TU Solution

60 - 70 4 65 2 8 16 32 64
81
70 - 80 1 75 3 3 9 27 582

50 -26 114 -116

Here, N = 50, A = 45, h = 10, Σfd' = -26, Σfd'2 = 144, Σfd'3 = -116, Σfd'4 = 582

Now, first four raw moments are

1′ = ′ × ℎ = −26 × 10 = −5.2
50

2′ = ′2 × ℎ2 = 114 × 102 = 228
50

3′ = ′3 × ℎ3 = −116 × 103 = −2320
50

4′ = ′4 × ℎ4 = 582 × 104 = 116400
50

Again, first four central moments are

1 = 0
μ2 = μ'2 – μ'12 = 228 – (-5.2)2 = 228 – 27.04 = 200.96

μ3 = μ'3 - 3μ'2μ'1 + 2(μ'1)3 = -2300 - 3×228×(-5.2) + 2 × (-5.2)3

= -2320 +3556.8 – 281.216 = 955.584

μ4 = μ'4- 4μ'3μ'1+ 6μ'2 (μ'1)2 – 3(μ'1)4

= 116400 – 4×(-2320)×(-5.2) + 6 × 228 × (-5.2)2 – 3 × (-5.2)4

= 116400 - 48256 + 36990.72 - 2193.4848=102941.235

β1 = μ23 = (955.584)2 = 913140.781 = 0.1125
μ23 200.963 8115753.84

γ1 = ± √ 0.1125 = ± 0.34

Since, μ3 is positive, γ1 would be positive, the distribution is positively skewed.

β2 = μ4 = 199453.235 = 102941.235 = 2.55
μ22 200.962 40384.9216

Since β2 = 2.55 < 3, the distribution is platykurtic.

QN. 25 (2073) The asset structure listed companies in Nepal is given below.

Asset (Rs. in millions) No of listed companies

150

0–5 20
5 – 10 25
10 – 15 50

15 -20 40

20 – 25 20

25 -30 15

Total 170

a) Compute the skewness and the kurtosis of the assets distribution by using the moment method.

b) Interpret the result of skewness.

c) Also calculate the coefficient of variance of asset distribution.

Solution:

QN. 26 (2072) The following is the daily wages of two factories.

No. of factories

Daily wages in Rs.

Factory X Factory Y

50 – 75 30 50
75 – 125 60 80
125 – 150 88 120
150 – 175 120 70

175 - 200 60 40

200 – 225 28 30

225 - 250 14 10

Find a) mean and standard deviation of each group and state which group of workers has greater
consistency in the wages b) Test the normality of the distribution.

Solution:

For factory X

Take d′ = X − A , where A = 162.5 and h = 25
h

Daily wages Frequency(f) Mid-value d' fd' fd'2 fd'3 fd'4
in Rs. (X)

151

50 – 75 30 62.5 -4 -120 480 -1920 7680
75 - 125 60 100
125 – 150 88 137.5 -2.5 -150 375 -937.5 2343.75
150 – 175 120 162.5
175 - 200 60 187.5 -1 -88 88 -88 88
200 – 225 28 212.5
225 - 250 14 237.5 0 0 00 0
400
1 60 60 60 60

2 56 128 256 512

3 42 126 378 1134

-200 1257 -2251.5 11817.75

Here, N = 400, A = 162.5, h = 25, Σfd' = -200, Σfd'2 = 1257, Σfd'3 = -2251.5, Σfd'4 = 11817.75

Now, mean(X̅) = A + Σfd′ × h = 162.5 + −200 × 25 = 162.5 − 12.5 = 150
N 400

Standard deviation(σ) = √ΣfNd′2 − (ΣNfd′) × h

= √1420507 − (−420000) × 25

= √3.1875 − 0.25 × 25

= √2.9375 × 25

= 42.85

C. V. (X) = σ × 100% = 42.84 × 100% = 28.58%
̅X 150

Now, first four raw moments are

μ1′ = Σfd′ × h = −200 × 25 = −12.5
N 400

μ2′ = Σfd′2 × h2 = 1257 × 252 = 1964.0625
N 400

μ′3 = Σfd′3 × h3 = −2251.5 × 253 = −87949.2188
N 400

μ4′ = Σfd′4 × h4 = 11817.75 × 254 = 11540771.5
N 400

152

Again, first four central moments are

μ1 = 0
μ2 = μ'2 – μ'12 = 1964.0625 – (-12.5)2 = 1964.0625 – 156.25 = 1807.8125

μ3 = μ'3 - 3μ'2μ'1 + 2(μ'1)3 = -87949.2188 - 3×1964.0625×(-12.5) + 2 × (-12.5)3

= -87949.2188 + 73652.3438 – 1953.125 = -16250

μ4 = μ'4- 4μ'3μ'1+ 6μ'2 (μ'1)2 – 3(μ'1)4

= 11540771.5 – 4×(-87949.2188)×(-12.5) + 6 × 1964.0625 × (-12.5)2 – 3 × (-12.5)4

= 11540771.5 – 4397460.94 + 1841308.59 – 73242.1875 = 8911376.96

β2 = μ4 = 8911376.96 = 8911376.96 = 2.73
μ22 1807.81252 3268186.04

Since β2 = 2.55 < 3, the distribution is platykurtic.

For factory Y (same as factory X)

153

Chapter - 6

Simple Correlation and Regression Analysis

Brief Question Answer

QN. 1) (2078) Calculate the coefficient of correlation between X and Y if the regression
coefficient of Y on X is -0.667 and regression coefficient X on Y is -0.75.

Solution: Here given regression coefficient of Y on X (bYX) = - 0.667

regression coefficient X on Y (bXY) = - 0.75.

correlation coefficient (r ) = ?

Now, correlation coefficient (r ) = √bYX. bXY = √−0.667 × −0.75 = 0.71

QN.2) (2076) Calculate the Pearson's coefficient of correlation from the following information:

N = 8, ΣX1 = 544, ΣX2 = 552, ΣX12 = 37028, ΣX22 = 38132, and ΣX1X2 = 37560

Solution: Pearson's coefficient of correlation

= √ 12 1 2 − 1 2 ( 2)2
− ( 1)2 √ 22 −

= √8 × 8 × 37560 − 544 × 552 − (552)2
37028 − (544)2 √8 × 38132

= √296224 300480 − 300288 − 304704
− 295936 √305056

= 192
√288 √352

= 192
16.96 × 18.76

= 0.60

QN.3) (2075) If bxy = -0.87 and byx = - 0.63, find the correlation coefficient.
Solution: Now, correlation coefficient

r = √byx × bxy = √−0.87 × −0.63 = 0.74

Since, bxy and byx both are negative, therefore correlation coefficient r = - 0.74
QN. 4) Determine the number of pair of observations, where
ΣD2 = Σ(R1 – R2)2 is 30 and rank correlation, R = 0.75
Solution: Now, rank correlation

154

= 1 − 6 2 1)
( 2 −

⇒ 0.75 = 1 − 6× 30
( 2 − 1)

⇒ 0.75 − 1 = 180
( 2 − 1)

⇒ −0.25 = − 180 1)
( 2 −

⇒ ( 2 − 1) = 180
0.25

⇒ ( − 1)( + 1) = 720

⇒ ( − 1) ( + 1) = 8 × 9 × 10

∴ = 9

QN. 5) (2073) The regression coefficient of y an x is 0.68 and correlation coefficient is 0.72, find
the regression coefficient x on y.

Solution: The regression coefficient y on x (byx) = 0.68, correlation coefficient (r) = 0.72
Now, correlation coefficient

= √ ×

⇒ 0.72 = √0.68 ×

⇒ 0.5184 = 0.68 ×

⇒ = 0.5184
0.68

∴ = 0.76

QN. 6) (2072) For the eight pair of observations on two variables sales (X) and pricing (Y), the
following results were obtained, ΣX = 156, ΣY = 132, ΣX2 = 4162, ΣY2 = 2434, ΣXY = 2844.

Find out if there exists any relationship between and sales and pricing.

