Least Square Method Questions and Answers

This set of Numerical Analysis Multiple Choice Questions & Answers (MCQs) focuses on “Approximation of Functions using Least Square Method”.

1. Fit the straight line to the following data.

x	1	2	3	4	5
y	1	2	3	4	5

a) y=x
b) y=x+1
c) y=2x
d) y=2x+1
View Answer

Answer: a
Explanation: The normal equation are:
Σy = aΣx + nb
and
Σxy = aΣx² + bΣx
Now,

x	y	x²	xy
1	1	1	1
2	2	4	4
3	3	9	9
4	4	16	16
5	5	25	25
Σx = 15	Σy = 15	Σx² = 55	Σxy = 55

Substituting in the equations,
15 = 15a + 4b and 55 = 55a + 15b
Solving these two equatons, we get a=1 and b=0,
Therefore the required straight line equation is y=x.

2. Fit the straight line to the following data.

x	0	5	10	15	20
y	7	11	16	20	26

a) y = 0.94x + 6.6
b) y = 6.6x + 0.94
c) y = 0.04x + 5.6
d) y = 5.6x + 0.04
View Answer

Answer: a
Explanation: First drawing the table,

x	y	x²	xy
0	7	0	0
5	11	25	55
10	16	100	160
15	20	225	300
20	26	400	520
Σx = 50	Σy = 26	Σx² = 750	Σxy = 1035

Subscribe Now: Numerical Methods Newsletter | Important Subjects Newsletters

The normal equation are:
Σy = aΣx + nb
and
Σxy = aΣx² + bΣx.
Substituting the values, we get,
80 = 50a + 5b
1035 = 750 a + 50 b
Solving them, we get a = 0.94 and b = 6.6.
Therefore the straight line equation is y=0.94x + 6.6.

3. Fit the straight line curve to the following data.

x	75	80	93	65	87	71	98	68	84	77
y	82	78	86	72	91	80	95	72	89	74

a) y = 0.9288x + 7.78155
b) y = 7.78155x + 0.9288
c) y = 0.8288x + 6.78155
d) y = 6.78155x + 0.8288
View Answer

Answer: a
Explanation: First drawing the table,

x	y	x²	xy
75	82	5625	6150
80	78	6400	6240
93	86	8349	7998
65	72	4225	4680
87	91	7569	7917
71	80	5041	5680
98	95	9605	9310
68	72	4624	4896
84	89	7056	7476
77	74	5929	5698
798	819	64422	66045

The normal equation are:
Σy = aΣx + nb
and
Σxy = aΣx² + bΣx.
Substituting the values, we get,
819 = 798a + 10b
66045 = 64422a + 798b
Solving, we get
a = 0.9288 and b = 7.78155
Therefore, the straight line equation is :
y = 0.9288x + 7.78155.

4. Fit a second degree parabola to the following data.

x	1	2	3	4	5	6	7	8	9
y	2	6	7	8	10	11	11	10	9

a) y = -0.2673x² + 3.5232x – 0.9286
b) y = 0.2673x² + 3.5232x – 0.9286
c) y = 0.2673x² + 3.5232x + 0.9286
d) y = -0.2673x² + 3.5232x + 0.9286
View Answer

Answer: a
Explanation: Here,

x	y	x²	x³	x⁴	xy	x²y
1	2	1	1	1	2	2
2	6	4	8	16	12	24
3	7	9	27	81	21	63
4	8	16	64	256	32	128
5	10	25	125	625	50	250
6	11	36	216	1296	66	396
7	11	49	343	2401	77	539
8	10	64	512	4096	80	640
9	9	81	729	6561	81	729
45	74	285	2025	15333	421	2771

The normal equations are:
Σy = aΣx² + bΣx + nc
Σxy = aΣx³ + bΣx² +cΣx
Σx²y = aΣx⁴ + bΣx³ + cΣx²
Substituting the values, we get
74 = 285a + 45b + 9c
421 = 2025 a + 285 b + 45 c
2771 = 15333a + 2025 b + 285 c
Solving them, we get the second order equation which is,
y = -0.2673x² + 3.5232x – 0.9286.

