 =
2399 a1
Substituting in (1) gives: 18.3 = 8 a0 + 83(0.216)
from which,
Hence, the equation of the regression line of Y on X is: i.e.
a0 =
517.9 = 0.216 2399
18.3 − 83(0.216) = 0.0477 8
Y= a0 + a1 X Y = 0.0477 + 0.216X
3. Determine the equation of the regression lines of X on Y, correct to 3 significant figures, for the data given in Problem 1
Substituting into and gives:
X ∑= = XY ∑
b0 N + b1 ∑ Y b0 ∑ Y + b1 ∑ Y 2
135 = 5 b0 + 9600 b1
(1) 1400
© 2014, John Bird
and
324 000 = 9600 b0 + 23 660 000 b1
(2)
1920 × (1) gives:
259 200 = 9600 b0 + 18 432 000 b1
(3)
(2) – (3) gives:
64 800 =
5 228 000 b1
from which, b1 =
64 800 = 5 228 000
0.012395 Substituting in (1) gives: 135 = 5 b0 + 9600(0.0124)
from which, b0 =
135 − 9600(0.012395) 5
= 3.20 Hence, the equation of the regression line of X on Y is:
X= b0 + b1Y
i.e.
X = 3.20 + 0.0124Y
4. Determine the equation of the regression lines of X on Y, correct to 3 significant figures for the data given in Problem 2
Substituting into and gives: and
X b N + b ∑Y ∑= = XY b ∑ Y + b ∑ Y ∑ 0
0
1
1
2
83 = 8 b0 + 18.3 b1
(1)
254.6 = 18.3 b0 + 56.07 b1
(2)
18.3 × (1) gives:
1518.9 = 146.4 b0 + 334.89 b1
(3)
8 × (2) gives:
2036.8 = 146.4 b0 + 448.56 b1
(4)
(4) – (3) gives:
517.9 =
113.67 b1
Substituting in (1) gives: 83 = 8 b0 + 18.3(4.56)
from which, b1 =
from which,
b0 =
517.9 = 4.56 113.67
83 − 18.3(4.56) = –0.056 8
Hence, the equation of the regression line of X on Y is:
X= b0 + b1Y
i.e.
X = –0.056 + 4.56Y
1401
© 2014, John Bird
5. The relationship between the voltage applied to an electrical circuit and the current flowing is as shown: Current (mA)
2
4
6
8
10
12
14
Applied voltage (V)
5
11
15
19
24
28
33
Assuming a linear relationship, determine the equation of the regression line of applied voltage, Y, on current, X, correct to 4 significant figures.
A table is produced as shown below, where current I = X and voltage V = Y X
Y
X2
XY
Y2
2
5
4
10
25
4
11
16
44
121
6
15
36
90
225
8
19
64
152
361
10
24
100
240
576
12
28
144
336
784
14
33
196
462
1089
∑ X = 56 ∑ Y = 135 ∑ X Substituting into and
Y ∑=
2
∑ XY = 1334
= 560
∑Y
2
= 3181
a0 N + a1 ∑ X
= ∑ XY a0 ∑ X + a1 ∑ X 2
135 = 7 a0 + 56 a1
(1)
and
1334 = 56 a0 + 560 a1
(2)
8 × (1) gives:
1080 = 56 a0 + 448 a1
(3)
gives:
(2) – (3) gives:
254 =
from which, a1 =
112 a1
Substituting in (1) gives: 135 = 7 a0 + 56(2.268)
from which,
Hence, the equation of the regression line of Y on X is: i.e.
254 = 2.268 112
a0 =
135 − 56(2.268) = 1.142 7
Y= a0 + a1 X Y = 1.142 + 2.268X
1402
© 2014, John Bird
6. For the data given in Problem 5, determine the equation of the regression line of current on applied voltage, correct to 3 significant figures. X ∑=
Substituting into and
= XY ∑
gives:
b0 N + b1 ∑ Y b0 ∑ Y + b1 ∑ Y 2
56 = 7 b0 + 135 b1
(1)
and
1334 = 135 b0 + 3181 b1
(2)
135 × (1) gives:
7560 = 945 b0 + 18 225 b1
(3)
7 × (2) gives:
9338 = 945 b0 + 22 267 b1
(4)
(4) – (3) gives:
1778 =
Substituting in (1) gives:
4042 b1
56 = 7 b0 + 135(0.43988) from which, b0 =
Hence, the equation of the regression line of X on Y is: i.e.