Solution: Pearson's coefficient of correlation
1 2 − 1 2
= √ 12 − ( 1)2 √ 22 − ( 2)2

= √8 × 8 × 2844 − 156 × 132 − (132)2
4162 − (156)2 √8 × 2434

= √33296 22752 − 20592 − 17424
− 24336 √19472

155

= 2160
√8960 √2048

= 2160
94.66 × 45.25

= 2160
4283.24

= 0.50

There exists medium degree of correlation between sales and pricing.

QN. 7) ( 2072) A student calculated the value of r = 0.81 from 10 observations and conclude that r
is highly significent. Approave his conclusion statistically.

Solution: Here, r = 0.81 and N = 10

Now, . . = 0.6745 (1√− 2)
= 0.6745 (1 −√01.0812)
= 0.6745 (1 −√615061)
= 0.6745 (03.3.41369)
= 0.6745 × 0.11
= 0.0734

Again, 6.× . . = 6 × 0.0734 = 0.44
Since, r > 6×P.E., so r is significant.

QN. 8) (2072) The rank correlation between the marks in English and Math obtained by 10
students was found to be 0.7. Later on it was found that the difference of rank in two subjects
obtained by one student was wrongly taken as 5 instead of 8. Find the correct rank correlation.

Solution: Here given Number of observations (N) = 10, Rank correlation (R) = 0.7, correct value
= 8 and wrong value = 5.

Now, rank correlation

= 1 − 6 2 1)
( 2 −

⇒ 0.7 = 1 − 6 × 2
10(102 − 1)

156

⇒ 0.7 − 1 = − 6 × 2
990

⇒ 297 = 6 × 2

⇒ 2 = 49.5

Again correct 2 = 49.5 − 52 + 82 = 88.5

Correct correlation

= 1 − 6 2 1)
( 2 −

= 1 − 6× 1)
10(102 −

= 1 − 531
990

= 1 − 0.5364

∴ = 0.4636

QN. 9) (2071) The sum of the squares of the differences of the ranks obtained from eight pairs of
observations of X and Y variables is calculated as 6. Calculate the correlation coefficient between
the variables.

Solution: Here given, number of observations (N) = 8,
sum of square of differences of ranks Σd2 = 6

Now, rank correlation

= 1 − 6 2 1)
( 2 −

= 1 − 6×6
8(82 − 1)

= 1 − 36
520

= 1 − 0.069

= 0.93

Descripetive Question Answer

QN. 10) (2077) A computer while calculating the correlation coefficient between X and Y from 25
pairs of observations obtained the following information:
n = 25, ΣX = 125. ΣX2 = 650, ΣY = 100, ΣY2 = 460, ΣXY = 508.

157

It was discovered later at the time of checking that it had copied down two pairs of wrong
observations as (6, 14) & (8, 6) while the correct values were (8, 12) & (6, 8). Calculate correct
correlation coefficient.
Solution: Here given n = 25, ΣX = 125. ΣX2 = 650, ΣY = 100, ΣY2 = 460, ΣXY = 508.

wrong observations are (6, 14) & (8, 6) and correct values are (8, 12) & (6, 8)

correct ΣX= 125 – 6 – 8 + 8 + 6 = 125

correct ΣY=100 – 14 – 6 + 12 + 8 = 100

Correct ΣXY=508 – 6×8 – 14×6 + 8×6 + 12×8

= 508 – 48 – 84 + 48 + 96 = 520
correct ΣX2 = 650 – 62 – 82 + 82 + 62 = 650
correct ΣY2 = 460 – 142 – 62 + 122 + 82

= 460 – 196 – 36 + 144 + 64 = 436

Correct Correlation coefficient
−
= √ 2 − ( )2 √ 2 − ( )2

= √25 × 25 × 520 − 125 × 100 − (100)2
650 − (125)2 √25 × 436

= √16250 13000 − 12500 − 10000
− 15625 √10900

= 13000 − 12500
√625 × √900

= 500
25 × 30

= 500
750

= 0.667

QN. 11) (2076) Calculate the Karl Pearson's Correlation Coefficient from the following data. Also

interpret the data based on the probable error.

X 67 65 68 64 66 70 63

Y 69 65 69 65 67 67 64

Solution: Calculation of correlation coefficient

X Y X2 Y2 XY
4623
67 69 4489 4761 4225
4692
65 65 4225 4225

68 69 4624 4761

158

64 65 4096 4225 4160

66 67 4356 4489 4422

70 67 4900 4489 4690

63 64 3969 4096 4032

463 466 30659 31046 30844

Here, N = 7, ΣX = 463, ΣY = 466, ΣX2 = 30659, ΣY2 = 31046, ΣXY = 30844

Now, correlation coefficient

= √ 2 − − ( )2
− ( )2 √ 2

= √7 × 7 × 30844 − 463 × 466 − (466)2
30659 − (463)2 √7 × 31046

= √214613 215908 − 215758 − 217156
− 214369 √217322

= 150
√244 √166

= 150
15.62 × 12.88

= 150
201.26

= 0.75

Again,

. . = 0.6745 (1 − 2
√
)

= 0.6745 (1 −√07.752)

= 0.6745 (1 −20.6.55625)

= 0.6745 (02.4.36755)

= 0.6745 × 0.165 = 0.111

Again, 6.× . . = 6 × 0.111 = 0.666

159

Since, r > 6×P.E., so r is significant.
QN. 12) (2072) Calculate the rank correlation coefficient from the following data and also
interpret the result.
Solution:
Marks in Statistics 20 30 40 40 50 60 70 80
Marks in Account 10 15 20 20 18 25 30 40

Solution: Calculation of rank correlation

Marks in Marks in RX RY d = RX - RY d2

Statistics (X) Accout (Y) 80 0
70 0
20 10 8 4.5 1 1
4.5 1 1
30 15 7 6 -2 4
30 0
40 20 5.5 20 0
10 0
40 20 5.5 6

50 18 4

60 25 3

70 30 2

80 40 1

Here, N = 8, Σd2 = 6, m1 = 2, m2 = 2

Now, rank correlation is ( 3 −
12
= 1 − 6 [ 2 + )]

( 2 − 1)

= 1 − 6 [6 + 8−2 + 8− 2]
12 12
8(64 − 1)

= 1 − 6[6 + 1]
= 1 − 42504
504

= 1 − 0.083
= 0.917
Therefore, marks in statistics and account havs highly positive correlation.

160

QN. 13) (2072) From the following bi-variate frequency distribution, determine the most
probable age of the wife whose husband's age is 75 years.

Age of wives Age of husbands age in years

age in years 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70
5 -
15 – 25 - 93 - -
25 – 35 - -
35 – 45 - 10 25 2 5
45 – 55 - 2
55 - 65 1 12 2

Solution: - 4 16

- -4

Let X and Y denotes the mid-point of age of husband and mid-point of age of wife respectively.

Then, dx = X−45 and dy = Y−40
10 10

Here,

= 100, = 45, ℎ = 10, = 40, = 10, = 8, = −34,
2 = 90, 2 = 154, = 8

̅Y = B, ̅+ =ΣfN d y+× k = 4×0ℎ+=(−45130+401)08×0 × 10 = 45 + 0.8 = 45.8
10 = 40 − 3.4 = 36.6

Again, regression coefficient Y on X

161

bYX = NΣfdxdy − (Σfdx)(Σfdy)
NΣfdx2 − (Σfdx)2

= 100 × 88 − 8 × −34
100 × 90 − (8)2

= 8800 + 272
9000 − 64

= 9072
8936

= 1.015

Regression equation Y on X is

Y – Y̅ =bYX(X-X̅ )
or Y – 36.6 = 1.015(X – 45.8)
or Y – 36.6 = 1.015X – 46.5
or Y = 1.015X – 9.9
when X = 75, Y = 1.015×75 – 9.9 = 66.225≈ 66

Therefore, most probable age of wife is 66 years when husbands age is 75 years.

QN. 14) (2071) The coefficient of rank correlation of the marks obtained by 10 students in English
and Statistics was found to be 0.8. It was later discovered difference in ranks obtained by one
student was wrongly taken as 7 instead of 9. Find the correct rank correlation.

