\(\overline {\overline{X}} = \frac {5.10+4.98+5.02+4.96+4.96+5.04+4.94+4.92+4.92+4.98}{10}\)
\(\overline {\overline{X}}\) = \(\frac {49.82}{10} = 4.982\)
\(\overline {\overline{R}} = \frac {0.3+.4+.2+.4+.1+.1+.8+.5+.3+.5}{10}\)
\(\overline {R}\) = \(\frac {3.6}{10}\) = 0.36
The control limits of mean chart are
UCL  = \(\overline {\overline{X}} + A_{2} \overline{R}\)
= 4.982 + .577(.36)
= 4.982+ .208 = 5.19
CL = 4.982
LCL =  \(\overline {\overline{X}} - A_{2} \overline{R}\)
= 4.982 - .208 = 4.774
The control limits of R-chart are
UCL = D₄\(\overline {R}\) = 2.114 (.36) = 0.761
CL = \(\overline {R}\)  = .36
LCL = D₃\(\overline {R}\)  = 0

Mean Chart

 Range chart

Conclusion: The above diagram shows all the control lines with the data points plotted.
Since one in Range chart, lie outside the control limits, we can say that the process is out of control.

\(\overline {\overline{X}} = \frac {1080+1390+1460+1380+1230+1370+1310+1630+1580+1510+1270+1200}{12}\)
\(\overline {\overline{X}}\) = \(\frac {16410}{12}\) = 1367.5
\(\overline {\overline{R}} = \frac {410+670+180+320+690+450+380+350+270+660+440+310}{12}\)
\(\overline {R}\) =  \(\frac {5130}{12}\) = 427.5
The control limits of mean chart are
UCL = \(\overline {\overline{X}} + A_{2} \overline{R}\)
= 1367.5 + 0.577(427.5)
= 1367.5 + 246.67 = 1614.17
CL =  \(\overline {\overline{X}}\) = 1367.5
LCL  =  \(\overline {\overline{X}} - A_{2} \overline{R}\)
= 1367.5 - 246.67 = 1120.83
The control limits of range chart are
UCL = D₄\(\overline {R}\) = 2.114 (427.5) = 903.74
CL = \(\overline {R}\)  = 427.5
[For n=5, A₂  = 0.577, D₃  = 0, D₄ = 2.114]
LCL = D₃\(\overline {R}\)  = 0
\(\overline {X}\) - chart

 
R-chart


Conclusion: The above diagram shows all the control lines with the data points plotted.
Since one point in mean chart, lie outside the control limits, we can say that the process is out of control.

Sample No.	Observation					Total	x	R = Xmax - Xmin
	1	2	3	4	5
1	50	55	52	49	54	260	52	55-49 = 6
2	51	50	53	50	46	250	50	53-46 = 7
3	50	53	48	52	47	250	50	53-47 = 6
4	48	53	50	51	53	255	51	53-48 = 5
5	46	50	44	48	47	235	47	50-44 = 6
6	55	51	56	47	51	260	52	56-47=9
7	45	48	53	48	51	245	49	53-45=8
8	50	56	54	53	57	270	54	57-50 = 7
9	47	53	49	52	54	255	51	54-47 = 7
10	56	53	55	54	52	270	54	56-52 = 4
							510	65

\(\overline {\overline{X}}\) = \(\frac {510}{10}\) = 51
 \(\overline {R}\)  = \(\frac {65}{10}\) = 6.5
The control limits of mean chart are
UCL =  \(\overline {\overline{X}} + A_{2} \overline{R}\) 
= 51 + 0.577 (6.5)
= 51 + 3.75 = 57.75
CL =  \(\overline {\overline{X}}\) = 51
LCL    =  \(\overline {\overline{X}} - A_{2} \overline{R}\)
= 51 - 3.75 = 47.25
The control limits of range chart are
UCL = D₄\(\overline {R}\) = 2.114 (6.3) = 13.32
CL =  \(\overline {R}\) = 6.5 
[For n=5, A₂  = 0.577, D₃  = 0, D₄ = 2.114]
LCL = D₃\(\overline {R}\)  = 0

Commodities	Price		Quantity
	Base year (p0)	Current year (p1)	q0	q1
Wheat	6	10	50	56
Ghee	2	2	60	60
Firewood	4	6	60	60
Sugar	10	12	30	24
Cloth	8	12	40	36

p0q0	p1q1	p0q1	p1q0
300	560	336	500
200	240	240	200
240	360	240	360
300	288	240	360
320	432	288	480
1360	1880	1344	1900

Fisher's price index number
\(P^{F}_{01}\) = \(\sqrt{ \frac {\sum p_{1}q_{0}}{\sum p_{0}q_{0}} \times ​\frac{​ {\sum p_{1}q_{1}} }{{\sum p_{0}q_{1}}}{}}\times 100\)
= \(\sqrt\frac {1900 \times 1880}{1360 \times 1344} \times 100\)
= \(\sqrt\frac {3572000}{1827840} \times 100\) 
= \(\sqrt {1.954} \times 100\)
\(P^{F}_{01}\)  = 139.8
Time reversal test:
\(P _{01} \times P_{10}\) = \(\sqrt{ \frac {\sum p_{1}q_{0}\times \sum p_{1}q_{1}\times \sum p_{0}q_{1}\times \sum p_{0}q_{0}}{\sum p_{0}q_{0}\times \sum p_{0}q_{1}\times \sum p_{1}q_{1} \times \sum p_{1}q_{0}} ​}\)
=  
=  \(\sqrt {1}\)  = 1
∴    \(P _{01} \times P_{10}\) = 1
Factor reversal test:
\(P _{01} \times Q_{01}\) = \(\sqrt{ \frac {\sum p_{1}q_{0}\times \sum p_{1}q_{1}\times \sum q_{1}p_{0}\times \sum q_{1}p_{1}}{\sum p_{0}q_{0}\times \sum p_{0}q_{1}\times \sum q_{0}p_{0} \times \sum q_{0}p_{1}} ​}\)

= \(\frac {1880}{1360} = \frac {\sum p_{1}q_{1}}{\sum p_{0}q_{0}}\)
Hence, it satisfies time reversal test and factor reversal test.

Commodities	Base year		Current year
	p0	q0	p1	q1
A	170	562	72	632
B	192	535	70	756
C	195	639	95	926
D	187	128	92	255
E	185	542	92	632
F	150	217	180	314
7	12.6	12.7	12.5	12.8
8	12.4	12.3	12.6	12.5
9	12.6	12.5	12.3	12.6
10	12.1	12.7	12.5	12.8

p0q0	p1q1	p0q1	p1q0
95540	45504	107440	40464
102750	52920	145152	37450
124605	87970	180570	60705
100270	58144	116920	49864
32550	56520	47100	39060
160.02	160	161.28	158.75
152.52	157.5	155	154.98
157.5	154.98	158.76	153.75
153.67	160	154.88	158.75
480244.7	325150.48	645496.92	239945.23

Laspeyre's pnce index number
\(P^{L}_{01}\) = \(\frac {\sum p_{1}q_{0}}{\sum p_{0}q_{0}} \times 100\)
= \(\frac {239945.23}{480244.71} \times 100 \)
Paasches index number
\(P^{p}_{01}\) =  \(\frac {\sum p_{1}q_{0}}{\sum p_{0}q_{0}} \times 100\)
= \(\frac {325150.48}{645496.92} \times 100 \)
= 50.37
Fisher's index number
\(P^{F}_{01}\)  = \(\sqrt{ \frac {\sum p_{0}q_{0}}{\sum p_{0}q_{0}} \times ​\frac{​ {\sum p_{0}q_{0}} }{{\sum p_{0}q_{0}}}{}}\times 100\)
= \(\sqrt\frac {239945.23 \times 325150.48}{480244.71 \times 645496.92} \times 100\)
= \(\sqrt\frac {489.84 \times 570.22}{693 \times 803.43} \times 100\)
= 50.1

Year X	Sales Y	X = \(\frac {x-1983.5}{0.5}\)	XY	X2
1980	50.3	-7	-352.1	49
1981	52.7	-5	-263.5	25
1982	49.3	-3	-147.9	9
1983	57.3	-1	-57.3	1
1984	56.8	1	56.8	1
1985	60.7	3	182.1	9
1986	62.1	5	310.5	25
1987	58.7	7	410.9	49
	447.9	0	139.5	168

Since \(\sum X = 0, a = \frac {\sum Y}{n} = \frac {447.9}{8} = 55.9875\)
b  = \(\frac {\sum XY}{\sum X^{2}} = \frac{139.5}{168}\)  = 0.83035
∴  The straight line trend is obtained by
Y = a + bX  ⇒ Y = 55.9875 + 0.830 \((\frac {x-1983.5}{0.5})\)
= 55.9875 + 0.8304 (-5) 
= 55.9875 - 4.152  
= 51.8355

Time	Weight in ml					Total	\(\overline {X}\)	R = x max - xmin
	1	2	3	4	5
8.00 AM	43	41	42	43	41	210	\(\frac {210}{5} = 42\)	43-41 = 2
9:00 AM	40	39	40	39	44	202	\(\frac {202}{5} = 40.4\)	43-41 = 2
10.00 AM	42	42	43	38	40	205	\(\frac {205}{5} = 41\)	43-38 = 5
11:00 AM	39	43	40	39	42	203	\(\frac {203}{5} = 40.6\)	43-39 = 4
							164	16

\(\overline {\overline{X}}\)  =  \(\frac {164}{4}\) = 41
\(\overline {R}\)  = \(\frac {16}{4}\) = 4
Control limits for \(\overline {X}\) - chart
UCL = \(\overline {\overline{X}} + A_{2} \overline {R}\)
= 41 + 0.577 (4)
= 41 + 2.308 = 43.31
CL =  \(\overline {\overline{X}} \) = 41
[when n = 5, A₂= 0.577, D₃ = 0, D₄ = 2.114]
LCL = \(\overline {\overline{X}} - A_{2} \overline {R}\)
= 41 - 2.308 = 38.69
The control limits of R-chart are
UCL =  D₄ \(\overline{R}\) = 2.114 (4) = 8.44
CL =  \(\overline{R}\) = 4
LCL = D3 \(\overline{R}=0\)

\(\overline {\overline {X}} = \frac{12+13+11+12+14+13+16+15}{8}\) = \(\frac {106}{8}\) = 13.25
\(\overline{R}\)  =  \( \frac{2+5+4+2+3+2+4+3}{8}\)
\(\overline{R}\)  = \(\frac {25}{8}\) = 3.12
Control limits for mean chart
UCL = \(\overline {\overline{X}} + A_{2} \overline {R}\) 
= 13.25+.729(3.12)
= 13.25+2.27 = 15.52
CL = \(\overline {\overline{X}}\)  = 13.25
LCL = \(\overline {\overline{X}} - A_{2} \overline {R}\)
= 13.25-2.27 =10.98
Control limits.for R-chart
UCL = D₄\(\overline {R}\) = 2.282 = (3.12)
= 7.12
CL = \(\overline {R}\) = 3.12
LCL = D₃\(\overline {R}\) = 0
Mean (\(\overline {X}\)) chart

R chart

Conclusion:  The above diagram shows all the control lines with the data points plotted.
Since one point in X - chart, lie out of the control lines, we can say that the process is out of control.

\(\overline {\overline {X}} = \frac{29+26+37+34+14+45+39+20+34+23}{10}\) = \(\frac {301}{10}\) = 30.1
\(\overline{R}\)  =  \( \frac{39+10+39+17+12+20+5+21+23+15}{10}\)
\(\overline{R}\)  = \(\frac {201}{10}\) = 20.1
Control limits for mean chart
UCL = \(\overline {\overline{X}} + A_{2} \overline {R}\) 
= 30.1+.729(20.1)
= 30.1 + 14.65 = 44.75
CL = \(\overline {\overline{X}}\)  = 30.1
LCL = \(\overline {\overline{X}} - A_{2} \overline {R}\)
= 30.1 - 14.65 =15.45
Control limits.for R-chart
UCL = D₄\(\overline {R}\) = 2.282 = (20.1)
= 45.86
CL = \(\overline {R}\) = 20.1
LCL = D₃ \(\overline {R}\) = 0
Mean (\(\overline {X}\)) chart

Mean (\(\overline {R}\)) chart

Conclusion: The above diagram shows all the control lines with the data points plotted.
Since one point in \(\overline {X}\) - chart, lie out of the control lines, we can say that the process is out of control.

Sample No	Observations				Total	\(\overline {X}\)	R = x max - xmin
	1	2	3	4
1	12.5	12.3	12.6	12.7	50.1	\(\frac {50.1}{4} = 12.5\)	12.7-12.3 = 0.4
2	12.8	12.4	12.4	12.8	50.4	\(\frac {50.4}{4} = 12.6\)	12.8-12.4 = 0.4
3	12.1	12.6	12.5	12.4	49.6	\(\frac {49.6}{4} = 12.4\)	12.6-12.1 = 0.5
4	12.2	12.6	12.5	12.3	49.6	\(\frac {49.6}{4} = 12.4\)	12.6-12.2 = 0.4
5	12.4	12.5	12.5	12.5	49.9	\(\frac {49.9}{4} = 12.5\)	12.6-12.4 = 0.1
6	12.3	12.4	12.6	12.6	49.9	\(\frac {49.9}{4} = 12.5\)	12.6-12.3 = 0.3
7	12.6	12.7	12.5	12.8	50.6	\(\frac {50.6}{4} = 12.7\)	12.6-12.5 = 0.3
8	12.4	12.3	12.6	12.5	49.8	\(\frac {49.8}{4} = 12.5\)	12.6-12.3 =.3
9	12.6	12.5	12.3	12.6	50	\(\frac {50}{4} = 12.5\)	12.6-12.3 =.3
10	12.1	12.7	12.5	12.8	50.1	\(\frac {50.1}{4} = 12.5\)	12.8-12.1 =.7
						125.1	3.7

\(\overline {\overline{X}} = \frac {125.1}{10}\) = 12.51
\(\overline{R} = \frac {3.7}{10}\) = 0.37
Control limits for \(\overline {X}\) - chart
UCL =  \(\overline {\overline{X}} + A_{2} \overline {R}\)
= [when n = 4, A₂ = .729]
= 12.51 + .729 (0.37)
= 12.51 + 0.27 = 12.78
CL = \(\overline {\overline{X}}\) = 12.51
LCL = \(\overline {\overline{X}} - A_{2} \overline {R}\)
= 12.51 - 0.27 = 12.24
Control limits for R-chart
UCL = D₄ \(\overline{R}\)
= 2.282(0.37) = 0.84
 [when n = 4, D₄ = 2.282]
CL = \(\overline{R}\) = 0.37
LCL = D₃ \(\overline{R}\) = 0
[When n = 4, D₃ = 0]
\(\overline {X}\) - chart

\(\overline {R}\) - chart

Conclusion: The above diagram shows all the control lines with the data points plotted.
Since all the points lie within the control limits, we can say that the process is in control.

\(\overline {\overline {X}} = \frac{11.2+ 11.8+ 10.8+ 11.6+ 11.0+ 9.6+ 10.4+9.6+ 10.6+ 10}{10}\) = \(\frac {106.6}{10}\) = 10.66
\(\overline {\overline {R}} = \frac{7+4+8+5+7+4+8+4+7+9}{10}\) = \(\frac {63}{10}\) = 6.3
Control limits for \(\overline {X}\) - chart
UCL = \(\overline {\overline{X}} + A_{2} \overline {R}\)
= 10.66 + 0.58 (6.3)
= 10.66 + 3.654 = 14.31
CL = \(\overline {\overline {X}}\) = 10.66
LCL = \(\overline {\overline{X}} - A_{2} \overline {R}\)
= 10.66 - 3.654 = 7.006
Control limits for R-chart
UCL = D₄\(\overline {R}\) = 2.115(6.3) = 13.3
CL = \(\overline {R}\) = 6.3
LCL = D₃\(\overline {R}\) = 0
Mean (\(\overline {X}\)) chart

R chart

Conclusion: The above diagram shows all the control lines with the data points plotted. Since all the points lie within the control limits, we can say that the process is in control.

Sample No	Observations			Total	\(\overline {X}\)	R = x max - xmin
	1	2	3
1	32	36	42	110	\(\frac {100}{3} = 36.7\)	42-32 = 10
2	28	32	40	100	\(\frac {100}{3} = 33.3\)	40-28 = 12
3	39	52	28	119	\(\frac {119}{3} = 39.7\)	52 -28 = 24
4	50	42	31	123	\(\frac {123}{3} = 41\)	50-31 = 19
5	42	45	34	121	\(\frac {123}{3} = 40.3\)	45-34 = 29
6	50	29	21	100	\(\frac {100}{3} = 33.3\)	50-21 = 29
7	44	52	35	131	\(\frac {131}{3} = 43.7\)	52-35 = 17
8	22	35	44	101	\(\frac {101}{3} = 33.7\)	44-22 = 22
					301.7	144

\(\overline {\overline{X}}\)   =  \(\frac {301.7}{8}\) = 37.71
\(\overline{R}\) = \(\frac {144}{8}\) = 18
Control limits for  \(\overline {X}\) - chart
UCl =  \(\overline {\overline{X}} + A_{2} \overline {R}\)
= 37.71 + 0.58 (18)
= 37.71 + 18.44
[when n = 3, A₂ = 1.023]
UCL = 56.12
CL =  \(\overline {\overline{X}}\) = 37.71
LCL =  \(\overline {\overline{X}} - A_{2} \overline {R}\)
= 37.71 - 18.44
LCL = 19.28
Control limits for R-chart
UCL = D₄ \(\overline {R}\) = 2.2574(18) = 46.33
CL = \(\overline {R}\) = 18
LC = D3 \(\overline {R}\) = 0
when n = 3, D₄ = 2.574]
\(\overline {X}\) - chart

.\(\overline {R}\) - chart


Conclusion: The above diagram shows all the control lines with the data points plotted. Since all the points lie within the control limits, we can say that the process is in control.

\(\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|} \hline \text { Sample number } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & \text { Total } \\ \hline \bar{X} & 49 & 45 & 48 & 53 & 39 & 47 & 46 & 39 & 51 & 45 & 462 \\ \hline \mathbf{R} & 7 & 5 & 7 & 9 & 5 & 8 & 8 & 6 & 7 & 6 & 68 \\ \hline \end{array}\)
\(\overline{\bar{X}}=\frac{\sum \bar{X}}{10}=\frac{462}{10}=46.2 \)
\(\bar{R}=\frac{\sum R}{10}=\frac{68}{10}=6.8 \)
\(\text { The control limits for } \overline{\mathrm{X}} \text { chart is }\)
\(\mathrm{UCL}=\overline{\overline{\mathrm{X}}}+\mathrm{A}_{2} \overline{\mathrm{R}}=46.2+(0.58)(6.8)=50.14\)
\(\mathrm{CL}=46.2\)
\(\mathrm{LCL}=\overline{\mathrm{X}}-\mathrm{A}_{2} \overline{\mathrm{R}}=46.2-(0.58)(6.8)=42.26 \) 

The control limits for range chart is
\(\mathrm{UCL}=\mathrm{D}_{4} \overline{\mathrm{R}}=(2.115)(6.8)=14.38 \)
\(\mathrm{CL}=\overline{\mathrm{R}}=6.8 \)
\(\mathrm{LCL}=\mathrm{D}_{3} \overline{\mathrm{R}}=0(6.8)=0\)

From the \(\overline X\) chart, we see that 4 points are outside the control limit lines. So we say that the process is out of control.
Conclusion: The above diagram shows all the three control lines with the data points plotted, since 2 points fall out of the control limits, we can say that the process is out of control.

Sample No	\(\overline {X}\)	R
1	15	7
2	17	7
3	15	4
4	18	9
5	17	8
6	14	7
7	18	12
8	15	4
9	17	11
10	16	5

\(\overline {\overline {X}} = \frac{15+17+15+18+17+14+18+15+17+15+17+16}{10}\) = \(\frac {162}{10}\) = 16.2
\(\overline {R}\) =  \(\frac{7 + 7 + 4+ 9+ 8+ 12 + 7 + 4+ 11+ 15}{10}\)  = \(\frac {74}{10}\) = 7.4
Control limits for mean chart
UCL = \(\overline {\overline{X}} + A_{2} \overline {R}\)
= 16.2  + 0.58(7.4)
= 16.2  + 4.292 = 20.492
[∴ A₂  = 0.58]
CL = \(\overline {\overline{X}}\) = 16.2
LCL =  \(\overline {\overline{X}} - A_{2} \overline {R}\)
= 16.2 - 0.58(7.4)
= 16.2 - 4.292 = 11.91

Control limits for R-chart
UCL = D₄  \(\overline{R}\) 
= 2.115 (7.4) = 15.65
[∴ D₄  = 2.115]
CL =  \(\overline{R}\) = 7.4
LCL = D₃ \(\overline{R}\)
= 0(\(\overline{R}\)) = 0
[∴ D₃  = 0]

Conclusion : The above diagram shows all the control lines with the data points plotted. Since all the points lie within the control limits, we can say that the process is in control.

Commodity	2016		2017
	p0	q0	p1	q1
Food	40	12	65	14
Fuel	72	14	78	20
Clothing	36	10	36	15
Wheat	20	6	42	4
Others	46	8	52	6

p0q0	p1q1	p1q0	p0q1
480	910	780	560
1008	1560	1092	1440
120	168	252	80
368	312	416	276
2336	3490	2900	2896

Fisher's price index number 
\(P^{F}_{01}\) = \(\sqrt \frac {\sum p_{1}q_{0}\times \sum p_{1}q_{1}}{{\sum p_{0}q_{0}\times \sum p_{0}q_{1}}}\) × 100
= \(\sqrt \frac {2900\times3490}{2336\times2896} \times 100\)
= \(\sqrt \frac{10121000}{6765056} \times{100}\) 
= \(\sqrt{1.496}\times{100}\)
\(P^{F}_{01}\) = 122.31
Time Reversal Test
P₀₁ × P₁₀  =  \(\sqrt \frac {\sum p_{1}q_{0}\times \sum p_{1}q_{1}\times{\sum p_{0}q_{1}\times \sum p_{0}q_{0}}}{{\sum p_{0}q_{0}\times \sum p_{0}q_{1}\times \sum p_{1}q_{1}\times \sum p_{1}q_{0}}}\)
=  \(\sqrt \frac {2900\times3490\times{2896\times2336}}{2336\times2896\times3490\times2900}\)
= \(\sqrt {1}\) = 1
Hence, it satisfies time reversal test.

Commodity	p0	p1	q0	q1
A	5	4	10	12
B	8	7	6	7
C	6	5	3	4

p0q0	p0q1	p1q0	p1q1
50	60	40	48
48	56	42	49
18	24	15	20
116	140	97	117

Fisher's price index number
\(P^{F}_{01}\) =  \(\sqrt \frac {\sum p_{1}q_{0}\times \sum p_{1}q_{1}}{{\sum p_{0}q_{0}\times \sum p_{0}q_{1}}}\) × 100
=  \(\sqrt \frac {97\times117}{116\times140} \times 100\)
= \(\sqrt \frac {11349}{16240} \times 100\)
= \(\sqrt {0.6988}\times100\)
\(P^{F}_{01}\)   =  83.59

Commodity	Price in Rupees per unit		Number of units
	p0	p1	q0	q1
A	6	10	50	56
B	2	2	100	120
C	4	6	60	60
D	10	12	50	24
E	8	12	40	36

p0q0	p0q1	p1q0	p1q1
300	560	500	336
200	240	200	240
240	360	360	240
500	288	600	240
320	432	480	288
1560	1880	2140	1344

Fisher's price index number 
\(P^{P}_{01}\) =  \(\sqrt \frac {\sum p_{1}q_{0}\times \sum p_{1}q_{1}}{{\sum p_{0}q_{0}\times \sum p_{0}q_{1}}}\) × 100
= \(\sqrt \frac{2140\times1880}{1560\times1344} \times 100\)
= \(\sqrt \frac{4023200}{2096640} \times 100\)
= \(\sqrt {1.9188} \times 100\)
\(P^{F}_{01}\) = 138.5
Time reversal test : 
P₀₁ × P₁₀ =  \(\sqrt \frac{\sum p_{1}q_{0}\times{p_{1}q_{1}}\times{p_{0}q_{1}}\times{p_{0}q_{0}}}{{\sum p_{0}q_{0}\times{p_{0}q_{1}}\times{p_{1}q_{1}}\times{p_{1}q_{0}}}}\)
= \(\sqrt \frac{2140\times1880 \times 1344\times1560}{1560\times1344 \times 1880\times2140 }\)
= \(\sqrt 1\) = 1
Time reversal test = 1
Factor reversal test: 
P₀₁ × Q ₀₁  =   \(\sqrt \frac{\sum p_{1}q_{0}\times{p_{1}q_{1}}\times{p_{0}q_{1}}\times{p_{0}q_{0}}}{{\sum p_{0}q_{0}\times{p_{0}q_{1}}\times{p_{1}q_{1}}\times{p_{1}q_{0}}}}\)
\(=\sqrt \frac{2140\times1880 \times 1344\times1560}{1560\times1344 \times 1880\times2140 }\)
= \(\sqrt{(\frac{1880}{1560})^2}\)
=  \(\frac {1880}{1560}\) 
P₀₁ × Q₀₁   =  \({\frac {\sum p_{1}q_{1}}{\sum p_{0}q_{0}}}\)

Commodity	Price		Quantity
	2000 (p0)	2010 (p1)	2000 (q0)	2010 (q1)
A	12	14	18	16
B	15	16	20	15
C	14	15	24	20
D	12	12	29	23

p0q0	p0q1	p1q0	p1q1
216	192	252	224
300	225	320	240
336	280	360	300
348	276	348	276
1200	973	1280	1040

(i) Laspeyre's index number
\(P^{L}_{01}\)  =  \({\frac {\sum p_{1}q_{0}}{\sum p_{0}q_{0}}} \times 100\)
=  \(\frac {1280}{1200} \times {100}\) = 106.66
(ii) Paasches index number
\(P^{P}_{01}\) =  \({\frac {\sum p_{1}q_{1}}{\sum p_{0}q_{1}}} \times 100\)
=  \(\frac {1040}{973} \times {100}\) = 106.88
(iii) Fisher's index number
\(P^{F}_{01}\) =  \(\sqrt \frac {\sum p_{1}q_{0}\times \sum p_{1}q_{1}}{{\sum p_{0}q_{0}\times \sum p_{0}q_{1}}}\) × 100
=  \(\sqrt \frac{1280\times1040}{1200\times973} \times 100\)
=\(\sqrt \frac {1331200}{1167600} \times {100}\)
= \(\sqrt {1.1401} \times {100}\)
= 1.067 × 100
\(P^{F}_{01}\) =  106.7

Commodity	1995		2005
	Price(p0)	(q0)	Price	Quantity
A	5	60	15	70
B	4	20	8	35
C	3	15	6	20

p0q0	p0q1	p1q0	p1q1
300	350	900	1050
80	140	160	280
45	60	90	120
425	550	1150	1450

Laspeyre's price index number
\(P^{L}_{01}\) = \({\frac {\sum p_{1}q_{0}}{\sum p_{0}q_{0}}} \times 100\) 
= \(\frac {1150}{425} \times {100}\) = 270.58
Paasche's index number
\(P^{L}_{01}\) =  \({\frac {\sum p_{1}q_{0}}{\sum p_{0}q_{1}}} \times 100\)
=  \(\frac {1450}{550} \times {100}\) = 263.63

Year (X)	No. of Salesmen Y	X = x -1994	X2	XY
1992	46	-2	4	-92
1993	48	-1	1	-48
1994	42	0	0	0
1995	56	1	1	56
1996	52	2	4	104
	244	0	10	20

Smce \(\sum\)X = 0, a =\(\frac {\sum Y}{n}\) = \(\frac {244}{5}\) = 48.8
b = \(\frac {\sum XY}{\sum X^2}\) = \(\frac {20}{10}\) = 2
∴ The required equation of the straight line trend is given by
Y = a + bX  \(\Rightarrow \) Y = 48.8 + 2X
\(\Rightarrow \) Y = 48.8 +2 (X - 1994) .... (1)
∴ Number of salesmen in 1997 is put X = 1997 in (1)
∴ Y = 48.8 + 2 (1997 - 1994) 
= 48.8 + 2 (3) 
= 48.8 + 6 = 54.8
∴  Number of salesmen in 1997 is 54.8

Year	I Quarter	II Quarter	III Quarter	IV Quarter
2008	72	68	62	76
2009	78	74	78	72
2010	74	70	72	76
2011	76	74	74	72
2012	72	72	76	68
Total	372	358	362	364
Average	74.4	71.6	72.4	72.8

Grand average = \(\frac {74.4+ 71.6+ 72.4 + 72.8}{4}\) = \(\frac {291.2}{4}\) = 72.8
S.I for I Quarter = \(\frac {Average \ for \ I \ Quarter}{Grand \ average}\) × 100
= \(\frac {74.4}{72.8}\) × 100 = 102.19
S.I for II Quarter = \(\frac {71.6}{72.8}\) × 100 = 98.35
S.I for III Quarter =   \(\frac {72.4}{72.8}\) × 100 = 99.45
S.I for IV Quarter =   \(\frac {72.8}{72.8}\) × 100  = 100

Year	Jan	Feb	Mar	Apr	May	Jun	Jul	Aug	Sep	Oct	Nov	Dec
2002	15	18	17	19	16	20	21	18	17	15	14	18
2003	20	18	16	13	12	15	22	16	18	20	17	15
2004	18	25	21	11	14	16	19	20	17	16	18	20
Monthly Total	53	61	54	43	42	51	62	54	52	51	49	53
Monthly Average	17.6	20.3	18	14.3	14	17	20.6	18	17.3	17	16.3	17.6

Grand Average =  \(= \frac {17.6+ 20.3+ 18+ 14.3 + 14+ 17 + 20.6+ 18+ 17.3 + 17 + 16.3+ 17.6}{12}\) = \(\frac {208}{12}\) = 17.33
S.I for Jan = \(\frac {17.6}{17.33}\) × 100 = 101.56
S.I for Feb = \(\frac {20.3}{17.33}\) × 100 = 117.14
S.I for Mar = \(\frac {18}{17.33}\) × 100 = 103.87
S.I for April = \(\frac {14.3}{17.33}\) × 100 = 82.51
S.I for May = \(\frac {14}{17.33}\) × 100 = 80.78
S.I for June = \(\frac {17}{17.33}\) × 100 = 98.10
S.I for July = \(\frac {20}{17.33}\) × 100 = 118.87
S.I for Aug = \(\frac {20.6}{17.33}\) × 100 = 103.87
S.I for Sep = \(\frac {17.3}{17.33}\) × 100 = 99.83
S.I for Oct = \(\frac {17}{17.33}\) × 100 = 98.10
S.I for Nov = \(\frac {16.3}{17.33}\) × 100 = 94.06
S.I for Dec = \(\frac {17.6}{17.33}\) × 100 = 101.56

Year 2010	Sales (in tones)
Jan	280
Feb	240
Mar	270
Apr	300
May	280
June	290

Average
\(\frac {280 + 240 + 270 + 300 + 280 + 290}{6}\) = \(\frac {1660}{6}\) = 276.6666

Year 2010	Sales (in tones)
Jul	210
Aug	200
Sep	230
Oct	200
Nov	230
Dec	210

\(\frac {210+ 200+ 230+ 200+ 230+ 210}{6}\) = \(\frac {1280}{6}\) = 213.333

Since the number of years is even (12), we can equally divide the given data into two equal parts and obtain the averages of first 6 months and last 6 months.

Year	Sales	X=x- 2002	X2	XY
2000	35	-2	4	-70
2001	36	-1	1	-36
2002	79	0	0	0
2003	80	1	1	80
2004	40	2	4	80
	270	0	10	54

 Since \(\sum X = 0, a = \frac{\sum Y}{n}\) = \(\frac {270}{5}\) = 54
b = \(\frac {\sum XY}{\sum X^2} = \frac {54}{10}\) = 5.4
∴  The required equation of the straight line trend is given by Y  = a +bX  
\(\Rightarrow \) Y = 54 + 5.4 (x - 2002)
The trend values can be obtained as follows: 
When X = 2000, Y_t = 54 + 5.4 (2000 - 2002)
= 54 + 5.4 (-2) = 43.2
When X = 2001, Y_t = 54 + 5.4 (2001 - 2002)
= 54 + 5.4(-1) = 48.6
When X = 2002, Y_t = 54 + 5.4 (2002 - 2002)
= 54
When X = 2003, Y_t = 54 + 5.4 (2003 - 2002) 
= 54 + 5.4 = 59.4
When X = 2004, Y_t = 54 + 5.4 (2004 - 2002)
= 54 + 5.4 (2) = 64.8

Year X	Production in tonnes (Y)	X = x-1998	X2	XY
1995	155	-3	9	-465
1996	162	-2	4	-324
1997	171	-1	1	-171
1998	182	0	0	0
1999	158	1	1	158
2000	180	2	4	360
20001	178	3	9	534
	1186	0	28	92

 Since \(\sum\)X = 0, a = \(\frac {\sum Y}{n}\) = \(\frac {1186}{7}\) = 169.428.
   b = \(\frac {\sum XY}{\sum X^2}\) \(\frac {92}{28}\) = 3.285
∴ The required equation of the straight line trend is given by Y = a + bX
 \(\Rightarrow \) Y= 69.428 + 3.285 (x - 1998)

Year	Quarterly production
	I	II	III	IV
2002	3.5	3.8	3.7	3.5
2003	3.6	4.2	3.4	4.1
2004	3.4	3.9	3.7	4.2
2005	4.2	4.5	3.8	4.4
2006	3.9	4.4	18.8	20.8
Quarterly Total	18.6	20.8	18.8	20.8
Average	3.72	4.16	3.76	20.8

Grand average = (3.72 + 4.16 + 3.76 + 4.16)/4 = 3.95
Seasonal Index (S.I) for I quarter = (Average of I quarter)/(Grand Average) \(\times\) 100
S.I. for I quarter = 3.72/3.95 \(\times\) 100 = 94.1772 
S.I. for II quarter = 4.16/3.95 \(\times\) 100 = 105.3165 
S.I. for III quarter = 3.76/3.95 \(\times\) 100 = 95.1899 
S.I. for IV quarter = 4.16/3.95 \(\times\) 100 = 105.3165 
Thus we obtain the average seasonal movement.

\(\overset { = }{ X } =\frac { \sum { \bar { X } } }{ 10 } =\frac { 442 }{ 10 } =44.2\)
\(\bar { R } =\frac { \sum { R } }{ n } =\frac { 58 }{ 10 } =5.8\)
\(UCL=\overset { = }{ X } +{ A }_{ 2 }\bar { R } \)
= 44.2 + 0.483(5.8) = 47.00
\(CL=\overset { = }{ X } =44.2\)
\(LCL=\overset { = }{ X } -{ A }_{ 2 }\bar { R } =44.2-0.483(5.8)=41.39\)
The above diagram shows all the three control lines with the data points plotted, since four points falls out of the control limits, we can say that the process is out of control.

Sample	1	2	3	4	5	6	7	8	9	10	Total
Mean	383	508	505	582	557	337	514	614	707	753	5460
Range	95	128	100	91	68	65	148	28	37	80	840

\(\overset { = }{ X } =\frac { \sum { \bar { X } } }{ 10 } =\frac { 5460 }{ 10 } =546\)
The control limits for \(\overset{-}{X}\) chart is
\(UCL=\overset { = }{ X } +{ A }_{ 2 }\bar { R } =546+0.483(84)=586.57\)
\(CL=\overset { = }{ X } =546\)
\(LCL=\overset { = }{ X } -{ A }_{ 2 }\bar { R } =546-0.483(84)=505.43\)
\(\bar { R } =\frac { \sum { R } }{ n } =\frac { 840 }{ 10 } =84\)
The control limits for Range chart is
\(UCL={ D }_{ 4 }\bar { R } =2.004(84)=168.336\)
\(CL=\bar { R } =84\)
\(LCL={ D }_{ 3 }\bar { R } =0(84)=0\)

Sample	1	2	3	4	5	6	7	8	9	10	Total
Mean	21	26	23	18	19	14	14	20	16	10	182
Range	5	6	9	7	4	6	8	9	4	7	65

\(\overset { = }{ X } =\frac { \sum { \overset { - }{ X } } }{ numberofsamples } =\frac { 182 }{ 10 } =18.2\quad \overset { - }{ R } =\frac { \sum { R } }{ n } =\frac { 65 }{ 10 } =6.5\)
The control limits for \(\overset{-}{X}\) chart is
\(UCL=\overset { = }{ X } +{ A }_{ 2 }\bar { R } =18.2+0.577(6.5)=21.95\)
\(CL=\overset { = }{ X } =18.2\)
\(LCL=\overset { = }{ X } -{ A }_{ 2 }\bar { R } =18.2-0.577(6.5)=14.5795\)
The control limits for Range chart is
\(UCL={ D }_{ 4 }\bar { R } =2.114(6.5)=13.741\)
\(CL=\bar { R } =65\)
\(LCL={ D }_{ 3 }\bar { R } =0(6.5)=0\)

Commodities Price Weights
(V) \(P=\frac { { P }_{ 1 } }{ { P }_{ 0 } } \times 100\) PV 2007
(P₀) 2011
(P₁) A 350 400 40 114.285 4571.4 B 175 250 35 142.857 4999.99 C 100 115 15 115 1725 D 75 105 20 114 2280 E 60 80 25 133.333 3333.33     Total 135   17429 76 Value of Cost of Living Index Number\(\frac { \sum { PV } }{ \sum { V } } =\frac { 17429 76 }{ 135 } =129.1093\)
Hence, the Cost of Living Index Number for a particular class of people for the year 2011 is increased by 129.1093 % as compared to the year 2007.

Here the base year quantities are given, therefore we can apply Aggregate Expenditure Method.

Commodities	Price		Quantity (q0)	p0q0	p1q0
	Base year (P0)	Current year (P1)
Rice	32	48	25	800	1200
Sugar	25	42	10	250	420
Oil	54	85	6	324	510
Coffe	250	460	1	250	460
Tea	175	275	2	350	550
			Total	1974	3140

Cost of Living Index Number=\(\frac { \sum { { p }_{ 1 }{ q }_{ 0 } } }{ \sum { { p }_{ 0 }{ q }_{ 0 } } } \times 100=\frac { 3140 }{ 1974 } \times 100 = 159.0679\)
Hence, the Cost of Living Index Number for a particular class of people for the year 2016 is increased by 59.0679 % as compared to the year 2011.

Here the base year quantities are given, therefore we can apply Aggregate Expenditure Method.

Commodities	Number of Units q0(2010)	Price (2010) p0	Price (2015) p1	p0q0	p1q0
Rice	5	1500	1750	7500	8750
Sugar	3.5	1100	1200	3850	4200
Pulses	3	800	950	2400	2850
Cloth	2	1200	1550	2400	3100
Ghee	0.75	550	700	412.5	525
Rent	12	2500	3000	30000	36000
Fuel	8	750	600	6000	4800
Misc	110	3200	3500	32000	35000
			Total	84562.5	95225

Cost of Living Index Number \(=\frac { \sum { { p }_{ 1 }{ q }_{ 0 } } }{ \sum { { p }_{ 0 }{ q }_{ 0 } } } \times 100=\frac { 95225 }{ 84562.5 } \times 100=112.609\)
Hence, the Cost of Living Index Number for a particular class of people for the year 2015 is increased by 12.61 % as compared to the year 2010.

Commodities	Quantity 2005(Q0)	Price		P1Q0	P0Q0
		2005 (P0)	2010 (P1)
A	10	7	9	90	70
B	12	6	8	96	72
C	17	10	15	255	170
D	19	14	16	304	266
E	15	12	17	255	180
			Total	1000	758

Cost of Living Index Number
= \(\frac { \sum { { p }_{ 1 }{ q }_{ 0 } } }{ \sum { { p }_{ 0 }{ q }_{ 0 } } } \times 100=\frac { 1000 }{ 758 } \times 100=131.926 \)

Commodities	Base Year		Current Year		p0q0	p0q1	p1q0	p1q1
	Price (p0)	Quantity (q1)	Price ((p0))	Quantity (q1)
Rice	40	5	48	4	200	160	240	192
Wheat	45	2	42	3	90	135	84	126
Rent	90	4	95	6	360	540	380	570
Fuel	85	3	80	2	255	170	240	160
Transport	50	5	65	8	250	400	325	520
Miscellaneous	65	1	72	3	65	195	72	216
				Total	1220	1600	1341	1784

Fisher’s price index number
\({ P }_{ 01 }^{ F }=\left( \sqrt { \frac { \sum { { p }_{ 1 }{ q }_{ 10 } } \times \sum { { p }_{ 1 }{ q }_{ 1 } } }{ \sum { { p }_{ 0 }{ q }_{ 0 } } \times \sum { { p }_{ 0 }{ q }_{ 1 } } } } \right) \times 100=\left( \sqrt { \frac { 1341\times 1784 }{ 1220\times 1600 } } \right) \times 100=110.706\)
Time Reversal Test:P₀₁x P₁₀=1
\({ P }_{ 01 }\times { P }_{ 10 }=\sqrt { \left( \frac { \sum { { p }_{ 1 }{ q }_{ 0 } } \times \sum { { p }_{ 1 }{ q }_{ 1 } } \times \sum { { p }_{ 0 }{ q }_{ 1 } } \times \sum { { p }_{ 0 }{ q }_{ 0 } } }{ \sum { { p }_{ 0 }{ q }_{ 0 } } \times \sum { { p }_{ 0 }{ q }_{ 1 } } \times \sum { { p }_{ 1 }{ q }_{ 1 } } \times \sum { { p }_{ 1 }{ q }_{ 0 } } } \right) } \)
\({ P }_{ 01 }\times { P }_{ 10 }=\sqrt { \left( \frac { 1341\times 1784\times 1600\times 1220 }{ 1220\times 1600\times 1784\times 1341 } \right) } \)
P₀₁x P₁₀ = 1
Factor Reversal Test
\(P_{01} \times Q_{01}=\frac{\sum p_{1} q_{1}}{\sum p_{0} q_{0}}\)
\(P_{01} \times Q_{01}=\sqrt{\left(\frac{\sum p_{1} q_{0} \times \sum p_{1} q_{1} \times \sum q_{1} p_{0} \times \sum q_{1} p_{1}}{\sum p_{0} q_{0} \times \sum p_{0} q_{1} \times \sum q_{0} p_{0} \times \sum q_{0} p_{1}}\right)}\)
\(P_{01} \times Q_{01}=\sqrt{\left(\frac{1341 \times 1784 \times 1600 \times 1784}{1220 \times 1600 \times 1220 \times 1341}\right)}\)
\(P_{01} \times Q_{01}=\sqrt{\left(\frac{1784 \times 1784}{1220 \times 1220}\right)}=\frac{1784}{1220}\)
\(\Rightarrow P_{01} \times Q_{01}=\frac{\sum p_{1} q_{1}}{\sum p_{0} q_{0}}\)

Commodities	Base Year		Current Year		p0q0	p0q1	p1q0	p1q1
	Price (p0)	Quantity (q1)	Price ((p0))	Quantity (q1)
Rice	10	5	11	6	50	60	55	66
Wheat	12	6	13	4	72	48	78	52
Rent	14	8	15	7	112	98	120	105
Fuel	16	9	17	8	144	128	153	136
Transport	18	7	19	5	126	90	133	95
Miscellaneous	20	4	21	3	80	60	84	63
				Total	584	484	623	517

Fisher’s price index number
\({ P }_{ 01 }^{ F }=\left( \sqrt { \frac { \sum { { p }_{ 1 }{ q }_{ 10 } } \times \sum { { p }_{ 1 }{ q }_{ 1 } } }{ \sum { { p }_{ 0 }{ q }_{ 0 } } \times \sum { { p }_{ 0 }{ q }_{ 1 } } } } \right) \times 100=\left( \sqrt { \frac { 623\times 517 }{ 584\times 484 } } \right) \times 100=106.74\)
Time Reversal Test: P₀₁ × P₁₀ = 1
\({ P }_{ 01 }\times { P }_{ 10 }=\sqrt { \left( \frac { \sum { { p }_{ 1 }{ q }_{ 0 } } \times \sum { { p }_{ 1 }{ q }_{ 1 } } \times \sum { { p }_{ 0 }{ q }_{ 1 } } \times \sum { { p }_{ 0 }{ q }_{ 0 } } }{ \sum { { p }_{ 0 }{ q }_{ 0 } } \times \sum { { p }_{ 0 }{ q }_{ 1 } } \times \sum { { p }_{ 1 }{ q }_{ 1 } } \times \sum { { p }_{ 1 }{ q }_{ 0 } } } \right) } \)
\({ P }_{ 01 }\times { P }_{ 10 }=\sqrt { \left( \frac { 623\times 517\times 484\times 584 }{ 584\times 487\times 517\times 623 } \right) } \)
P₀₁ x P₁₀ = 1
Factor Reversal Test
\({ P }_{ 01 }\times { Q }_{ 01 }=\frac { \sum { { p }_{ 1 }{ q }_{ 1 } } }{ \sum { { p }_{ 0 }{ q }_{ 0 } } } \)
\({ P }_{ 01 }\times { P }_{ 01 }=\sqrt { \left( \frac { \sum { { p }_{ 1 }{ q }_{ 0 } } \times \sum { { p }_{ 1 }{ q }_{ 1 } } \times \sum { { p }_{ 1 }{ q }_{ 0 } } \times \sum { { p }_{ 1 }{ q }_{ 1 } } }{ \sum { { p }_{ 0 }{ q }_{ 0 } } \times \sum { { p }_{ 0 }{ q }_{ 1 } } \times \sum { { p }_{ 0 }{ q }_{ 0 } } \times \sum { { p }_{ 0 }{ q }_{ 1 } } } \right) } \)
\({ P }_{ 01 }\times { P }_{ 10 }=\sqrt { \left( \frac { 623\times 517\times 484\times 517 }{ 584\times 484\times 584\times 623 } \right) } \)
\({ P }_{ 01 }\times { P }_{ 01 }=\sqrt { \left( \frac { 517\times 517 }{ 585\times 584 } \right) } =\frac { 517 }{ 584 } \)
\({ P }_{ 01 }\times { P }_{ 01 }=\frac { \sum { { p }_{ 1 }{ q }_{ 1 } } }{ \sum { { p }_{ 0 }{ q }_{ 0 } } } \)

Commodities	Price		Quandity		p0q0	p0q1	p1q0	p1q1
	2003 (p0)	2009 (q1)	2003 (p0)	2009 (q1)
Rice	10	13	4	6	40	60	52	78
Wheat	125	18	7	8	105	120	126	144
Rent	25	29	5	9	125	225	145	261
Fuel	11	14	8	10	88	110	140	140
Miscellaneous	14	17	6	7	84	98	102	119
Total					442	613	537	742

Fisher’s price index number
\({ P }_{ 01 }^{ F }=\left( \sqrt { \frac { \sum { { p }_{ 1 }{ q }_{ 10 } } \times \sum { { p }_{ 1 }{ q }_{ 1 } } }{ \sum { { p }_{ 0 }{ q }_{ 0 } } \times \sum { { p }_{ 0 }{ q }_{ 1 } } } } \right) \times 100=\left( \sqrt { \frac { 537\times 742 }{ 442\times 613 } } \right) \times 100=121.2684\)
Time Reversal Test:P₀₁\(\times\) P₁₀ = 1
\({ P }_{ 01 }\times { P }_{ 10 }=\sqrt { \left( \frac { \sum { { p }_{ 1 }{ q }_{ 0 } } \times \sum { { p }_{ 1 }{ q }_{ 1 } } \times \sum { { p }_{ 0 }{ q }_{ 1 } } \times \sum { { p }_{ 0 }{ q }_{ 0 } } }{ \sum { { p }_{ 0 }{ q }_{ 0 } } \times \sum { { p }_{ 0 }{ q }_{ 1 } } \times \sum { { p }_{ 1 }{ q }_{ 1 } } \times \sum { { p }_{ 1 }{ q }_{ 0 } } } \right) } \)
\({ P }_{ 01 }\times { P }_{ 10 }=\sqrt { \left( \frac { 537\times 742\times 613\times 442 }{ 442\times 613\times 742\times 537 } \right) } \)
\({ P }_{ 01 }\times { P }_{ 10 }=1\)
Factor Reversal Test
\({ P }_{ 01 }\times { Q }_{ 01 }=\frac { \sum { { p }_{ 1 }{ q }_{ 1 } } }{ \sum { { p }_{ 0 }{ q }_{ 0 } } } \)
\({ P }_{ 01 }\times { P }_{ 01 }=\sqrt { \left( \frac { \sum { { p }_{ 1 }{ q }_{ 0 } } \times \sum { { p }_{ 1 }{ q }_{ 1 } } \times \sum { { p }_{ 1 }{ q }_{ 0 } } \times \sum { { p }_{ 1 }{ q }_{ 1 } } }{ \sum { { p }_{ 0 }{ q }_{ 0 } } \times \sum { { p }_{ 0 }{ q }_{ 1 } } \times \sum { { p }_{ 0 }{ q }_{ 0 } } \times \sum { { p }_{ 0 }{ q }_{ 1 } } } \right) } \)
\({ P }_{ 01 }\times { P }_{ 01 }=\sqrt { \left( \frac { 537\times 742\times 613\times 742 }{ 442\times 613\times 442\times 537 } \right) } \)
\({ P }_{ 01 }\times { P }_{ 01 }=\sqrt { \left( \frac { 742\times 742 }{ 442\times 442 } \right) } =\frac { 742 }{ 442 } \Rightarrow { P }_{ 01 }\times { P }_{ 01 }=\frac { \sum { { p }_{ 1 }{ q }_{ 1 } } }{ \sum { { p }_{ 0 }{ q }_{ 0 } } } \)

Commodities	Base Year		Current Year		p0q0	p0q1	p1q0	p1q1
	Price (p0)	Quantity (q1)	Price ((p0))	Quantity (q1)
Rice	15	5	16	8	75	120	80	128
Wheat	10	6	18	9	60	90	108	162
Rent	8	7	15	8	56	64	105	120
Fuel	9	5	12	6	45	54	60	72
Transport	11	4	11	7	44	77	44	77
Miscellaneous	16	6	15	10	96	160	90	150
				Total	376	565	487	709

Laspeyre’s price index number
\({ P }_{ 01 }^{ L }=\frac { \sum { { p }_{ 1 }{ q }_{ 0 } } }{ \sum { { p }_{ 0 }{ q }_{ 0 } } } \times 100=\frac { 487}{ 376} \times 100=129.5212\)
Paasche’s price index number
\({ P }_{ 01 }^{ P }=\frac { \sum { { p }_{ 1 }{ q }_{ 1 } } }{ \sum { { p }_{ 0 }{ q }_{ 1 } } } \times 100=\frac { 709 }{ 565 } \times 100=125.4867\)
Fisher’s price index number
\({ P }_{ 01 }^{ F }=\sqrt { \frac { \sum { { p }_{ 1 }{ q }_{ 0 } } \times \sum { { p }_{ 1 }{ q }_{ 1 } } }{ \sum { { p }_{ 0 }{ q }_{ 0 } } \times \sum { { p }_{ 0 }{ q }_{ 1 } } } } \times 100=\sqrt { \frac { 487\times 709}{ 376\times 565} } \times 100=127.4879\)
On an average, there is an increase of 29.52%, 25.48% and 27.48% in the price of the commodities by Laspeyre’s, Paasche’s, Fisher’s price index number respectively, when the base year compared with the current year.

Commodities	Price		Quandity		p0q0	p0q1	p1q0	p1q1
	2000 (p0)	2010 q1	2000 (p0)	2010 (q1)
Rice	38	35	6	7	228	266	210	245
Wheat	12	18	7	10	84	120	126	180
Rent	10	15	10	15	100	150	150	225
Fuel	25	30	12	16	300	400	630	480
Miscellaneous	30	33	8	10	240	300	264	330
				Total	952	1236	1110	1460

Laspeyre’s price index number
\({ P }_{ 01 }^{ L }=\frac { \sum { { p }_{ 1 }{ q }_{ 0 } } }{ \sum { { p }_{ 0 }{ q }_{ 0 } } } \times 100=\frac { 1110 }{ 952 } \times 100=116.60\)
On an average, there is an increase of 16.60 % in the price of the commodities when the year 2000 compared with the year 2010.
Paasche’s price index number
\({ P }_{ 01 }^{ P }=\frac { \sum { { p }_{ 1 }{ q }_{ 1 } } }{ \sum { { p }_{ 0 }{ q }_{ 1 } } } \times 100=\frac { 1460 }{ 1236 } \times 100=118.12\)
On an average, there is an increase of 18.12 % in the price of the commodities when the year 2000 compared with the year 2010.
Fisher’s price index number
\({ P }_{ 01 }^{ F }=\sqrt { \frac { \sum { { p }_{ 1 }{ q }_{ 0 } } \times \sum { { p }_{ 1 }{ q }_{ 1 } } }{ \sum { { p }_{ 0 }{ q }_{ 0 } } \times \sum { { p }_{ 0 }{ q }_{ 1 } } } } \times 100=\sqrt { \frac { 1110\times 1460 }{ 952\times 1236 } } \times 100=117.36\)
On an average, there is an increase of 17.36 % in the price of the commodities when the year 2000 compared with the year 2010.

Computation of Seasonal Index by the method of simple averages.

Year	I Quarter	II Quarter	III Quarter	IV Quarter
2005	255	351	425	400
2006	269	310	396	410
2007	291	332	358	395
2008	198	289	310	357
2009	200	290	331	359
2010	250	300	350	400
Quarterly Total	1463	1872	2170	2321
Quarterly Averages	243.83	312	361.67	386.83

S.I for I Quarter = \(\frac{Average\ of \ I\ quarter }{Grand\ average} \times100\)
Grand Average = \(\frac{1304.333}{4}=326.0833\)
S.I for I Q = \(\frac{243.8333}{326.0833} \times100= 74.77;\) 
S.I for II Q = \(\frac{312}{326.0833} \times100=95.68;\)
S.I for III Q = \(\frac{361.6667}{326.0833} \times 100=110.91; \) 
S.I for IV Q = \(\frac{386.833}{326.0833} \times 100=118.63\)

Computation of seasonal Indices by method of simple averages.
Months Jan Feb Mar Apr May June July Aug Sep Oct Nov Dec Year 2001 15 41 25 31 29 47 41 19 35 38 40 30 2002 20 21 27 19 17 25 29 31 35 39 30 44 2003 18 16 20 28 24 25 30 34 30 38 37 39 Monthly Total 53  78 72 78 70 97 100 84 100 115 107 113 Monthly Averages 17.67 26 24 26 23.33 32.33 33.33 28 33.33 38.33 35.67 37.67 S.I for Jan = \(\frac{Monthly\ Average\ (for\ Jan)}{Grand\ average}\times100\)
Grand Average = \(\frac{355.582}{12}=29.63\)
S.I for Jan = \(\frac{17.666}{29.361}\times100=59.62; \) 
S.I for Feb = \(\frac{26}{29.361}\times 100=87.72\)
Similarly other seasonal index values can be obtained.
 
\(\begin{array}{|c|c|} \hline \text { Months } & \text { Seasonal Index } \\ \hline \text { Jan } & 59.62 \\ \hline \text { Feb } & 87.72 \\ \hline \text { Mar } & 80.97 \\ \hline \text { Apr } & 87.72 \\ \hline \text { May } & 78.71 \\ \hline \text { June } & 109.08 \\ \hline \text { July } & 112.45 \\ \hline \text { Aug } & 94.47 \\ \hline \text { Sep } & 112.45 \\ \hline \text { Oct } & 129.32 \\ \hline \text { Nov } & 120.34 \\ \hline \text { Dec } & 127.09 \\ \hline \end{array}\)

Computation of trend values by the method of least squares.
In case of EVEN number of years, let us consider
\(X=\frac{\text{(x-Arithimetic mean of two middle years)}}{0.5}\)

Year(x)	Sales(Y)	X=\(\frac{(x-1998.5)}{0.5}\)	XY	X2	Trend Values (Yt)
1995	6.7	-7	-46.9	49	5.6166
1996	5.3	-5	-26.5	25	5.7190
1997	4.3	-3	-12.9	9	5.8214
1998	6.1	-1	-6.1	1	5.9238
1999	5.6	1	5.6	1	6.0261
2000	7.9	3	23.7	9	6.1285
2001	5.8	5	29.0	25	6.2309
2002	6.1	7	42.7	49	6.3333
N = 8	47.8	\(\sum X\) = 0	8.6	168

\(a=\frac { \sum { Y } }{ n } =\frac { 47.8 }{ 8 } =5.975;\quad b=\frac { \sum { XY } }{ { \sum { X } }^{ 2 } } =\frac { 8.6 }{ 168 } =0.05119\)
Therefore, the required equation of the straight line trend is given by
Y = a + bX; Y = 5.975 + 0.05119 X.
When X = 1995, Y_t = 5.975 + 0.05119\(\left( \frac { 1995-1998.5 }{ 0.5 } \right) =5.6166\)
When X = 1996, Y_t = 5.975 + 0.05119\(\left( \frac { 1996-1998.5 }{ 0.5 } \right) =5.7190\)
similarly other values can be obtained.

Computation of trend values by the method of least squares (ODD Years).

Year(x)	Production of Sugarcane(Y)	X=(x–2003)	X2	XY	Trend values(Yt)
2000	40	-3	9	-120	42.04
2001	45	-2	4	-90	43.07
2002	46	-1	1	-46	44.11
2003	42	0	0	0	45.14
2004	47	1	1	47	46.18
2005	50	2	4	100	47.22
2006	46	3	9	138	48.25
N=7	\(\sum Y\)=316	\(\sum X\)=0	\(\sum X\)2=8	\(\sum XY\)=29	\(\sum Yt\)=316

\(a=\frac { \sum { Y } }{ n } =\frac { 316 }{ 7 } =45.143;\quad b=\frac { \sum { XY } }{ { \sum { X } }^{ 2 } } =\frac { 29 }{ 28 } =1.036\)
Therefore, the required equation of the straight line trend is given by
Y = a + bX
Y = 45.143 + 1.036 (x - 2003)
The trend values can be obtained by
When X = 2000 , Yt = 45.143 + 1.036(2000–2003) = 42.035
When X = 2001, Yt = 45.143 + 1.036(2001–2003) = 43.071,
similarly other values can be obtained.