from which, b1 =
1778 = 0.43988 4042
56 − 135(0.43988) = –0.483 7
X= b0 + b1Y
X = –0.483 + 0.440Y, correct to 3 significant figures
7. Draw the scatter diagram for the data given in Problem 5 and show the regression lines of applied voltage on current and current on applied voltage. Hence determine the values of (a) the applied voltage needed to give a current of 3 mA and (b) the current flowing when the applied voltage is 40 volts, assuming the regression lines are still true outside of the range of values given.
A scatter diagram is shown below. Current I = X and voltage V = Y The regression line of voltage on current, Y = 1.142 + 2.268X, and current on voltage, X = –0.483 + 0.440Y are shown and to the scale drawn are seen to coincide
1403
© 2014, John Bird
(a) When current, X = 3 mA, then applied voltage, Y = 1.142 + 2.268(3) = 7.95 V (b) When the applied voltage, Y = 40 V, current, X = –0.483 + 0.440(40) = 17.1 mA
8. In an experiment to determine the relationship between force and momentum, a force, X, is applied to a mass by placing the mass on an inclined plane, and the time, Y, for the velocity to change from u m/s to v m/s is measured. The results obtained are as follows: Force (N)
11.4
18.7
11.7
12.3
14.7
18.8
19.6
Time (s)
0.56
0.35
0.55
0.52
0.43
0.34
0.31
Determine the equation of the regression line of time on force, assuming a linear relationship between the quantities, correct to 3 significant figures.
Let force F = X and time = Y. A table is produced as shown below
1404
© 2014, John Bird
X
Y
X2
XY
Y2
11.4
0.56
129.96
6.384
0.3136
18.7
0.35
349.69
6.545
0.1225
11.7
0.55
136.89
6.435
0.3025
12.3
0.52
151.29
6.396
0.2704
14.7
0.43
216.09
6.321
0.1849
18.8
0.34
353.44
6.392
0.1156
19.6
0.31
384.16
6.076
0.0961
∑X = 107.2
∑Y
∑X
= 1721.52
= 3.06 Y ∑=
Substituting into
∑ XY
2
∑Y
= 44.549
2
= 1.4056
a0 N + a1 ∑ X
= ∑ XY a0 ∑ X + a1 ∑ X 2
and gives:
3.06 = 7 a0 + 107.2 a1
and
(1)
44.549 = 107.2 a0 + 1721.52 a1
(2)
107.2 × (1) gives:
328.032 = 750.4 a0 + 11491.84 a1
(3)
7 × (2) gives:
311.843 = 750.4 a0 + 12050.64 a1
(4)
(3) – (4) gives:
16.189 =
–558.8 a1
from which, a1 =
−16.189 = –0.0290 558.8
Substituting in (1) gives: 3.06 = 7 a0 + 107.2(–0.0290) from which,
a0 =
3.06 − 107.2(−0.0290) = 0.881 7
Hence, the equation of the regression line of Y on X is:
Y= a0 + a1 X
i.e.
Y = 0.881 – 0.0290X
9. Find the equation for the regression line of force on time for the data given in Problem 8, correct to 3 decimal places.
Substituting into and
X ∑=
= XY ∑
b0 N + b1 ∑ Y
b0 ∑ Y + b1 ∑ Y 2
1405
© 2014, John Bird
gives:
107.2 = 7 b0 + 3.06 b1
(1)
44.549 = 3.06 b0 + 1.4056 b1
(2)
3.06 × (1) gives:
328.032 = 21.42 b0 + 9.3636 b1
(3)
7 × (2) gives:
311.843 = 21.42 b0 + 9.8392 b1
(4)
and
(3) – (4) gives:
16.189 =
–0.4756 b1
from which, b1 =
16.189 = –34.039 −0.4756
Substituting in (1) gives: 107.2 = 7 b0 + 3.06(–34.039) from which,
b0 =
107.2 + 3.06(34.039) = 30.194 7
Hence, the equation of the regression line of X on Y is:
X= b0 + b1Y
i.e.
X = 30.194 – 34.039Y
10. Draw a scatter diagram for the data given in Problem 8 and show the regression lines of time on force and force on time. Hence, find (a) the time corresponding to a force of 16 N, and (b) the force at a time of 0.25 s, assuming the relationship is linear outside of the range of values given.
A scatter diagram is shown below. The regression line of force on time, Y = 0.881 – 0.0290X, and force on time, X = 30.194 – 34.039Y are shown and to the scale drawn are seen to coincide
(a) When force X = 16 N then time, Y = 0.881 – 0.0290(16) = 0.417 s 1406
© 2014, John Bird
(b) When time Y = 0.25 s, force, X = 30.194 – 34.039(0.25) = 21.7 N
1407
© 2014, John Bird
1. Determine the equation of the regression line of Y on X, correct to 3 significant figures: X
14
18
23
30
50
Y
900
1200
1600
2100
3800
X
Y
X2
XY
Y2
14
900
196
12 600
810 000
18
1200
324
21 600
1 440 000
23
1600
529
36 800
2 560 000
30
2100
900
63 000
4 410 000
50
3800
2500
190 000
14 440 000
∑ Y = 9600 ∑ X
∑ X = 135
Y ∑=
Substituting into
2
∑ XY = 324 000
= 4449
∑Y
2
= 23 660 000
a0 N + a1 ∑ X
= ∑ XY a0 ∑ X + a1 ∑ X 2
and gives:
9600 = 5 a0 + 135 a1
(1)
and
324 000 = 135 a0 + 4449 a1
(2)
27 × (1) gives:
259 200 = 135 a0 + 3645 a1
(3)
804 a1
from which, a1 =
64 800 = 80.6 804
Substituting in (1) gives: 9600 = 5 a0 + 135(80.6)
from which, a0 =
9600 − 135(80.6) = –256 5
(2) – (3) gives:
64 800 =
Hence, the equation of the regression line of Y on X is: i.e.
Y= a0 + a1 X Y = –256 + 80.6X
2. Determine the equation of the regression line of Y on X, correct to 3 significant figures: X
6
3
9
15
2
14
21
13
Y
1.3 0.7 2.0 3.7 0.5
2.9
4.5 2.7
1399
© 2014, John Bird
X
Y
X2
XY
Y2
6
1.3
36
7.8
1.69
3
0.7
9
2.1
0.49
9
2.0
81
18.0
4.0
15
3.7
225
55.5
13.69
2
0.5
4
1.0
0.25
14
2.9
196
40.6
8.41
21
4.5
441
94.5
20.25
13
2.7
169
35.1
7.29
∑ X = 83 ∑ Y = 18.3 ∑ X Substituting into and gives: and
Y ∑=
2
∑ XY = 254.6 ∑ Y
= 1161
2
= 56.07
a0 N + a1 ∑ X
= ∑ XY a0 ∑ X + a1 ∑ X 2
18.3 = 8 a0 + 83 a1
(1)
254.6 = 83 a0 + 1161 a1
(2)
83 × (1) gives:
1518.9 = 664 a0 + 6889 a1
(3)
8 × (2) gives:
2036.8 = 664 a0 + 9288 a1
(4)
(4) – (3) gives:
517.9 =
from which, a1 =
2399 a1
Substituting in (1) gives: 18.3 = 8 a0 + 83(0.216)
from which,
Hence, the equation of the regression line of Y on X is: i.e.
a0 =
517.9 = 0.216 2399
18.3 − 83(0.216) = 0.0477 8
Y= a0 + a1 X Y = 0.0477 + 0.216X
3. Determine the equation of the regression lines of X on Y, correct to 3 significant figures, for the data given in Problem 1
Substituting into and gives:
X ∑= = XY ∑
b0 N + b1 ∑ Y b0 ∑ Y + b1 ∑ Y 2
135 = 5 b0 + 9600 b1
(1) 1400
© 2014, John Bird
and
324 000 = 9600 b0 + 23 660 000 b1
(2)
1920 × (1) gives:
259 200 = 9600 b0 + 18 432 000 b1
(3)
(2) – (3) gives:
64 800 =
5 228 000 b1
from which, b1 =
64 800 = 5 228 000
0.012395 Substituting in (1) gives: 135 = 5 b0 + 9600(0.0124)
from which, b0 =
135 − 9600(0.012395) 5
= 3.20 Hence, the equation of the regression line of X on Y is:
X= b0 + b1Y
i.e.
X = 3.20 + 0.0124Y
4. Determine the equation of the regression lines of X on Y, correct to 3 significant figures for the data given in Problem 2
Substituting into and gives: and
X b N + b ∑Y ∑= = XY b ∑ Y + b ∑ Y ∑ 0
0
1
1
2
83 = 8 b0 + 18.3 b1
(1)
254.6 = 18.3 b0 + 56.07 b1
(2)
18.3 × (1) gives:
1518.9 = 146.4 b0 + 334.89 b1
(3)
8 × (2) gives:
2036.8 = 146.4 b0 + 448.56 b1
(4)
(4) – (3) gives:
517.9 =
113.67 b1
Substituting in (1) gives: 83 = 8 b0 + 18.3(4.56)
from which, b1 =
from which,
b0 =
517.9 = 4.56 113.67
83 − 18.3(4.56) = –0.056 8
Hence, the equation of the regression line of X on Y is:
X= b0 + b1Y
i.e.
X = –0.056 + 4.56Y
1401
© 2014, John Bird
5. The relationship between the voltage applied to an electrical circuit and the current flowing is as shown: Current (mA)
2
4
6
8
10
12
14
Applied voltage (V)
5
11
15
19
24
28
33
Assuming a linear relationship, determine the equation of the regression line of applied voltage, Y, on current, X, correct to 4 significant figures.
A table is produced as shown below, where current I = X and voltage V = Y X
Y
X2
XY
Y2
2
5
4
10
25
4
11
16
44
121
6
15
36
90
225
8
19
64
152
361
10
24
100
240
576
12
28
144
336
784
14
33
196
462
1089
∑ X = 56 ∑ Y = 135 ∑ X Substituting into and
Y ∑=
2
∑ XY = 1334
= 560
∑Y
2
= 3181
a0 N + a1 ∑ X
= ∑ XY a0 ∑ X + a1 ∑ X 2
135 = 7 a0 + 56 a1
(1)
and
1334 = 56 a0 + 560 a1
(2)
8 × (1) gives:
1080 = 56 a0 + 448 a1
(3)
gives:
(2) – (3) gives:
254 =
from which, a1 =
112 a1
Substituting in (1) gives: 135 = 7 a0 + 56(2.268)
from which,
Hence, the equation of the regression line of Y on X is: i.e.
254 = 2.268 112
a0 =
135 − 56(2.268) = 1.142 7
Y= a0 + a1 X Y = 1.142 + 2.268X
1402
© 2014, John Bird
6. For the data given in Problem 5, determine the equation of the regression line of current on applied voltage, correct to 3 significant figures. X ∑=
Substituting into and
= XY ∑
gives:
b0 N + b1 ∑ Y b0 ∑ Y + b1 ∑ Y 2
56 = 7 b0 + 135 b1
(1)
and
1334 = 135 b0 + 3181 b1
(2)
135 × (1) gives:
7560 = 945 b0 + 18 225 b1
(3)
7 × (2) gives:
9338 = 945 b0 + 22 267 b1
(4)
(4) – (3) gives:
1778 =
Substituting in (1) gives:
4042 b1
56 = 7 b0 + 135(0.43988) from which, b0 =
Hence, the equation of the regression line of X on Y is: i.e.
from which, b1 =
1778 = 0.43988 4042
56 − 135(0.43988) = –0.483 7
X= b0 + b1Y
X = –0.483 + 0.440Y, correct to 3 significant figures
7. Draw the scatter diagram for the data given in Problem 5 and show the regression lines of applied voltage on current and current on applied voltage. Hence determine the values of (a) the applied voltage needed to give a current of 3 mA and (b) the current flowing when the applied voltage is 40 volts, assuming the regression lines are still true outside of the range of values given.