Solution: Here given Number of students (N) = 10, correlation = 0.8,

wrong value = 7, correct value = 9

Now, rank correlation

= 1 − 6 2 1)
( 2 −

⇒ 0.8 = 1 − 6 × 2
10(102 − 1)

⇒ 0.8 − 1 = − 6 × 2
990

⇒ 198 = 6 × 2

⇒ 2 = 33

Again correct 2 = 33 − 72 + 92 = 65

Correct correlation

= 1 − 6 2 1)
( 2 −

162

= 1 − 6 × 65 1)
10(102 −

= 1 − 390
990

= 1 − 0.3939

∴ = 0.61

QN. 15) (2070) Find whether there exists any relationship between the age of driver and the no. of
accidents from the following bi-variate table.

No. of Age of drivers age in years

accidents 20 - 22 22 - 24 24 - 26 26 - 28 28 - 30

0 5 7 13 - -

1 - 10 15 2 -

2 - 3 12 6 -

3 - - 4 12 5

4 - - -42

Solution: Let X denotes the mid-point of age of drivers and Y denotes no. of accidents.

Then, dx = X−25
2

Here,
N = 100, Σfdx = 8, Σfy = 156, Σfdx2 = 92, Σfy2 = 396, Σfdxy = 102

163

Now, correlation coefficient
NΣfdxy − (Σfdx)(Σfy)
r = √NΣfdx2 − (Σfdx)2 √NΣfy2 − (Σfy)2

= √100 × 100 × 102 − 8 × 156 − (156)2
92 − (8)2 √100 × 396
10200 − 1248
= √9200 − 64 √39600 − 24336

= 8952
√9136 √15264

= 8952
95.58 × 123.55

= 8952
11809.21

= 0.76

Comprehencive Question Answer
QN. 16) (2078) The income and expenditure od 100 families is given below:

Income (Rs.) 0 - 500 Expenditure (Rs.) 1500 - 2000
- 500 - 1000 1000 - 1500 3
0 - 1000 - 4
1000 - 2000 3 -- 8
2000 - 3000 7 49 5
3000 - 4000 9 10 19 -
4000 - 5000 6 12
7-

Find (a) Two regression co-efficient.

(b) Co-efficient of correlation between income and expenditure.

(c) Estimate the expenditure when income is Rs.10,000.

Solution: Let X denotes the mid-point of family income and Y denotes expenditure.
X − 2500 − 1250
Then, dx = 1000 , = 500

164

Here, N = 106, Σfdx = 41, Σfdy = -45, Σfdx2 = 172, Σfdy2 = 123, Σdxdy = - 71

(a) Two regression co-efficient

Now, regression coefficient Y on X

bYX = NΣfdxdy − (Σfdx)(Σfdy)
NΣfdx2 − (Σfdx)2

= 106 × −71 − 41 × −45
106 × 172 − (41)2

= −7526 + 1845
18232 − 1681

= − 5681
16551

= −0.34

And, regression coefficient X on Y

bXY = NΣfdxdy − (Σfdx)(Σfdy)
NΣfdy2 − (Σfdy)2

= 106 × −71 − 41 × −45
106 × 123 − (45)2

= −7526 + 1845
13038 − 2025

165

= − 5681
11013

= − 0.52

(b) Co-efficient of correlation between income and expenditure.

r = √bYX × bXY = √−0.34 × −0.52 = −0.42

(c) Estimate the expenditure when income is Rs.10,000.

̅ = + × ℎ = 2500 + 1−49041865××1500000== 2500 + 386.79 = 2886.79
̅ = + × = 1250 + 1250 − 112.26 = 1137.74


Regression equation Y on X is

Y – Y̅ =bYX(X-X̅ )
or Y – 1137.74 = - 0.34(X – 2886.79)
or Y – 1137.74 = - 0.34X + 981.51
or Y = - 0.34X + 981.51 + 1137.74

or Y = - 0.34X + 2119.25

when X = 10,000, Y = - 0.34×10000 + 2119.25 = -3400 + 2119.25 = 1280.75

Therefore, estimated expenditure when income is Rs.10,000 is Rs.1280.75

QN. 17) (2077) A family income and its percentage expenditure on food for 100 families gave the
following bi-variate frequency distribution. Find the correlation coefficient between them.
Estimate the percentage expenditure on food when family income = Rs.90,000. Also, find the
probable error and interpret the result.

Food Family income (000Rs.)

expenditure 20 - 40 40 - 60 60 - 80 80 - 100 100 - 120

(%)

15 - - - 3 7

20 - 4 9 4 3

25 7 6 12 3 -

30 3 10 19 8 -

Solution: Let X denotes the mid-point of family income and Y denotes food expenditure in
percentage. Then, dx = X−2025, dy = Y – 25

166

Here,

N = 98, A = 25, h = 20, B = 25, Σfdx = −2, Σfdy = 0,
Σfdx2 = 118, Σfdy2 = 100, Σfdxdy = −18

Now, correlation coefficient NΣfdxy − (Σfdx)(Σfdy)
√NΣfdx2 − (Σfdx)2 √NΣfdy2 −
r = (Σfdy)2

= √98 × 98 × −18 − (−2) × 0 − (0)2
118 − (−2)2 √98 × 100
−1764 − 0
= √11564 − 4 √9800 − 0

= −1764
√11560 √9800

= −1764
107.54 × 98.99

= −1764
10645.92

= −0.17

Again,

̅ = + × ℎ = 25 + −990882××1200 = 25 − 0.41 = 24.6
̅ = + × = 25 + = 25 + 0 = 25


Again, regression coefficient Y on X

167

= − ( )( )
2 − ( )2

= 98 × −18 − (−2) × 0
98 × 118 − (−2)2

= −1764
11564 − 44

= −1764
11560

= −0.15

Regression equation Y on X is
Y – Y̅ =bYX(X-X̅ )

or Y – 25 = -0.15(X – 24.6)
or Y – 25 = -0.15X + 3.75
or Y = - 0.15X + 3.75 + 25

or Y = - 0.15X + 28.75

when X = 90 , Y = - 0.15×90 +28.75 = -13.5 + 28.75 = 15.25

Therefore, food expenditure is 15.25% when family income is Rs.90, 000.

QN. 18) (2076) The following table shows the number of motor registration and the sales of

"Gorkhali Motor Tires" by the whole-sales dealer in Kathmandu for the terms of 10 years.

Years 1 2 3 4 5 6 7 8 9 10

Motor registration (000 nos) 60 62 65 70 48 53 73 65 82 72

No of tiers sold in (000 nos) 68 60 62 80 40 52 62 60 81 85

i) Find the coefficient of correlation between the motor registration and number of tyres

solds.

ii) Interpret your result.

iii) Test the significance of the result.

iv) Compute two regression coefficients.

v) Estimate the number of tyres to be sold when the expected motor registration is 92,000.

Solution:
Year Motor registration (X) No of tiers sold (Y) X2 Y2 XY

1 60 68 3600 4624 4080

2 62 60 3844 3600 3720

3 65 62 4225 3844 4030

4 70 80 4900 6400 5600

5 48 40 2304 1600 1920

6 53 52 2809 2704 2756

7 73 62 5329 3844 4526

168

8 65 60 4225 3600 3900

9 82 81 6724 6561 6642

10 72 85 5184 7225 6120

650 650 43144 44002 43294

Here, N = 10, ΣX = 650, ΣY = 650, ΣX2 = 43108, ΣY2 = 44002, ΣXY = 43294

i) Now, correlation coefficient

= √ 2 − − ( )2
− ( )2 √ 2

= √10 × 10 × 43294 − 650 × 650 − (6500)2
43108 − (650)2 √10 × 44002

= √ 431440 432940 − 422500 − 422500
− 422500 √440020

= 10440
√8940 √17520

= 10440
94.55 × 132.36

= 10440
12514.85

= 0.8342

ii) There is highly positive correlation between the motor registration and number of tyres solds.
iii) Probable error of correlation coefficient
. . ( ) = 0.6745 (1√− 2)
= 0.6745 (1 − (√0.180342)2)
= 0.6745 (1 −30.1.66959)
= 0.6745 (03.3.01461)
= 0.6745 × 0.096 = 0.065

Again, 6.× . . = 6 × 0.065 = 0.3894

Since, r > 6×P.E., so r is significant.
iv) regression coefficient Y on X

169

= − ( )( )
2 − ( )2

= 1043294 − 650 × 650
10 × 43144 − (650)2

= 432940 − 422500
431440 − 422500

= 10440
8940

= 1.1678

regression coefficient X on Y

= − ( )( )
2 − ( )2

= 1043294 − 650 × 650
10 × 44002 − (650)2

= 432940 − 422500
440020 − 422500

= 10440
17520

= 0.5959

v) Estimated number of tyres sold when the expected motor registration is 92,000.

̅ = = 650 = 65 & ̅ = = 650 = 65
10
Regression equation y on x is Y – Y̅ = bYX(X-X̅ )

or Y – 65 = 1.1678(X - 65)

or Y – 65 = 1.1678X – 75.907

or Y = 1.1678X – 75.907 + 65

or Y = 1.1678X – 10.907

When X = 92, Y = 1.1678×92 – 10.907 = 107.4376 – 10.907 = 96.5306
∴ required tyre sold is 96530.6

QN. 19) (2075) From the following bi-variate table, compute coefficient of correlation. Also test

the significance of the result. Estimate sales revenue when advertisement cost is Rs. 70 thousand.

Sales Advertisement expenditure (000 Rs.)

revenue 5 - 15 15 - 25 25 - 35 35 - 45 Total

(000Rs.) 4 1- - 5
75 – 125 7 62 1 16
125 – 175 1 34 2 10
175 – 225
170

225 - 275 1 1 3 49
Total 13 11 9 7 40

Solution:

Here, N = 40, Σfdx = -30, Σfdy = -17, Σfdx2 = 70, Σfdy2 = 45, Σdxdy = 38
Now, correlation coefficient

r = NΣfdxfdy − ΣfdxΣfdy (Σfdy)2
√NΣfdx2 − (Σfdx)2 √NΣfdy2 −

= 40 × 38 − (−17) × (−30)
√40 × 70 − (−30)2√40 × 45 − (−17)2

= 1520 − 510
√2800 − 900√ 1800 − 289

= 1010
√1900√1511

= 1010
43.59 × 38.87

= 1010
1694.3433

= 0.596

171

Probable error of correlation coefficient
. . ( ) = 0.6745 (1√− 2)
= 0.6745 (1 − (√04.5096)2)
= 0.6745 (1 −6.302.3555)
= 0.6745 (06..362455)
= 0.6745 × 0.102 = 0.069

Again, 6.× . . = 6 × 0.069 = 0.413

Since, r > 6×P.E., so r is significant.
iv) regression coefficient Y on X

bYX = NΣfdxfdy − ΣfdxΣfdy
NΣfdx2 − (Σfdx)2

= 40 × 38 − (−17) × (−30)
40 × 70 − (−30)2

= 1520 − 510
2800 − 900

= 1010
1900

= 0.53

X̅ = A + Σfdx × h = 23000++−4−3401000××1500==3020−07−.51=2.252=.5187.5
̅Y = B + ΣfNdy × k =

N
Regression equation Y on X is

Y – Y̅ =bYX(X-X̅ )
or Y – 187.5 = 0.53(X – 22.5)
or Y – 187.5 = 0.53X – 11.925
or Y = 0.53X – 11.912 + 187.5

or Y = 0.53X + 178.575

when X = 70 , Y = 0.53×70 +178.575 = 37.1 + 178.575 = 215.675

Therefore, sales revenue when advertisement cost is Rs. 70 thousand is Rs.215675

QN. 20) (2074) You are given the sales and advertisement expenditure in crores:

Sales Advertisement expenditure

20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 Total

172

10 - 15 - - - 3 7 10

15 - 20 - 4 9 4 3 20

20 - 25 7 6 12 5 - 30

25 - 30 3 10 19 8 - 40

Total 10 20 40 20 10 100

Is there significance relationship between advertisement expenditure and sales if significance

relationship exists between given variables, estimate the sales when advertisement expenditure is

82 crores.

Solution:

Here, N = 100, Σfdx = 0, Σfdy = 0, Σfdx2 = 120, Σfdy2 = 100, Σdxdy = - 48

Now, correlation coefficient

r = NΣfdxfdy − ΣfdxΣfdy (Σfdy)2
√NΣfdx2 − (Σfdx)2 √NΣfdy2 −

= √100 × 100 × −48 − 0 × 0 − (0)2
120 − (0)2√100 × 100

= −4800 − 0
√12000 − 0√ 10000 − 0

= −4800
√12000√10000

173

= −4800 100
109.5445 ×

= −4800
10954.45

= −0.438

Probable error of correlation coefficient
. . ( ) = 0.6745 (1√− 2)
(1 − (−0.438)2
= 0.6745 √100 )

= 0.6745 (1 − 01.01918)

= 0.6745 (0.810082)

= 0.6745 × 0.08082 = 0.05451

Again, 6.× . . = 6 × 0.05451 = 0.327

Since, |r| > 6×P.E., so r is significant.
iv) regression coefficient Y on X

= −
2 − ( )2

= 100 × −48 − 0×0
100 × 120 − (0)2

= −4800
12000 − 0

= −4800
12000

= −0.4

̅ = + × ℎ = 4252.+5 +1004000××150== 45 − 0 = 45
̅ = + × = 22.5 − 0 = 22.5



Regression equation Y on X is
Y – Y̅ =bYX(X-X̅ )

or Y – 22.5 = -0.4(X – 45)
or Y – 22.5 = -0.4X +18
or Y = -0.4X +18 + 22.5

or Y = -0.4X +40.5

174

when X = 82 , Y = -0.4× 82 + 40.5 = -32.5 + 40.5 = 7.7

Therefore, sales revenue when advertisement cost is Rs. 70 thousand is 193476.7

QN. 21) (2073) A researcher is seeing how accurately a new job performance index measure what

is important for corporation. One way to check is to look at the relationship between job

evaluation index and an employee's salary is significance or not. A sample of digits employees

was taken, and information about salary in thousand rupees and job performance index (1 – 10, 10

is best) was collected.

Job performance index 9 7 8 4 7 5 5 6

Salary '000' Rs. 36 25 33 15 28 19 20 22

Develop an estimating equation that best describe those data and estimate the salary of an

employees whose job performance is 10 and 2. Also find whether there exists significance

relationship between the given variables.

Solution:

Job performance Salary '000' Rs. X2 Y2 XY

index (X) (Y)

9 36 81 1296 324

7 25 49 625 175

8 33 64 1089 264

4 15 16 225 60

7 28 49 784 196

5 19 25 361 95

5 20 25 400 100

6 22 36 484 132

51 198 345 5264 1346

Here, N = 8, ΣX = 51, ΣY = 198, ΣX2 = 345, ΣY2 = 5264, ΣXY = 1346
Now, regression coefficient Y on X

= − ( )( )
2 − ( )2

= 8 × 1346 − 51 × 198
8 × 345 − (51)2

= 10768 − 10098
2760 − 2601

= 670
159

= 4.2138

̅ = = 51 = 6.375 & ̅ = = 198 = 24.75
8 8

175

Regression equation y on x is Y – Y̅ = bYX(X-X̅ )
or Y – 24.75 = 4.2138 (X – 6.375)
or Y – 24.75 = 4.1238X – 26.863
or Y = 4.1238X – 26.863 + 24.75
or Y = 4.1238X – 2.1130
Now, correlation coefficient

= √ 2 − − ( )2
− ( )2 √ 2

= √8 × 8 × 1346 − 51 × 198 (198)2
345 − (51)2 √8 × 5264 −

= √ 2760 10768 − 10098 − 39204
− 2601 √42112

= 670
√159 √2908

= 814
12.61 × 53.93

= 670
680

= 0.9853

There is highly positive correlation between salary and job performance.
Again, Probable error of correlation coefficient
. . ( ) = 0.6745 (1√− 2)
= 0.6745 (1 − (0√.98853)2)
= 0.6745 (1 −2.803.97)
= 0.6745 (20..8033)
= 0.6745 × 0.01 = 0.007

Again, 6.× . . = 6 × 0.007 = 0.0048

Since, r > 6×P.E., so r is significant.
Estimating salary when employee's performance is 10 and 2.
when X = 2, Y = 4.1238 × 2 – 2.1130 = 8.2476 – 2.1130 = 6.1346
when X = 10, Y = 4.1238 × 10 – 2.1130 = 41.38 – 2.1130 = 39.267

176

QN. 22) (2073, old) In a study, the following information were obtained

No. of observations 9

Sum of product of two variables income(X) and expenditure(Y) 731

Sum of square of income (X) 2085

Sum of squares of expenditure(Y) 285

Mean income 15

Mean expenditure 5

Find,

a) Two regression coefficients

b) Whether there exists any relationship between income and expenditure

c) Two regression lines

d) The most probable amount of expenditure when income is Rs. 200

Solution:

Here given

No. of observations(N) = 9

Sum of product of two variables income and expenditure (ΣXY) = 731
Sum of square of income (ΣX2) = 2085
Sum of squares of expenditure(ΣY2) = 285

Mean income (X̅ ) = 15

Mean expenditure (Y̅ ) = 5 ΣX
N
Now, we have mean income(̅X) =

or 15 = ΣX
9
∴ ΣX = 135
ΣX
Now, mean(̅X) = N

or 15 = ΣX
9
∴ ΣX = 135
ΣY
and mean expenditure(̅Y) = N

5 =
9
∴ = 45

a) Two regression coefficients

The regression coefficient Y on X is

= − ( )( )
2 − ( )2

= 9 × 731 − 135 × 45
9 × 2085 − (135)2

177

= 6579 − 6075
18765 − 18225

= 504
540

= 0.9333

The regression coefficient X on Y is

= − ( )( )
2 − ( )2

= 9 × 731 − 135 × 45
9 × 285 − (45)2

= 6579 − 6075
2565 − 2025

= 504
540

= 0.9333

b) Correlation coefficient

= √ 2 − − ( )2
− ( )2 √ 2

= √9 × 9 × 731 − 135 × 45 − (45)2
2085 − (135)2 √9 × 285

= √ 18765 6579 − 6075 − 2025
− 18225 √ 2565

= 504
√504 √540

= 504
540

= 0.9333

There is highly positive correlation between income and expenditure.

Again, Regression line Y on X is

Y – Y̅ = byx(X – X̅ )
or Y – 5 = 0.9333(X – 15)

or Y – 5 = 0.9333X – 13.9995

178

or Y = 0.9333X – 13.9995 + 5
or Y = 0.9333X – 8.9995

and Regression line X on Y is

X – X̅ = bxy(Y – Y̅ )
or X – 15 = 0.9333(Y – 5)
or X – 15 = 0.9333Y – 4.6665
or X = 0.9333Y – 4.6665 + 15
or X = 0.9333Y – 10.3335

d) Most probable expenditure when income is Rs.200

Y = 0.9333 × 200 – 8.9995 = 186.66 – 8.9995 = 177.6605

Most probable expenditure is Rs. 177.6605 when income is Rs.200

QN. 23) (2072) The data sales and promotion expenditure on a newly lunched product for 6 years

are given below

Year 2003 2004 2005 2006 2007 2008

Sales (in Rs.00, 000) 16 20 18 24 20 22

Promotion expansion (Rs. '000) 4 4 6 10 10 12

a) Calculate the two regression coefficients from the above data of sales and expenses.

b) Compute the correlation coefficient between the sales and promotional expenditure and

interpret the result.

c) Test the significance of the correlation coefficient.

d) Develop the estimating equation that describe the effect if promotional expanse on sales.

Estimate the expected sales if the promotional expanses is Rs. Rs. 20000

e) Explain the meaning of each parameter of the equation in terms of above information.

Solution: Let X denotes the promotional expanses and Y denotes the sales.

Promotion expansion(X) Sales(Y) X2 Y2 XY

4 16 16 256 64

4 20 16 400 80

6 18 36 324 108

10 24 100 576 240

10 20 100 400 200

12 22 144 484 264

179

ΣX = 46 ΣY = 120 ΣX2 = 412 ΣY2 = 2440 ΣXY = 956

a) Two regression coefficients

The regression coefficient Y on X is

= − ( )( )
2 − ( )2

= 6 × 956 − 46 × 120
6 × 412 − (46)2

= 5736 − 5520
2472 − 2116

= 216
356

= 0.6067

The regression coefficient X on Y is

= − ( )( )
2 − ( )2

= 6 × 956 − 46 × 120
6 × 2440 − (120)2

= 5736 − 5520
14640 − 14400

= 216
240

= 0.90

b) Correlation coefficient

= √ 2 − − ( )2
− ( )2 √ 2

= √6 × 6 × 956 − 46 × 120 − (120)2
412 − (46)2 √6 × 2440

= √ 2472 − 5736 − 5520 − 14400
2116 √ 14640

= 216
√356 √240

= 216
18.87 × 15.49

= 216
292.30

180

= 0.738

There is highly positive correlation between promotional expenditure and Sales.

c) Test the significance of the correlation coefficient.

Probable error of correlation coefficient
. . ( ) = 0.6745 (1√− 2)
= 0.6745 (1 − (√0.6738)2)
= 0.6745 (1 −20.4.55446)
= 0.6745 (02.4.4555)
= 0.6745 × 0.1857 = 0.125

Again, 6.× . . = 6 × 0.125 = 0.75

Since, r > 6×P.E., so nothing can be concluded.

d) Estimating the expected sales

̅X = ΣX = 46 = 7.667 and ̅Y = ΣY = 120 = 20
N 6 N 6

Regression equation Y on X

Y – Y̅ = byx(X – X̅ )

or Y – 20 = 0.6067(X – 7.667)

or Y – 20 = 0.6067X – 4.65

or Y = 0.6067X – 4.65 + 20

or Y = 0.6067X + 15.35

When X = 20,000

Y = 0.6067×20000 + 15.35 = 121340 + 15.35 = 121355.35
∴ Expected sales is Rs. 121355.35

e) byx= 0.6067 indicates the of change in sales when the unit change in the promotional

expenditure.

a = 15.35 indicates the sales when promotional expenditure is zero.

QN. 24) (2072) The data on the sales and frequency of advertisement of the company available for
the past 30 days were as follows.

Frequency of advertisement in electric media (per day)

181

Volume of sales per 0 2 4 6 8

day (unit in thousand)

0–5 2- - - -

5 – 10 - 4534

10 – 15 - - -46

15 - 20 - - - -2

a) Is there any significance relationship between frequency of advertisement in electronic

media and volume of sales?

b) Should be increase or decrease frequency of advertisement an electronic media to promote

his sales?

c) If he found relationship is significance is significance, what will be his sales per day

when the frequency of advertisement per day is 7?

Solution:

QN. 25) (2071) From the following bi-variate table, compute the correlation coefficient between

advertisement expenditure and sales revenue and test the significance of the result. Also, estimate

sales revenue when advertisement expenditure is Rs. 400,000.

Advertisement Sales revenue (in Rs. 000)

0 - 50 50 - 100 100 - 150 150 - 200 200 - 250

0 - 40 12 6 8 - -

40 – 80 2 18 4 5 1

80 - 120 - 8 10 2 4

120 - 160 - 1 10 2 1

140 - 180 - - 1 23

Solution:
QN. 26) (2071, Old) From the following bi-variate table, compute the regression coefficient,
coefficient of variance, and correlation coefficient. Estimate the sales revenue when the
advertisement cost is Rs. 90 thousand.

Sales revenue 5 - 15 Advertisement cost (in Rs. '000) 45 - 55
(in Rs. '000) 4 15 - 25 25 - 35 35 - 45 -
7 3
75 – 125 1- -
125 – 175 621

182

175 – 225 1 3 4 2 2

225 – 275 1 1 3 4 -

275 - 325 3 1 1 - -

Solution:

183

Chapter– 7

Time Series Analysis

Brief Answer Question

QN. 1) (2078) What are the components of a time series.
Solution: A time series is a sequence of observations of a variable made at regular points or
interval of time and arranged in chronological order. Time series analysis is one quantitative
method use to determine patterns in data collected over time. A time series depicts the relationship
between two variables, one of them is time. Mathematically, a time series is defined as the
functional relationship Y = f(t), where Y is the value of the variable under consideration at time t.
The there are four components of time series, they are;

d) Secular trend
The general tendency of the time series data to increase or decrease or stagnate
during a long period of time is called the secular trend or simply trend.

e) Seasonal variation
The patterns of changes caused by seasonal factors and custom such as climate,
weather conditions, festivals, etc. within that trend to be repeated from year to year
is called seasonal variation.

f) Cyclical fluctuation
Cyclic variation is the component of a time series that tend to oscillate above and
below the secular trend line for periods greater than one year. The cyclic movement
do not follow any regular pattern but move in somewhat unpredictable manner.

g) Irregular variation.
A condition in time series when the value of the variable is completely
unpredictable is called irregular or random variation. show variation do not show
any definite pattern like trend, cyclic, or seasonal.

QN.2) (2076) The quarterly sales of a certain product of four quarters for year 1982 to 1986 are

75.25, 177, 96.8, 177.5 respectively. What are the seasonal indices of different quarters?

Solution: Now average sales of four quarters

75.25 + 117 + 96.8 + 177.5 = 466.55 = 116.6375
4 4

Seasonal index of first quarter = 75.25 × 100 = 64.52
116.6375

Seasonal index of first quarter = 117 × 100 = 100.31
116.6375

Seasonal index of first quarter = 96.8 × 100 = 82.99
116.6375

Seasonal index of first quarter = 177.5 × 100 = 152.18
116.6375

QN.3) (2075) The following figures are giving relating to the output in a factory. Find the trend

values with the help of 3 yearly moving average method.

184

Year 2001 2002 2003 2004 2005 2006 2007

Output 5 8 10 7 12 11 14

Solution: Calculation of 3-yearly moving average and trend values

Year Output(Y) 3-yearly moving 3-yearly moving

total average

2001 5 - -

2002 8 23 7.67

2003 10 25 8.33

2004 7 29 9.67

2005 12 30 10.00

2006 11 37 12.33

2007 14 - -

QN.4) (2074) The personal direct Nepal Drug Limited recorded the average percentage absentee

rate for each quarter for a 4 years periods are 55, 67.5, 62.5, and 53, find the seasonal index.

Solution: Now average absentee rate

55 + 67.5 + 62.5 + 53 = 238 = 59.5
4 4

Seasonal index of first quarter = 55 × 100 = 92.44
59.5

Seasonal index of first quarter = 67.5 × 100 = 113.45
59.5

Seasonal index of first quarter = 62.5 × 100 = 105.04
59.5

Seasonal index of first quarter = 53 × 100 = 55.50
95.5

QN.5) (2073) The sales manager of Media home appliances has collection the data regarding unit

sales during the last 5 years 2009 to 2013 and calculated the values as a = 52.4 and b = 9.2 taking

deviation 2011. Estimate the sales for years 2015.

Solution: let the trend line be Y = a + bx where x = X – 2011

Y = 52.4 + 9.2X

Now, x = 2015 - 2011 = 4

Estimated sales for 2015 is Y = 52.4 + 9.2 × 4 = 89.2

185

QN.4) (2072) Consider the following straight line trend equation obtained from the following data
of annual profit (in Rs. 000') Y = 90 + 2X. Interpret the meaning of coefficient of 'X' of this
model. What is monthly increase in profit?
Solution: The coefficient X indicates the of change in annual profit.
The annual profit is increased by Rs. 2000 and monthly increase is Rs. 166.67
QN.6) (2072) The year of the origin of the following straight line trend equation of the profits in
000 Rs. is 2008.
The equation y = 6.5 + 3x
Estimate the profit for the the year 2014.
Solution: Now, x = X – 2008

= 2014 – 2008 = 6
y = 6.5 + 3 × 6 = 24.5
Estimate the profit for the the year 2014 is Rs. 24500
QN.7) (2071) In the following straight line trend equation of production (in 000 Rs.), year of the
origin is considered as 2004.
Y = 20 + 3.5X
Estimate the production for the year 2009.
Solution: Now, x = X – 2004

= 2009 – 2004 = 5
y = 20 + 3.5 × 5 = 37.5
Estimate the profit for the the year 2014 is Rs. 37500.
Descriptive Answer Question
QN.8) (2077) Fit straight line by the method of least square to the data given below. Also, find the
trend values and predict the sales for 2019.
Year 2012 2013 2014 2015 2016 2017 2018
Sales (000 units) 12 13 14 16 19 21 30

Solution:
Let the trend line be y = a + b x, where x = X - Middle-year
Calculation of trend line

Year(X) Sales(y) x=X- 2 xy Trend values
2015 = 89 + 2

186

2012 12 -3 9 -36 11.86
2013 13 -2 4 -26 13.86
2014 14 -1 1 14 15.86
2015 16 0 0 0 17.86
2016 19 1 1 19 19.86
2017 21 2 4 42 21.86
2018 30 3 9 90 23.86
=125 = 0 2 =28 =103

, = = 125 = 17.86 = = 103 = 3.68
7 2 28
Thus, the trend line is y = a + bx

y = 17.86 + 2x

Again, trend values

when x = -3, y = 17.86 + 2 × -3 = 17.86 – 6 = 11.86

when x = -2, y = 17.86 + 2 × -2 = 17.86 – 4 = 13.86

when x = -1, y = 17.86 + 2 × -1 = 17.86 – 2 = 15.86

when x = 0, y = 17.86 + 2 × 0 = 17.86 – 0 = 17.86

when x = 1, y = 17.86 + 2 × 1 = 17.86 + 2 = 19.86

when x = 2, y = 17.86 + 2 × 2 = 17.86 + 4 = 21.86

when x = 3, y = 17.86 + 2 × 3 = 17.86 + 6 = 23.86

Estimated sales for 2019
when x = 4, x = 2019 – 2015 = 4
yc = 17.86 + 2 × 4 = 17.86 + 8 = 25.86
Therefore, required estimated sales for 2019 is Rs.25860.
QN.9) (2076) From the following annual data of sales (in 000 Rs.), find the trend values by using
least square method. Also estimate the sales for 2014.
Solution:
Let the trend line be y = a + b x, where x = X - Middle-year
Calculation of trend line

Age of Sales(y) x=X- 2 xy Trend values = 89 + 2
students 2015
12 9 -36
2012 13 -3 4 -26
2013 14 -2 1 14
2014 16 -1 0 0
2015 19 0 1 19
2016 1

187

2017 21 2 4 42
2018 30 3 90
=125 = 0 9 =103
2 =28

, = = 125 = 17.86 = = 103 = 3.68
7 2 28
Thus, the trend line is y = a + bx

y = 17.86 + 2x

Again, trend values

when x = -3, y = 17.86 + 2 × -3 = 17.86 – 6 = 11.86

when x = -2, y = 17.86 + 2 × -2 = 17.86 – 4 = 13.86

when x = -1, y = 17.86 + 2 × -1 = 17.86 – 2 = 15.86

when x = 0, y = 17.86 + 2 × 0 = 17.86 – 0 = 17.86

when x = 1, y = 17.86 + 2 × 1 = 17.86 + 2 = 19.86

when x = 2, y = 17.86 + 2 × 2 = 17.86 + 4 = 21.86

when x = 3, y = 17.86 + 2 × 3 = 17.86 + 6 = 23.86

Estimated sales for 2019
when x = 4, x = 2019 – 2015 = 4
yc = 17.86 + 2 × 4 = 17.86 + 8 = 25.86

Therefore, required estimated sales for 2019 is Rs.25860.

QN. 10) (2074) Find the seasonal index using simple index method and comment the result.

Year Spring Summer Fail Winter

2012 8 10 7 5

2013 9 10 7 6

2014 10 11 7 6

2015 10 12 8 7

2016 11 13 9 8

Solution: Calculation of seasonal indices

Year Spring Summer Fail Winter

2012 8 10 7 5

2013 9 10 7 6

2014 10 11 7 6

2015 10 12 8 7

2016 11 13 9 8

Total 48 56 38 32

188

Average 9.6 11.2 7.6 6.4
S.I. 110.34 128.74 87.56 73.56

Now, average of averages = 9.6 + 11.2 + 7.6 + 6.4 = 8.7
4

Seasinal index of Spring = Average of Spring × 100 = 9.6 × 100 = 110.34
Average of averages 8.7

Seasinal index of Spring = Average of Summer × 100 = 11.2 × 100 = 128.74
Average of averages 8.7

Seasinal index of Spring = Average of Fail × 100 = 7.6 × 100 = 87.36
Average of averages 8.7

Seasinal index of Spring = Average of Winter × 100 = 6.4 × 100 = 73.56
Average of averages 8.7

QN. 11) (2073) Bhat Bhateni Departmental store has been expending market share during the past

7 years, posting the following gross sales in million rupees.

Year: 2006 2007 2008 2009 2010 2011 2012

Sales: 148 207 246 329 378 476 517

a) Find the linear estimating equation that describe the data.

b) Estimate the sales of the year 2014.

c) Do these figure show a rising trend or falling trend? How do you arrive at your

conclusion?

Solution:
Let the trend line be y = a + b x, where x = X - Middle-year
Calculation of trend line

Year(X) Sales(y) x = X - 2009 2 xy
2006 148 -3 9 - 444
2007 207 -2 4 - 414
2008 246 -1 1 - 246
2009 329 0 0
2010 378 1 1 0
2011 476 2 4 378
2012 517 3 9 952
2 =28 1551
= 2301 = 0 = 1777
189

, = = 2301 = 328.71 = = 1777 = 63.46
7 2 28
a) Equation of linear estimating is y = a + bx

y = 328.71 + 63.46x

b) Estimate the sales of the year 2014

x = 2009 – 2015 = 5

y = 328.71 + 63.46 × 5 = 646.01

Therefore, required estimated sales for 2014 is Rs.646.01.

c) As b = 63.46 > 0, which shows the increasing trend of sales.

QN. 12) (2073, old) Kathmandu Metropolitan Police is studying the number of traffic fatalities

resulting from truck driving during 2005 – 2011.

Year: 2005 2006 2007 2008 2009 2010 2011

No of deaths 75 90 94 91 85 90 98

Compute the trend value using least square method and estimate the number of deaths in the year

2015.

Solution:
Let the trend line be y = a + b x, where x = X - Middle-year
Calculation of trend line

Year(X) Sales(y) x = X - 2009 2 xy
2006 75 -3 9 - 225
2007 90 -2 4 - 180
2008 94 -1 1 - 94
2009 91 0 0
2010 85 1 1 0
2011 90 2 4 85
2012 98 3 9 180
2 =28 294
= 623 = 0 = 60

Now, a = Σy = 623 = 89 and b = Σxy = 60 = 2.143
n 7 Σx2 28
Now, the trend line is y = a + bx

y = 89 + 2.14x

Again, trend values
when x = -3, y = 89 + 2.143 × -3 = 89 – 6.42 = 82.57
when x = -2, y = 89 + 2.143 × -2 = 89 – 4.28 = 84.712
when x = -1, y = 89 + 2.143 × -1 = 89 – 2.143 = 86.857

190

when x = 0, y = 89 + 2.143 × 0 = 89 – 0 = 89.00

when x = 1, y = 89 + 2.143 × 1 = 89 + 2.1429 = 91.1429

when x = 2, y = 89 + 2.143 × 2 = 89 + 4.2858 = 93.2858

when x = 3, y = 89 + 2.143 × 3 = 89 + 6.447 =95.447

Estimate the sales of the year 2015

x = 2009 – 2015 = 65

y = 89 + 2.1429 × 6 = 101.86

Therefore, required estimated sales for 2015 is Rs.101.59

QN. 13) (2072)

Compute the index for the following data, assuming that there is no need to adjust the data for the

trend.

Quarters/Years 1988 1989 1990 1991

I 3.5 3.5 3.5 4.0

II 3.9 4.1 3.9 4.6

III 3.4 3.7 3.7 3.8

IV 3.6 4.8 4.0 4.5

Solution: Computation of indices

Years/Quarters I II III IV
1988 3.5 3.9 3.4 3.6
1989 3.5 4.1 3.7 4.8
1990 3.5 3.9 3.7 4.0
1991 4.0 4.6 3.8 4.5
Total 14.5 16.5 14.6 16.9
Average 3.625 4.125 3.65 4.225

Now, average of averages = 3.625 + 4.125 + 3.65 + 4.225 = 15.625 = 3.90625
4 4

Seasinal index of I quater = Average of I quater × 100 = 3.625 × 100 = 92.80
Average of averages 3.90625

Seasinal index of II quater = Average of II quater × 100 = 4.125 × 100 = 105.60
Average of averages 3.90625

Seasinal index of III quater = Average of III quater × 100 = 3.65 × 100 = 93.44
Average of averages 3.90625

Seasinal index of IV = Average of IV quater × 100 = 4.225 × 100 = 108.16
Average of averages 3.90625

QN. 14) (2072, old) Compute the seasonal index by using for the following data and state which

quarter is seasonally high, justify the solution.

Quarters/Years 2008 2009 2010 2011
I - 3.5 3.5 4.0
II 3.9 4.1 3.9 4.6
III 3.4 3.7 3.7 3.8

191

IV 3.6 4.8 4.0 -

Solution: Computation of indices III IV
3.4 3.6
Years/Quarters I II 3.7 4.8
2008 - 3.9 3.7 4.0
2009 3.5 4.1 3.8 -
2010 3.5 3.9 14.6 12.4
2011 4.0 4.6 3.65 4.133
Total 11.0 16.5
Average 3.667 4.125

Now, average of averages = 3.667 + 4.125 + 3.65 + 4.133 = 15.575 = 3.894
4 4

Seasinal index of I quater = Average of I quater × 100 = 3.667 × 100 = 94.17
Average of averages 3.894

Seasinal index of II quater = Average of II quater × 100 = 4.125 × 100 = 105.93
Average of averages 3.894

Seasinal index of III quater = Average of III quater × 100 = 3.65 × 100 = 93.73
Average of averages 3.894

Seasinal index of IV = Average of IV quater × 100 = 4.133 × 100 = 106.14
Average of averages 3.894

QN. 15) (2072, old) Calculate the four yearly centered moving and short-term fluctuation for the

following data.

Year 2001 2002 2003 2004 2005 2006 2007 2008

Profit (000) 506 620 673 588 696 1116 738 663

Solution: Calculation of four yearly moving average

Year Profit 4-yearly 4-yearly moving Central moving Short term
fluctuation ( −
(Rs'000) moving total average average )

2001 506 - - - -

-- - -

2002 620 - - - -

2,387 596.75 - -

2003 673 - - 620.49 52.51

2,577 644.25 - -

192

2004 588 - - 706.25 -118.25
2005 696 3073 768.25 - -
2006 1116 - - 776.375 -80.375
2007 738 3138 784.5 - -
2008 663 - - 793.875 322.125
3213 803.25 - -
- - - -
- - - -
- - - -

193

QN. 15) (2071) From the data given below, fit the straight trend by the method of least square.

Also calculate trend value.

Year: 2006 2007 2008 2009 2010

Profit (in Rs.'000) 12 18 20 23 27

Also, estimate the profit for the year 2012.

Solution:
Let the trend line be y = a + b x, where x = X - Middle-year
Calculation of trend line

Year(X) Profit (y) x = X - 2008 2 xy
2006 12 -2 4 -24
2007 18 -1 1 - 18
2008 20 0 0 0
2009 23 1 1 23
2010 27 2 4 54
2 =10 = 35
= 100 = 0

, = = 100 = 20 = = 35 = 3.5
5 2 10
Now, the trend line is y = a + bx

y = 20 + 3.5x

Again, trend values
when x = -2, y = 20 + 3.5 × -2 = 20 – 7 = 13
when x = -1, y = 20 + 3.5 × -1 = 20 – 3.5 = 16.5
when x = 0, y = 20 + 3.5 × 0 = 20 – 0 = 20
when x = 1, y = 20 + 3.5 × 2 = 20 + 7 = 27

when x = 2, y = 20 + 3.5 × 1 = 20 + 3.5 = 23.5

Estimate the sales of the year 2012
x = 2008 – 2012 = 4
y = y = 20 + 3.5 × 4 = 20 + 14 = 34

Therefore, required estimated sales for 2012 is Rs.34000.

194

QN. 16) (2071, old) Fit the straight trend by the method of least square and obtain the trend values

from the following data.

Year: 2005 2006 2007 2008 2009

Production in '000 tons 100 120 110 140 80

Also, predict the production for the year 2015.

Solution:
Let the trend line be y = a + b x, where x = X - Middle-year
Calculation of trend line

Year(X) Production(y) x = X - 2007 2 xy
2005 100 -2 4 - 200
2006 120 -1 1 - 120
2007 110 0 0
2008 140 1 1 0
2009 80 2 4 140
2 =10 160
= 550 = 0 = -20

, = = 550 = 110 = = −20 = −2
5 2 10
Now, the trend line is y = a + bx

y = 110 - 2x

Again, trend values

when x = -2, y = 110 – 2 × -2 = 110 + 4 = 114
when x = -1, y = 110 – 2 × -1 = 110 + 42= 112
when x = 0, y = 110 – 2 × 0 = 110 + 0 = 110
when x = 1, y = 110 – 2 × 1 = 110 - 2 = 108
when x = 2, y = 110 – 2 × 2 = 110 - 4 = 106
Estimate the production of the year 2015

x = 2007 – 2015 = 8
y = 110 – 2 × 8 = 110 + 16 = 94
Therefore, required estimated production for 2015 is Rs.94000.

QN. 17) (2070) Assuming that trend is absent, determine the seasonality, if any in the data given

below and state which quarter of the year is seasonality high.

Years 1st 2nd 3rd 4th

2006 172 168 180 170

2007 176 170 190 174

2008 174 166 300 180

2009 176 174 200 178

195

2010 178 174 250 182

Total 876 852 1120 884

Average 219 213 280 221

Solution:

Now, average of averages = 219 + 213 + 280 + 221 = 933 = 233.25
4 4

Seasinal index of I quater = Average of I quater × 100 = 219 × 100 = 93.89
Average of averages 233.25

Seasinal index of II quater = Average of II quater × 100 = 213 × 100 = 91.32
Average of averages 233.25

Seasinal index of III quater = Average of III quater × 100 = 280 × 100 = 120.04
Average of averages 233.25

Seasinal index of IV = Average of IV quater × 100 = 221 × 100 = 94.75
Average of averages 233.25

Therefore 3rd season has high seasonality.

Analytical Answer Question

QN. 18) (2075) The following are the profit in thousands in a certain business.

Year 2001 2002 2003 2004 2005 2006 2007

Profit (000Rs.) 60 72 75 65 80 85 97

a) Fit a straight line trend by the method of least square and tabulate the trend values and

short time fluctuation.

b) What is the monthly increase in profit?

c) Estimate the profit for 2014.

Solution: Let the trend line be y = a + b x, where x = X - Middle-year i.e. X - 2004
Calculation of trend line

Short term fluctuation

Year(X) Profit(y) x 2 xy Trend values Multiplicative
model ( × 100)
(yc) Additive
model (y – yc)

2001 60 -3 9 -180 61.08 98 - 1.08
2002 72 -2
2003 75 -1 4 -144 66.15 108.84 5.85
2004 65 0
2005 80 1 1 -75 71.83 104.41 3.17

00 76.29 103.34 -11.29

1 80 81.99 97.57 -1.99

196

2006 85 2 4 170 86.43 98.34 -1.43
2007 97 106.01 5.5
3 9 291 91.5

Σy= 534, Σx=0, Σx2 =28, Σxy= 142
534 2 162
, = = 7 = 76.29 = = 28 = 5.07

a) Now, the trend line is y = a + bx

y = 76.29 + 5.43x

Again, trend values

when x = -3, y = 76.29 + 5.07 × -3 = 61.08

when x = -2, y = 76.29 + 5.07 × -2 = 66.15

when x = -1, y = 76.29 + 5.07 × -1 = 71.83

when x = 0, y = 76.29 + 5.07 × 0 = 76.29

when x = 1, y = 76.29 + 5.07 × 1 = 81.99

when x = 2, y = 76.29 + 5.07 × 2 = 86.43

when x = 3, y = 76.29 + 5.07 × 3 = 91.5

b) Here yearly increase in profit is Rs. 5070

c) Hence, monthly increase in profit = Rs. 50700/12 = 422.50

Estimate the production of the year 2014

x = 2004 – 2014 = 10

y = 76.29 + 5.07 × 10 = 126.99

Therefore, required estimated production for 2014 is Rs.126990.

QN. 19) (2074) The tourist industry to enormous seasonal variation. The Fulbari Resort in

Pokhara has recorded its occupancy rate percentage of total rooms for each quarter during last 4

years. These data are shown in the following table.

Year Q1 Occupancy Rate % Q4
2008 56 Q2 Q3 60
70 80

2009 57 73 86 62

2010 60 74 90 65

2011 66 83 87 87

Analyze the quarterly time series to determine the effect of seasonal component by ratio moving

average method using multiplication Model.

Solution: Calculation of ratio to moving average

Year Quarters Occupancy 4-quar. 4-quart. 2-period 4-quar. Ratio to
M.T. M.A. = ×
rate (y) M.T. M.A. Centered M.A. 100
( )

2008 I 56

197

II 57 239 59.75 97.561
III 60 123 61.5 101.1494
IV 66 101.4493
2009 I 70 253 63.25 100.1715
II 73 130.5 65.25 97.0492
III 74 104.8973
IV 83 269 67.25 96.6767
2010 I 80 138 69 100.8798
II 86 108.1081
III 90 283 70.75 111.8971
IV 87 145.75 72.875 83.7696
2011 I 60 90.5109
II 62 300 75
III 65 152.5 76.25
IV 87
310 77.5
158.25 79.125

323 80.75
165.5 82.75

339 84.75
170.5 85.25

343 85.75
166.5 83.25

323 80.75
155.5 77.75

299 74.75
143.25 71.625

274 68.5
137 68.5

274 68.5

198

Calculation of Seasonal Index

Trend eliminated values

year 1st quart. 2nd quart. 3rd quart. 4th quart Total
2008 97.561 101.1494
2009 -- 97.0492 104.8973 398.04
2010 108.1081 111.8971 399.99
2011 101.4493 100.1715 - -
Total 302.7183 317.9438
S.I. 96.6767 100.8798 100.9061 105.9813
Adjusting S.I. 101.40 106.50
83.7696 90.5109

281.8956 291.5622

93.9652 97.1874

94.43 97.66

Since S.I. is not equal to 400, so adjustment is necessary.
Adjusting factor k= 400/398.04=1.0049

QN. 20) (2072) The following table represents the annual trend of profit of two companies seeking

investment for their development project. In which company would you invest money. Justify

your solution by using necessary statistical tool.

Net profit in million Rs.

Year Company A Company B

2001 16 16

2002 32 16

2003 40 22

2004 24 36

2005 40 40

2006 32 44

2007 88 48

Solution: Calculation of growth rate and coefficient of variance
Let x = X – Middle year i. e. X = 2004

Company A Company B

Year(X) x x2 Net Profit 12 1 Net Profit 22 2
2001 -3 (y1) (y2) - 48

9 16 256 - 48 16 256

199