Maharashtra Board Class 12 Maths Part 2 Chapter 4 Time Series Miscellaneous Solutions

(I) Choose The Correct Alternative.

Question 1. Which of the following can't be a component of a time series?
(a) Seasonality
(b) Cyclical
(c) Trend
(d) Mean
Answer: (d) Mean
In simple words: A time series typically consists of trend, cyclical, seasonal, and irregular components, while the mean is a statistical measure of central tendency for a dataset, not a component describing patterns over time.

🎯 Exam Tip: Understanding the fundamental components of a time series is crucial for identifying and classifying different types of patterns in data over time.

Question 2. The first step in time series analysis is to
(a) Perform regression calculations
(b) Calculate a moving average
(c) Plot the data on a graph
(d) Identify seasonal variation
Answer: (c) Plot the data on a graph
In simple words: The initial step in time series analysis involves visually representing the data, usually by plotting it on a graph, to observe any obvious patterns, trends, or anomalies before applying more complex analytical methods.

🎯 Exam Tip: Visual inspection of time series data through a graph provides immediate insights into its behavior and helps in choosing appropriate forecasting models.

Question 3. Time-series analysis is based on the assumption that
(a) Random error terms are normally distributed.
(b) The variable to be forecast and other independent variable are correlated.
(c) Past patterns in the variable to be forecast will continue unchanged into the future.
(d) The data do not exhibit a trend.
Answer: (c) Past patterns in the variable to be forecast will continue unchanged into the future.
In simple words: Time series forecasting fundamentally assumes that historical patterns and relationships within the data will persist and be observable in future data points, allowing past behavior to predict future outcomes.

🎯 Exam Tip: This core assumption is critical for all time series forecasting; if past patterns do not hold, forecasts will be unreliable.

Question 4. Moving averages are useful in identifying
(a) Seasonal component
(b) Irregular component
(c) Trend component
(d) Cyclical component
Answer: (c) Trend component
In simple words: Moving averages help to smooth out short-term fluctuations and irregular variations in a time series, thereby making the underlying long-term trend more apparent.

🎯 Exam Tip: Moving averages are a basic but effective tool for trend analysis, especially when dealing with noisy data that obscures the general direction of movement.

Question 5. We can use regression line for past data to forecast future data. We then use the line which
(a) Minimizes the sum of squared deviations of past data from the line.
(b) Minimizes the sum of deviations of past data from the line.
(c) Maximizes the sum of squared deviations of past data from the line.
(d) Maximizes the sum of deviation of past data from the line.
Answer: (a) Minimizes the sum of squared deviations of past data from the line
In simple words: The method of least squares, commonly used to fit a regression line, determines the line that best fits the data by ensuring the sum of the squared vertical distances (deviations) from each data point to the line is as small as possible.

🎯 Exam Tip: The principle of least squares is fundamental in regression analysis for finding the "best fit" line, as squaring deviations prevents positive and negative errors from canceling out and heavily penalizes larger errors.

Question 6. Which of the following is a major problem for forecasting, especially when using the method of least squares?
(a) The past cannot be known
(b) The future is not entirely certain
(c) The future exactly follows the patterns of the past
(d) The future may not follow the patterns of the past
Answer: (d) The future may not follow the patterns of the past
In simple words: A significant challenge in forecasting, particularly with methods like least squares that rely on historical data, is the inherent uncertainty that past patterns or trends may not perfectly continue into the future due to unforeseen changes or events.

🎯 Exam Tip: While least squares provides an optimal fit to historical data, its predictive power diminishes if the underlying data generation process changes, highlighting a key limitation of all forecasting models.

Question 7. An overall upward or downward pattern in an annual time series would be contained in which component of the time series
(a) Trend
(b) Cyclical
(c) Irregular
(d) Seasonal
Answer: (a) Trend
In simple words: The trend component of a time series represents the long-term, underlying direction or general movement of the data, indicating whether it is consistently increasing, decreasing, or remaining stable over a prolonged period.

🎯 Exam Tip: Distinguishing the trend from other components is vital for understanding the fundamental long-term changes in a variable, separating it from shorter-term seasonal or cyclical fluctuations.

Question 8. The following trend line equation was developed for annual sales from 1984 to 1990 with 1984 as base or zero year. Y1 = 500 + 60X (in 1000 Rs.) The estimated sales for 1984 (in 1000 Rs) is:
(a) Rs. 500
(b) Rs. 560
(c) Rs. 1,040
(d) Rs. 1100
Answer: (a) Rs. 500
In simple words: Since 1984 is designated as the base or zero year, setting X=0 in the given trend line equation Y = 500 + 60X directly yields the estimated sales for that year as 500 (in 1000 Rs).