A scatter diagram is shown below. Current I = X and voltage V = Y The regression line of voltage on current, Y = 1.142 + 2.268X, and current on voltage, X = –0.483 + 0.440Y are shown and to the scale drawn are seen to coincide
1403
© 2014, John Bird
(a) When current, X = 3 mA, then applied voltage, Y = 1.142 + 2.268(3) = 7.95 V (b) When the applied voltage, Y = 40 V, current, X = –0.483 + 0.440(40) = 17.1 mA
8. In an experiment to determine the relationship between force and momentum, a force, X, is applied to a mass by placing the mass on an inclined plane, and the time, Y, for the velocity to change from u m/s to v m/s is measured. The results obtained are as follows: Force (N)
11.4
18.7
11.7
12.3
14.7
18.8
19.6
Time (s)
0.56
0.35
0.55
0.52
0.43
0.34
0.31
Determine the equation of the regression line of time on force, assuming a linear relationship between the quantities, correct to 3 significant figures.
Let force F = X and time = Y. A table is produced as shown below
1404
© 2014, John Bird
X
Y
X2
XY
Y2
11.4
0.56
129.96
6.384
0.3136
18.7
0.35
349.69
6.545
0.1225
11.7
0.55
136.89
6.435
0.3025
12.3
0.52
151.29
6.396
0.2704
14.7
0.43
216.09
6.321
0.1849
18.8
0.34
353.44
6.392
0.1156
19.6
0.31
384.16
6.076
0.0961
∑X = 107.2
∑Y
∑X
= 1721.52
= 3.06 Y ∑=
Substituting into
∑ XY
2
∑Y
= 44.549
2
= 1.4056
a0 N + a1 ∑ X
= ∑ XY a0 ∑ X + a1 ∑ X 2
and gives:
3.06 = 7 a0 + 107.2 a1
and
(1)
44.549 = 107.2 a0 + 1721.52 a1
(2)
107.2 × (1) gives:
328.032 = 750.4 a0 + 11491.84 a1
(3)
7 × (2) gives:
311.843 = 750.4 a0 + 12050.64 a1
(4)
(3) – (4) gives:
16.189 =
–558.8 a1
from which, a1 =
−16.189 = –0.0290 558.8
Substituting in (1) gives: 3.06 = 7 a0 + 107.2(–0.0290) from which,
a0 =
3.06 − 107.2(−0.0290) = 0.881 7
Hence, the equation of the regression line of Y on X is:
Y= a0 + a1 X
i.e.
Y = 0.881 – 0.0290X
9. Find the equation for the regression line of force on time for the data given in Problem 8, correct to 3 decimal places.
Substituting into and
X ∑=
= XY ∑
b0 N + b1 ∑ Y
b0 ∑ Y + b1 ∑ Y 2
1405
© 2014, John Bird
gives:
107.2 = 7 b0 + 3.06 b1
(1)
44.549 = 3.06 b0 + 1.4056 b1
(2)
3.06 × (1) gives:
328.032 = 21.42 b0 + 9.3636 b1
(3)
7 × (2) gives:
311.843 = 21.42 b0 + 9.8392 b1
(4)
and
(3) – (4) gives:
16.189 =
–0.4756 b1
from which, b1 =
16.189 = –34.039 −0.4756
Substituting in (1) gives: 107.2 = 7 b0 + 3.06(–34.039) from which,
b0 =
107.2 + 3.06(34.039) = 30.194 7
Hence, the equation of the regression line of X on Y is:
X= b0 + b1Y
i.e.
X = 30.194 – 34.039Y
10. Draw a scatter diagram for the data given in Problem 8 and show the regression lines of time on force and force on time. Hence, find (a) the time corresponding to a force of 16 N, and (b) the force at a time of 0.25 s, assuming the relationship is linear outside of the range of values given.
A scatter diagram is shown below. The regression line of force on time, Y = 0.881 – 0.0290X, and force on time, X = 30.194 – 34.039Y are shown and to the scale drawn are seen to coincide
(a) When force X = 16 N then time, Y = 0.881 – 0.0290(16) = 0.417 s 1406
© 2014, John Bird
(b) When time Y = 0.25 s, force, X = 30.194 – 34.039(0.25) = 21.7 N
1407
© 2014, John Bird