5. The normal equations for a straight line y = ax + b are:
a) Σy = aΣx + nb and Σxy = aΣx² + bΣx
b) Σxy = aΣx + nb and Σy = aΣx² + bΣx
c) Σy = aΣx + nb and Σxy = aΣx² + bΣxy
d) Σy = aΣx + nb and Σx2y = aΣx² + bΣx
View Answer

Answer: a
Explanation: Let the sum of residues be E.
E = Σd_i² = Σ(y_i – (ax_i+b))²
Here \(\frac{∂E}{∂a}\)=0 and \(\frac{∂E}{∂b}\)=0
Solving these two equations, we get the normal equations as
Σy = aΣx + nb and Σxy = aΣx² + bΣx.

6.The normal equations for a second degree parabola y = ax² + bx + c are Σy = aΣx² + bΣx + nc, Σxy = aΣx³ + bΣx² + cΣx and Σx²y = aΣx⁴ + bΣx³ + cΣx².. Is it true or false?
a) True
b) False
View Answer

Answer: a
Explanation: The second order parabola is given by
y = ax² + bx +c.
Let the sum of residues be E.
E = Σd_i² = Σ(y_i – (ax_i² + bxi +c))².
Here \(\frac{∂E}{∂a}\)= 0, \(\frac{∂E}{∂b}\)= 0 and \(\frac{∂E}{∂c}\)= 0.
Solving these three equations, we get the normal equations as
Σy = aΣx² + bΣx + nc, Σxy = aΣx³ + bΣx² + cΣx and Σx²y = aΣx⁴ + bΣx³ +cΣx².

7. If the equation y = ae^bx can be written in linear form Y=A + BX, what are Y, X, A, B?
a) Y = logy, A = loga, B=b and X=x
b) Y = y, A = a, B=b and X=x
c) Y = y, A = a, B=logb and X=logx
d) Y = logy, A = a, B=logb and X=x
View Answer

Answer: a
Explanation: The equation is
y = ae^bx.
Taking log to the base e on both sides,
we get logy = loga + bx.
Which can be replaced as Y=A+BX,
where Y = logy, A = loga, B = b and X = x.

8. If the equation y=ab^x can be written in linear form Y=A+BX, what are Y, X, A, B?
a) Y=logy, X=x, A=loga and B=logb
b) Y=y, A=a, B=b and X=x
c) Y=y, A=a, B=logb and X=logx
d) Y=logy, A=a, B=logb and X=x
View Answer

Answer: a
Explanation: The given curve is y=ab^x.
Taking log on bothe the sides, we get,
logy = loga + x logb.
This can be written in the format of Y=A+BX
where
Y = logy, X = x, A = loga and B = logb.

9. If the equation y=ax^b can be written in the linear form Y=A+BX, what are Y, X, A, B?
a) Y=logy, A=loga, B=b and X=logx
b) Y=y, A=a, B=b and X=x
c) Y=y, A=a, B=logb and X=logx
d) Y=logy, A=a, B=logb and X=x
View Answer

Answer: a
Explanation: The given curve is y=ax^b.
Taking log on bothe the sides, we get,
logy = loga + blogx.
This can be written as Y=A+BX,
where
Y=logy, A=loga, B=b and X=logx.

10. The parameter E which we use for least square method is called as ____________
a) Sum of residues
b) Residues
c) Error
d) Sum of errors
View Answer

Answer: a
Explanation: E is given by
E = Σdi² = Σ(yi – f(xi))².
Where the term inside the summation is called as residues and the sum is said to be a sum of residues.
Therefore, E is said to be the sum of residues.

Sanfoundry Global Education & Learning Series – Numerical Methods.

To practice all areas of Numerical Methods, here is complete set of 1000+ Multiple Choice Questions and Answers.

If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]

« Prev - Numerical Analysis Questions and Answers – Newton-Gregory Forward Interpolation Formula

Related Posts:

Practice Engineering Mathematics MCQ
Apply for 1st Year Engineering Internship
Check Numerical Methods Books
Practice Probability and Statistics MCQ

Recommended Articles: