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English: y = 50 , y = 10 Regression equation of x on y is:  x x – x = r y y – y 15 ? x – 80 = 0.4 × 10 (y – 50) = 0.6 (y – 50) x – 80 = 0.6y – 30 x = 0.6y – 30 + 80 x = 0.6y + 50 CU IDOL SELF LEARNING MATERIAL (SLM)

Regression Analysis 343 If y = 60, then x = 0.6×60+50 = 36+50 ? x = 86 Marks obtained by the student in Maths = 86. Relation between Correlation and Regression = Whereas the correlation is the relationship between two or more variables, when the movement of one tend to be corresponding to the other, the Regression is the return to the average value and is the mathematical average relationship between the two variables. = Correlation need not imply cause and effect relationship, between the variables under study, whereas Regression clearly establishes this relationship. There is a definite cause and effect relationship between dependent and independent variable. = Correlation coefficient between two variables is the measure of direction and degree of the linear relationship between the two variables, which is mutual and symmetric, i.e., rxy = ryx. But in case of regression, the dependent and independent variables have a definite direction. Hence there are two distinct lines of regression, i.e., y on x or x on y and hence byx ? bxy. = Whereas the correlation coefficient rxy is a relative measure of the linear relationship of x and y and is independent of the unit of measurement, the regression coefficient bxy, and byx are absolute measures representing the change in the values of variable y(x) for a unit change in x(y). = There can be a non-sense correlation between the two variables; there is no such non- sense regression.

= Correlation analysis centres around only linear relationship between the variables, whereas in regression, there can be linear and non-linear relationship. Probable Error Probable error of Co-efficient of Correlation gives us the two limits within which the co- efficient of correlation of Series selected at random from the same universe is likely to fall. The formula for the probable error of r is as follows. CU IDOL SELF LEARNING MATERIAL (SLM)

344 Business Mathematics and Statistics 1 r2 PE= 0.6745 N Where, r = Co-efficient of Correlation N = Number of pairs of observations Significance of Probable Error Probable error is useful in the following interpretation: ` If the value of ‘N’ is less than the probable error there is no evidence of correlation, i.e., The value of ‘r’ is not at all significant. ` If the value of ‘r’ is more than 6 times the probable error, the Co-efficient of Correlation is practically certain, i.e., The value of ‘r’ is Significant. ` By adding and subtracting the value of P.E from the value of ‘r,’ we get the upper limit and the lower limit, respectively within which the correlation of coefficient is expected to lie. Symbolically, it can be expressed: P - r ± P.E. r Where U (rho) denotes Correlation in the population The Conditions necessary for the use of probable error The measure of probable error can be Properly used only When the following three conditions exist:

= The data must approximate a normal frequency curve, i.e. bell shaped curve. = The Statistical measure for which the P.E is computed must have been calculated from a Sample. = The sample must have been Selected in an unbiased manner and the individual items must be independent. Practical Problems on Correlation Analysis Illustration 1: If r = 0.6 and N = 64 of a distribution, Find out the probable error. Solution: 1r2 1(0.6)2 PE = 0.6745 = 0.6745 = 0.6745 × 0.08 = 0.06 N 64 CU IDOL SELF LEARNING MATERIAL (SLM)

Regression Analysis 345 Illustration: 2: For a given distribution the value of correlation is 0.64 and its probable error is 0.13274. Find the number of items in the series. 1r2 Solution: P.E. =0.6745 × Given, PE = 0.13274 Correlation (r) = 0.64 n n= ? 1(0.64)2 PE = 0.6745 × n 0.6745(10.4096) 0.13274 = n 0.13274 = 0.6745 x 0.5904 n 0.3982 n = 2.999  n = 9 (Approx) n= 0.13274 Illustration: 3: Compute Karl Pearson’s Co-efficient of Correlation from the following data and calculate its probable error and interpret the result. Marks in Accountancy 77 54 27 52 14 35 90 25 56 60 Marks in English 35 58 60 40 50 40 35 56 34 42 Solution: Calculation of Karl Pearson’s Co-efficient of Correlation

X X–49 2 Y Y–45 2 x y xy xy 77 + 28 784 35 – 10 100 – 280 169 + 65 54 + 5 25 58 + 13 225 – 330 27 – 22 484 60 + 15 25 – 15 25 – 175 52 + 3 9 40 –5 25 + 70 100 – 410 14 – 35 1,225 50 +5 121 – 264 121 – 77 35 – 14 196 40 –5 9 – 33 90 + 41 1,681 35 – 10 ¦y2 = 920 ¦xy = – 1,449 25 – 24 576 56 + 11 56 + 7 49 34 – 11 60 + 11 121 42 –3 ¦X = 490 ¦x2 = 5,150 ¦Y = 450 r CU IDOL SELF LEARNING MATERIAL (SLM)

346 Business Mathematics and Statistics ¦xy 1449 0.666 = ¦ x 2 u ¦y2 = 5150 u 920 1r2 1(0.666)2 0.3753 PE = 0.6745 = 0.6745 = 0.119 N 10 3.162 Illustration 9: Calculate Karl Pearson’s coefficient of correlation between percentage of pass and failures from the following data. Also obtain probable error: No. of Students: 80 60 90 70 50 40 45 56 45 30 No. of Students passed: 48 30 Solution: % of pass % of fail x=x–a x2 y=y–b y2 xy (x) (y) a = 50 b = 50 –100 60 40 10 100 –10 100 0 00 0 50 50 00 00 –900 50 50 00 –30 900 –1,600 –40 1,600 80 20 30 900 –25 625 –625 90 10 40 1,600 75 25 25 625

405 195 105 3,225 -105 3,225 –3,225 405 195 X= = 67.5 = = 32.5 66 N.6xy6x. 6y r = N6x 2 6 x)2 N6 y 2 ( 6y)2 6(3,225)(105)(105) = 6(3,225)(105)2 6(3,225)(105)2 19,35011,025 8,325 8,325 = 19,35011,025 19,35011,025 = (8,325)2 = 8,325 = –1 2 0.6745 1(1) 2 0.6745× 0 1r PE = 0.6745 = = =0 n 6 6 CU IDOL SELF LEARNING MATERIAL (SLM)

Regression Analysis 347 15.6 Summary Regression: The term “regression” was used by Sir Francis Galton to describe a hereditary phenomenon that he observed in his study of the heights of sons and fathers. His main observation was that though tall fathers usually had tall sons, the average height of the sons of tall fathers was less than the average height of the fathers. In short, the average height of the sons of tall fathers will regress or go back to the general average height. Galton called this backward or downward tendency in the average height as regression. At phenomena — whether economic business or social. A line that is drawn as close as possible to the plotted points of the scatter diagram shows the average tendency of the plotted points. This line is known as the regression line and its equation is called the regression equation. Unlike the coefficient of correlation, which indicates the extent of the relationship between two sets of figures, a regression equation enables us to calculate the amount of change in one variable corresponding to a change in the other. 15.7 Key Words/Abbreviations Regression Equation, Regression Coefficients 15.8 Learning Activity ` Form a frequency distribution of the following test scores of 50 individuals using a class interval of 10: 75 130 135 90 118 92 80 142 97 147

98 94 115 109 154 109 111 117 120 91 124 101 97 98 126 94 109 109 94 110 82 96 119 92 99 114 104 169 107 93 102 83 117 98 77 133 87 145 91 CU IDOL SELF LEARNING MATERIAL (SLM)

348 Business Mathematics and Statistics ? The following data represent weights, recorded to the nearest kilogram, of 30 students selected from a school of 500 students: 21 30 40 25 26 22 39 31 29 36 38 35 34 33 30 23 27 27 29 31 33 22 21 36 40 31 33 30 37 36 In the above data, state what is the: Population (ii) Sample and (iii) Variate. = Draft a blank table to show: Sex. Three ranks — supervisors, assistants and clerks. Years 1998 and 2003. Age-groups: 18 years and under, over 18 but less than 55 years, over 55 years. = Draw up the proforma of suitable table (complete with title, rulings, columnar heading, sub-headings, source, note etc.) showing the number of students in your university in various classes, classified according to sex, residence, domicile and medium of instruction. = Draw up a table to show the distribution of workers in a certain cloth mill according to: (i) Salary: Below ` 150, ` 150 to 300, ` 300 to 450, ` 450 and over,

Three years: 1967, 68, 69. Age groups: Below 20, 20 and under 30, 30 and under 40, 40 and under 50,50 and over. Sex. = Present the following information in a suitable tabular form, filling the gaps if any. “The total population of India is 3,566 lakhs, of which 2,948 lakhs belong to rural areas. In rural areas, 2,404 lakhs belong to agricultural classes; while in urban areas 531 lakhs belong to the non-agricultural classes. Of the rural agricultural classes, 687 lakhs are self-supporting persons, 1,414 lakhs are non-earning dependents and the remaining CU IDOL SELF LEARNING MATERIAL (SLM)

Regression Analysis 349 are earning dependents. Of the rural non-agricultural population, 170 lakhs are self- supporting, 326 lakhs are non-earning dependants and the remaining are earning dependants. In the urban agricultural classes 23 lakhs are self-supporting, 56 lakhs are non-earning dependants and the remaining are earning dependants. In the urban non- agricultural population, 153 lakhs are self-supporting, 347 lakhs are non-earning dependants and the remaining are earning dependants.” 15.9 Unit End Questions (MCQ and Descriptive) A. Descriptive Type: Short Answer Type Questions ` Explain the concept of ‘regression’ and comment on its utility. ` Explain the terms (i) coefficient of regression and (ii) lines of regression. ` Show that the correlation coefficient is the geometric mean between regression coefficients. ` Explain the concept of linear regression. Why are there two regression lines? Do they cut each other? ` Compare and contrast the roles of correlation and regression in studying the interdependence of two variates. ` Define ‘regression’. Why are there two regression lines? Under what conditions can there be only one regression line? ` Distinguish between coefficient of correlation and coefficient of regression and state briefly why the latter is often more useful in statistics than the former. ` Find the regression coefficients and the regression equations for the following data:

X : 76 50 61 45 27 49 32 54 70 36 Y : 58 60 72 78 56 40 43 65 22 36 + From the undermentioned data, find out Karl Pearson’s coefficient of correlation, the two regression equations, CU IDOL SELF LEARNING MATERIAL (SLM)

350 Business Mathematics and Statistics ? the two regression coefficients, ? the best mean value of X when Y is 50, ? the best mean value of Y when X is 45. Y: 55 59 63 68 56 73 82 76 64 74 X : 60 62 55 54 63 72 78 79 65 82 = Calculate the coefficient of correlation and obtain the lines of regression for the following data: X: 1 2 3 4 5 6 7 8 9 Y: 9 8 10 12 11 13 14 16 15 Obtain an estimate of y which should correspond on the average to x = 6.2 = In a partially destroyed laboratory record of an analysis of correlation data, the following results only are legible: Variance of x = 9. Regression equations: 8x – 10y + 66 = 0, 40x – 18y = 214. What were the mean values of x and y, the standard deviation of y and the coefficient of correlation between x and y? = The following are marks in Statistics (x) and Mathematics (y) of ten students. x : 56 55 58 58 57 56 60 54 59 57 y : 68 67 67 70 65 68 70 66 68 66

Calculate the coefficient of correlation and estimate marks in Mathematics of a student who scores 62 marks in statistics. ? The following data give the height in inches (X) and the weight in lb (Y) of random sample of 10 students from a large group of students of age 17 years. x : 61 68 68 64 65 70 63 62 64 67 y : 112 123 130 115 110 125 100 113 116 126 Estimate the weight of a student of height 69 inches. ? From the following data calculate the coefficient of correlation and estimate x when y is equal to 105. CU IDOL SELF LEARNING MATERIAL (SLM)

Regression Analysis 351 x : 44 58 49 46 58 56 48 46 48 47 y : 88 114 102 113 91 89 102 93 114 94 15. For a bivariate data the mean value of x = 53.2 the mean value of y = 27.9 the regression coefficient of x on y = – 0.2 and the regression coefficient of y on x = – 1.5 Find (i) the most probable value of y when x = 60, (ii) r the coefficient of correlation. ? You are given the following results for the heights (x) and weights (y) of 1000 workers of a factory. x = 68 inches, x = 2.5 inches y = 150 lb, y = 20 lb rxy = 0.6. Estimate from the above data. The height fo a particular factory worker whose weight is 200 lb The weight of a particular factory worker who is 5 feet tall. ? Find the most likely price in Bombay (x) corresponding to the price of Rs.70 at Calcutta

(y) from the following data: Average price at Calcutta = 55 Average prie at Bombay = 67 S. D. at Calcutta = 2.5 S. D. at Bombay = 3.5 + .8 r The equation of two regerssion lines obtained in a correlation analysis are as follows: 4y = 9x + 15 25x = 6y + 7 CU IDOL SELF LEARNING MATERIAL (SLM)

352 Business Mathematics and Statistics Obtain the value of correlation coefficient, the ratio of the coefficient of variability of x to that of y and also mean values of x and y. 19. Following is the information relating to per capita consumption of wheat per year and the retail price of wheat: Year (Y) (X) Per Capita Consumption Retail Price 1960 1961 (Kg) (Paise) 1962 35 63 1963 34 65 1964 32 68 1965 36 64 1966 37 64 1967 34 68 33 70 30 75 Fit a straight line of per capita consumption (Y) on retail price (X). ? From the data given below find: = The two regression equations. = The coefficient of correlation between marks in Economics and Statistics. = The most likely marks in Statistics when the marks in Economics are 30. Marks in Economics : 25 28 35 32 31 36 29 38 34 32

Marks in Statistics : 43 46 49 41 36 32 31 30 33 39 B. Multiple Choice/Objective Type Questions 1. Given the 2 Keynesian equation 3x – 2y + 1 = 0 62y – 5x + 3 = 0 (a) 0.45 (b) 0.53 (c) 0.63 (d) None of these CU IDOL SELF LEARNING MATERIAL (SLM)

Regression Analysis 353 2. Given x = 1, y = 43, r = 0.62 Vx = 8.3, Vy = 6/4. like 2 Regression equations are x = 0.8y – 3.57 and y = 0.48 + 28.18 (a) Wrong (b) Correct i None of these Ÿ The term “regression” was used by Sir Francis Galton to describe the average ____________ of father and son. (a) Weight (b) Complexion (c) Height (d) size 4. The equation of regression line is called as __________. (a) Regression equation (b) Coefficient of correlation (c) Correlation (d) All the above Answers: (1) (b); (2) (b); (3) (c); (4) (a) 15.10 References References of this unit have been given at the end of the book.

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354 Business Mathematics and Statistics UNIT 16 TIME SERIES ANALYSIS Structure 16.0 Learning Objectives 16.1 Introduction 16.2 Time Series: Its meaning 16.3 Time Series: Analysis 16.4 Measurement of General Trend 16.5 Solved Problems 16.6 Summary 16.7 Key Words/Abbreviations 16.8 Learning Activity 16.9 Unit End Questions (MCQ and Descriptive) 16.10 References

16.0 Learning Objectives After completing this unit, you would be able to: Ÿ Explain the meaning of a time series and the various types of influences that include, General trend, Seasonal Changes, Cyclical fluctuations and irregular variations. Ÿ Elaborate the methods of measurement of the general trend that include: Geographical method, Method of moving averages and the method of least squares. CU IDOL SELF LEARNING MATERIAL (SLM)

Time Series Analysis 355 ? The procedure for finding 3 yearly, moving averages as explained in the solved problems ? Differentiate the Merits and Demerits of Moving Averages. ? Analyse the method of least-squares with reference to the solved problem. ? Explain the histogram. ? Elaborate the Parabolic Trend Equation with reference to the solved problem. ? Explain the Exponentist Trend Method: and as well as various solved problems. 16.1 Introduction A orderly set of number corresponding to time is a time series. Time series corresponding to variables such as sales, projects, investment etc., show a number of changes done to the influence of various factory. These details have been explained in 16.3. For calculation of these changes Cartesian methods are used. These methods are explained in the pages that follows. 16.2 Time Series: Its Meaning In the world of business and economics we find variables of one kind or the other. For instance, sales, profits, capital, investments, prices, costs, rates of interest, savings, demand for commodities etc. These are said to be variables because they vary over a period of time. Managements are very often concerned in closely examining these variables over different periods of time. An orderly set of numbers written in the order corresponding to time is called a time series. For instance, a series of terms showing the profits of a business concern over a number of years is

a time series because the terms of series representing profits are written in relation to time. A careful scrutiny of a time series along with a detailed analysis of condition in the past enables one to forecast future changes and estimates. The following is a time series showing the production of cereals in India in million tonnes. CU IDOL SELF LEARNING MATERIAL (SLM)

356 Business Mathematics and Statistics Net Production (Million Tonnes)* of Cereals in India Year Production 1951 40.01 1952 40.59 1953 45.36 1954 53.44 1955 51.59 1956 50.33 1957 52.67 1958 49.35 1959 57.29 1960 56.76 1961 60.66 1962 62.08 1963* 58.43 * Provisional. Source: Ministry of Food & Agriculture. Published in the Quarterly Bulletin of the Eastern Economist. New Delhi. Vol. 16, No. 2. 16.3 Time Series: Analysis

A detailed analysis of the above time series enables us to find the different types of movements influencing its course. Usually in any given time series the different types of movements are characterized by the following component: ? General Trend ? Seasonal changes ? Cyclical Fluctuations ? Irregular Variations 16.3.1 General Trend The time series of the net productions of cereals in India shows that there is certainly a gradual increase in the production of cereals even though the increase is not regular. This gradual increase is called the “General Trend” or secular trend. This trend, which means here a gradual increase in the production of cereals, is brought about by the proper use of modern agricultural machinery and CU IDOL SELF LEARNING MATERIAL (SLM)

Time Series Analysis 357 techniques, better manure and chemical fertilizers and also the farmer’s efforts. Maximum yields can be expected when farmers know exactly the proper proportions of nutrients that are ideal for crops. An hectare of agricultural land would need several hundred kilograms of nitrogen, phosphorous and potash and a fair sparkling of secondary and micro elements like iron copper. In other words this gradual increase can be attributed to the growing realization among farmers about the advantages of profit-oriented farming, modern equipment, improved seeds, greater use of fertilizers, pesticides and modern methods of farming. The general trend is influenced by a group of factors which include seasonal fluctuations, cyclical fluctuations and also unforeseen events. The trend can therefore be clearly studied when these different influences are clearly eliminated. 16.3.2 Seasonal Fluctuations Eliminating seasonal influences is no easy task. The occurrence of seasons is responsible for these fluctuations. Natural phenomena are always unavoidable beyond any measure of control. For instance, variations in food grain production over time depend on whether the season is favorable or not. It is quite possible that production may be very high and beyond expectations in a certain season and very low at other times. In short seasonal influences are more or less regularly present in any time series. 16.3.3 Cyclical Fluctuations These are due to the occurrence of business or trade cycles in the economic world. For instance; there may be a period of depression followed by a boom after a few years. It is at such times that there is considerable variability in the production of cereals. Trade cycles cause up and down movements in any time series of economic data. 16.3.4 Irregular Variations

In many time series of economic data, we observe sudden upward or downward movements in the trend. Such movements are irregular and are brought about by unforeseen events and sudden changes in economic activities and business conditions. Further, political factors too contribute to uncertain and significant changes. Other irregular and unforeseen events may be wars, earthquakes, revolutions, fires and floods, which cause sudden and unexpected increase in prices in general. CU IDOL SELF LEARNING MATERIAL (SLM)

358 Business Mathematics and Statistics Further technological progress may lead to an increase in industrial activities. Sometimes agricultural prosperity might be ushered in by a bounty of nature - favorable rainfall etc. agricultural prosperity in turn may bring about falling prices in foodgrains and other commodities. In short, irregular movements are changes by ‘fits and starts’. 16.4 Measurement of General Trend A general trend can be measured by the following methods: ? Graphical method-freehand curve. ? Method of moving averages. ? Method of least squares. 16.4.1 Graphical Method This method consists in drawing the curve of the time series on graph paper. First of all, certain points representing the time series are plotted on the graph paper and these plotted points are joined by a smooth curve. Such a curve gives a picture of the different up and down movements in the time series. This curve, when it is drawn smoothly, fairly and regularly, shows the general direction of the trend. This method gives a rough indication of the nature of the trend and is especially useful to all those who do not know other techniques of measuring trends. The main drawback of this graphical method of indicating a trend is that different curves may be drawn by different persons for the same data and may provide a misleading picture. 16.4.2 Method of Moving Averages

Moving Averages are calculated on the basis of a given period, such as, 3-yearly or 4-yearly or 5-yearly. For instance, 3-yearly moving averages are calculated by taking the average of the values of 3 years beginning from the first year. This method is especially useful in the case of a time series with regular cyclical movements. If the periodicity of a time series is known, moving averages can be calculated easily. The following examples make the method quite clear. 16.4.3 Solved Problems Problem 1: Calculate the three yearly moving average of the net production (million tonnes) of cereals. CU IDOL SELF LEARNING MATERIAL (SLM)

Time Series Analysis 359 Year: 1981 1982 1983 1984 1985 1986 1987 1988 1989 1990 1991 1992 1993 53.4 57.2 Production: 40.01 40.59 45.36 4 51.59 50.33 52.67 49.35 9 56.76 60.66 62.08 58.63 Year Production Three Three yearly yearly Total Moving Average 1981 40.01 — — 1982 40.59 125.96 41.99 1983 45.36 139.39 46.46 1984 53.44 150.39 50.13 1985 51.59 155.36 51.79 1986 50.33 154.59 51.53 1987 52.67 152.35 50.78 1988 49.35 159.31 53.10 1989 57.29 163.40 54.47 1990 56.76 174.71 58.24 1991 60.66 179.50 59.83 1992 62.08 181.37 60.46 1993 58.63 — — Problem 2: Calculate four-yearly moving averages and five yearly moving averages for the following data:

Year: 1982 1983 1984 1985 1986 1987 1988 1989 1990 1991 1992 1993 1994 Profits: 112 114 119 111 106 101 113 118 121 119 123 124 126 Year Profits Four-yearly Total of Four-yearly Moving Two Moving Average 1982 112 Total 4-yearly 1983 114 Centered 1984 456 1985 119 906 113.25 1986 450 110.87 111 887 108.50 437 106 431 868 CU IDOL SELF LEARNING MATERIAL (SLM)

360 Business Mathematics and Statistics 1987 101 869 108.62 1988 438 111.37 1989 115.50 1990 113 891 119.00 1991 453 121.00 1992 122.37 1993 118 924 1994 471 Year 121 952 481 1982 1983 119 968 1984 487 1985 123 979 492 124 126 Profits Four-yearly Total of Four-yearly Moving Two Moving Average 112 Total 114 4-yearly Centered 119 562 111 551 906 112.4 887 110.2

1986 106 550 868 110 1987 101 549 869 1098 1988 113 559 891 111.8 1989 118 572 924 114.4 1990 121 594 952 118.8 1991 119 605 968 121 1992 123 613 979 122.6 1991 124 1994 126 16.4.4 Merits of the Moving Average Method 9. It is particularly useful in the case of a time series with regular cyclical movements. 10. Moving averages can be easily calculated and readily understood. 11. The trend either increasing or decreasing that may exist in a time series is clearly brought out by this method. CU IDOL SELF LEARNING MATERIAL (SLM)

Time Series Analysis 361 11. It is far better than the freehand curve method. 12. If the period of the moving averages and the duration of the cyclical variation happen to be the same, then the moving average method completely eliminates the variations. 16.4.5 Demerits of the Moving Averages Method 14. It is applicable to only such time series that show periodic fluctuations. 15. The trend values at the two extremities of a time series cannot be determined. 16. It is not useful in the case of a time series with regular variations. 16.4.6 Method of Least Squares Instead of drawing an observation-trend-line for a time series freehand, it is better to fit a straight line trend more accurately using the mathematical method of least squares. This straight line so drawn is called the ‘Line of Best Fit’. It is a straight line drawn so that the sum of the squares of the vertical deviations of the plotted points of the time series is the least. It is for this reason that the method is called the method of least squares. If the equation to the line of best fit is written as Y = a +bx, then the values of the constants a and b are determined by using the following two normal equations: 6y = Na + b6x 6XY = a6X + b6x2 Where X represents the year (time).

15. represents the actual values of the terms in the series. N is the number of terms in the time series. This method can be best explained by the following example: Problem 3: Fit a straight line trend using the method of least squares to the/scale^ of X, XZ Industries as given below: CU IDOL SELF LEARNING MATERIAL (SLM)

362 Business Mathematics and Statistics Year Sales in Thousands of Rupees 1988 1989 204 1990 1230 1991 192 1992 ?~n X 274 Year 1 Y XY X2 2 1988 3 204 204 1 1989 4 230 460 4 1990 5 192 576 9 1991 250 1000 16 1992 15 274 1370 25 1150 3610 55 Substituting the values N = 5, XY = 1150, 6X = 15, 6XY = 3610, 6X2 = 55 in the two. Normal equations: 6Y = Na + b6x

6XY = a6X + b6X2 We get 1150 = 5a + 15b … (1) 3610 = 15a + 55b … (2) Now, 3450 = 15a + 45b … (1) That is multiplying equation (1) by 3 and subtracting the resulting equation (1) from equation (2) we get: 160 = 10b 160 ? b = 10 = 16 Now 5a = 1150 – 240 910 ? a = 5 = 1182 The equation to the line of best fit is: Y = 182 + 16x CU IDOL SELF LEARNING MATERIAL (SLM)

Time Series Analysis 363 By putting X = 1, 2, 3, 4, 5 in the above equation, we obtain the trend values 198, 214, 230, 246 and 262 respectively. X=1,Y=182+16=198 X=2,Y=182+32=214 X=3,Y=182 +48=230 X=4,Y=182 +64=246 X=5,Y=182 +80=262 Short Cut Method We can simplify above calculations by using a short cut method. This method consists of numbering the years (time) under the column heading X in a different manner. We put zero at the centre and write negative and positive values above and below it — that is, on either side of zero. By doing so, we get X = 0 in the two normal equations we obtain the following simplified equations: 6Y = Na6XY = b6X2 ¦Y ¦XY a = N b = ¦x2 We now use the short cut method to fit a straight line trend (least squares) to the data given in the previous example. Let the equation to the line of best fit be: Y = a + bX

To find the values of the constants a & b we prepare the following table: Year YX X2 XY 1988 204 –2 4 -408 1989 230 –1 1 -230 1990 192 0 00 1991 250 +1 1 250 1992 274 +2 4 548 1150 10 160 For the table we have N = 5, 6x2 = 10, 6X = 1150, 6XY = 1600.

CU IDOL SELF LEARNING MATERIAL (SLM) 364 Business Mathematics and Statistics Substituting these values in the two simplified equations it follows that: ¦Y 1150 a = N = 5 = 230 And, ¦ XY 160 b = ¦x2 = 10 = 16 Therefore the equation to the line of best fit is – Y=230+16X By putting x = –2, –1, 0, 1 and 2 in this equation, at each stage, we get the values of the trend ordinates 198, 214, 230, 246 and 262 respectively. The following graph shows vividly actual sales trend, both in freehand and by the method of least squares.

SALES OF X, Y, Z INDUSTRIES 300 250 200 ACTUAL SALES AND FORECAST SALES ACTUAL SALES 150 TREND (FREE·HAND) TREND (LEAST·SQUARES) 100 50 0 1988 1989 1990 1991 1902

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Time Series Analysis 365 16.5 Solved Problems Problem 5: Fit an exponential trend Y = abx to the following data by the method of least squares and find the trend values for the years 1994-98. Year(X): 1994 1995 1996 1997 1998 1999 2000 Sales(Y) : 87 97 113 129 202 195 193 The exponential curve, is given by the equation Y = abx which can be written in the logarithmic form Log Y = log a + X log b … (1) If the deviations (X) of the given years are calculated from the middle year 1945, then 6X = 0 and the values of log a and log b is given by: ¦log Y ¦(Xlog Y) log a = N and, log b = ¦X2 Year Sales X Log Y XLogY X2 1994 87 –3 1.9395 –5.8185 9 1995 97 –2 1.9868 –3.9736 4 1996 113 –1 2.0531 –2.0531 1 1997 0 2.1106 0 0 1998 129 1 2.3054 1 202 2.3054

1999 195 2 2.2900 4.58 4 2000 193 3 2.2856 6.8568 9 0 14.9710 1.8970 28 N = 7, 6logY = 14.9710, 6(X log Y) = 1.8970, 6X2 =28. ¦log Y 14.9710 log a = = 2.1387 N 7 And 6(X log Y) = log b (6X2) ¦(XlogY) 1.9870 log b = ¦X2 28 = 0.0678 Substituting the values of log a and log b in equation (I), we obtain the equation for the exponential trend in the logarithmic form CU IDOL SELF LEARNING MATERIAL (SLM)

366 Business Mathematics and Statistics log Y = 2.1387 + 0.0678X. Further a = Antilog (2.1387) = 137.6 b = Antilog (0.0678) = 1.169. Therefore the equation of the exponential curve is: Y = 137.6(1.169)x Substituting the relevant values of X (deviations from 1945 in the equation log Y = 2.1387 + 0.0678 X we get the logarithms of the trend values for the different years. By finding the anti- logarithms of these we obtain the trend values. The trend values for the years 1944 to 1948 are as follows: Year X log Y 1994 –1 2.1387 + (0.0678) (–1) = 2.0709 1995 0 2.1387 + (0.0678) (0) = 2.1387 1996 1 2.1387 + (0.0678) (1) = 2.2065 1997 2 2.1387 + (0.0678) (2) = 2.2743 1998 3 2.1387 + (0.0678) (3) = 2.3421 Year Trend values 1994 Antilog (2.0709) = 117.7 1995 Antilog (2.1387) = 137.6 1996 Antilog (2.2065) = 160.9 1997 Antilog (2.2743) = 188.0

1998 Antilog (2.3421) = 219.9 Problem 6: Use the method of moving averages (three yearly) to determine the trend in the following series showing the index numbers for values of imports into India during 1914-1928. 87, 62, 47, 24, 45, 57, 96, 97, 84, 79, 77, 80, 92, 106, and 113. CU IDOL SELF LEARNING MATERIAL (SLM)