🎯 Exam Tip: For trend equations where X represents time relative to a base year, the estimated value for the base year itself is simply the intercept (constant term) of the equation, as X will be zero.

Question 9. What is a disadvantage of the graphical method of determining a trend line?
(a) Provides quick approximations
(b) Is subject to human error
(c) Provides accurate forecasts
(d) Is too difficult to calculate
Answer: (b) Is subject to human error
In simple words: The graphical method for fitting a trend line is subjective because it relies on an individual's visual judgment to draw the line, which can lead to inconsistencies and varying results among different analysts.

🎯 Exam Tip: While simple and quick for initial insights, graphical methods lack the objectivity and precision of mathematical or statistical approaches, making them less suitable for formal analysis or critical decision-making.

Question 10. Which component of time series refers to erratic time series movements that follow no recognizable or regular pattern.
(a) Trend
(b) Seasonal
(c) Cyclical
(d) Irregular
Answer: (d) Irregular
In simple words: The irregular component of a time series captures unpredictable, random fluctuations that cannot be explained by trend, cyclical, or seasonal patterns, often resulting from unforeseen or rare events.

🎯 Exam Tip: The irregular component is often considered "noise" in the data; understanding its presence helps differentiate actual patterns from random variations.

(II) Fill In The Blanks.

Question 1. __________ components of time series is indicated by a smooth line.
Answer: Trend
In simple words: The long-term direction of a time series, known as the trend, is typically represented by a smoothed line that filters out short-term variations, showcasing the overall upward, downward, or stable movement.

🎯 Exam Tip: A smooth line helps to visually isolate the underlying trend from the noise of seasonal, cyclical, and irregular fluctuations, providing a clearer picture of long-term development.

Question 2. __________ component of time series is indicated by periodic variation year after year.
Answer: Seasonal
In simple words: Seasonal variations are repetitive, predictable patterns in a time series that occur within a year, such as monthly, quarterly, or daily, and repeat annually.

🎯 Exam Tip: Recognizing seasonal components is crucial for accurate forecasting, as they represent predictable intra-year fluctuations that can be accounted for in models.

Question 3. __________ component of time series is indicated by a long wave spanning two or more years.
Answer: Cyclical
In simple words: The cyclical component describes fluctuations in a time series that extend over several years, characterized by periods of expansion and contraction (like economic cycles), and are not fixed in their duration or amplitude.

🎯 Exam Tip: Cyclical variations are generally longer in duration and less predictable in their exact timing than seasonal variations, making them more challenging to model accurately.

Question 4. __________ component of time series is indicated by up and down movements without any pattern.
Answer: Irregular
In simple words: The irregular component accounts for random, unpredictable fluctuations in a time series that are not attributable to trend, seasonality, or cyclical patterns, often due to unpredictable events.

🎯 Exam Tip: Irregular components represent the residual variation after accounting for systematic patterns; they are typically assumed to be random and unpredictable.

Question 5. Additive models of time series __________ independence of its components.
Answer: assume
In simple words: In an additive time series model, it is presumed that the components (trend, seasonal, cyclical, irregular) are independent of each other and simply sum up to form the observed data.

🎯 Exam Tip: Additive models are suitable when the magnitude of seasonal or irregular variations does not change with the level of the time series, assuming a constant absolute effect.

Question 6. Multiplicative models of time series __________ independence of its components.
Answer: does not assume
In simple words: A multiplicative time series model posits that the components interact with each other in a proportional way, meaning the magnitude of variations (like seasonality or irregularity) changes with the level of the trend.

🎯 Exam Tip: Multiplicative models are used when the fluctuations' size increases as the series' value increases, implying a percentage-based effect of components rather than an absolute one.

Question 7. The simplest method of measuring the trend of time series is __________
Answer: graphical method
In simple words: The graphical method involves simply plotting the time series data and then drawing a freehand smooth curve or line through the points to visually estimate the underlying trend.

🎯 Exam Tip: While easy and quick, the graphical method is subjective and lacks precision, often used for initial exploratory data analysis rather than rigorous forecasting.

Question 8. The method of measuring the trend of time series using only averages is __________
Answer: moving average method
In simple words: The moving average method calculates the average of data points over a specific period and then shifts this window forward, smoothing out short-term fluctuations to reveal the trend.

🎯 Exam Tip: Moving averages are effective for smoothing data and identifying trends, but they introduce a lag and lose data points at the beginning and end of the series.

Question 9. The complicated but ancient method of measuring the trend of time series is __________
Answer: least-squares method
In simple words: The least-squares method is a statistical technique that fits a straight line or curve to time series data by minimizing the sum of the squared differences between the observed data points and the fitted line, providing an objective measure of the trend.

🎯 Exam Tip: Despite its computational intensity (compared to graphical or simple moving averages), the least-squares method provides a statistically rigorous and objective way to determine the trend, essential for more accurate modeling.

Question 10. The graph of time series clearly shows __________ of it is monotone.
Answer: trend
In simple words: A monotone trend in a time series graph indicates a consistent increase or decrease over time without significant reversals in direction, showing a clear, steady progression.

🎯 Exam Tip: Observing a monotone trend visually simplifies initial model selection, as it suggests a stable underlying direction without complex oscillations.

(III) State Whether Each Of The Following Is True Or False.

Question 1. The secular trend component of the time series represents irregular variations.
Answer: False
In simple words: The secular trend represents the long-term, underlying direction of a time series, while irregular variations are unpredictable, random fluctuations, meaning they are distinct components.

🎯 Exam Tip: It is crucial to differentiate between trend (long-term direction) and irregular variation (random noise) as they describe different aspects of time series behavior.

Question 2. Seasonal variation can be observed over several years.
Answer: True
In simple words: Seasonal variation refers to patterns that recur within a year (e.g., monthly, quarterly) but these patterns repeat consistently year after year.

🎯 Exam Tip: While seasonal patterns are intra-year, their recurrence over multiple years is what makes them a predictable and identifiable component of a time series.

Question 3. Cyclical variation can occur several times in a year.
Answer: False
In simple words: Cyclical variations are fluctuations that typically span two or more years, unlike seasonal variations which complete their pattern within a single year.

🎯 Exam Tip: The distinguishing characteristic of cyclical patterns is their duration, which is longer than a year, separating them from the shorter, regular seasonal movements.

Question 4. Irregular variation is not a random component of time series.
Answer: False
In simple words: Irregular variation is inherently random and unpredictable, representing the residual fluctuations in a time series that cannot be explained by any systematic patterns like trend, seasonality, or cycles.

🎯 Exam Tip: Irregular variations are considered the "error" or "noise" component, making them by definition random and distinct from the systematic components of a time series.

Question 5. The additive model of time series does not require the assumptions of independence of its components.
Answer: False
In simple words: The additive model assumes that the components of a time series (trend, seasonal, cyclical, irregular) are independent of each other and simply sum up to form the observed data, meaning their effect is constant regardless of the series' level.

🎯 Exam Tip: A core assumption of the additive model is that components contribute their effects independently; if there's interaction or proportional effect, a multiplicative model is more appropriate.

Question 6. The multiplicative model of time series does not require the assumption of independence of its components.
Answer: True
In simple words: The multiplicative model assumes that the components of a time series interact with each other proportionally, meaning, for instance, that seasonal variations increase in magnitude as the trend increases, thus they are not independent.

🎯 Exam Tip: Multiplicative models are used when the amplitude of seasonal or irregular fluctuations scales with the trend level, reflecting an interdependence rather than simple addition of components.

Question 7. The graphical method of finding trends is very complicated and involves several calculations.
Answer: False
In simple words: The graphical method is actually the simplest approach to finding trends, as it involves merely plotting data and visually drawing a line, requiring no complex calculations.

🎯 Exam Tip: While straightforward, the graphical method's simplicity comes at the cost of subjectivity and lack of statistical rigor, making it less suitable for precise analysis.

Question 8. Moving the average method of finding trends is very complicated and involves several calculations.
Answer: False
In simple words: The moving average method, while involving some arithmetic calculations, is generally considered straightforward and less complicated than statistical methods like least squares, as it primarily involves successive averaging.

🎯 Exam Tip: Moving averages are computationally simpler than regression models, making them an accessible initial method for trend smoothing without extensive statistical background.

Question 9. The least-squares method of finding trends is very simple and does not involve any calculations.
Answer: False
In simple words: The least-squares method is a statistical technique that requires complex mathematical calculations, including sums of squares and solving simultaneous equations, to determine the trend line that best fits the data.

🎯 Exam Tip: Despite its analytical power, the least-squares method is computationally intensive, especially for large datasets or complex models, distinguishing it from simpler, non-calculative methods.

Question 10. All three methods of measuring trends will always give the same results.
Answer: False
In simple words: Different methods of measuring trends-such as graphical, moving average, and least squares-employ distinct methodologies and assumptions, leading to potentially different trend lines and results for the same dataset.

🎯 Exam Tip: The choice of trend measurement method impacts the outcome; methods vary in objectivity, smoothing capabilities, and mathematical rigor, so results will differ.

(IV) Solve The Following Problems.

Question 1. The following table shows the productivity of pig-iron and ferro-alloys ('000 metric tonnes)

Fit a trend line to the above data by graphical method.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र 1974 से 1982 तक पिग-आयरन और फेरो-अलॉयज़ के उत्पादन डेटा को दर्शाता है। इसमें प्रत्येक वर्ष के लिए उत्पादन मूल्यों को बिंदुओं के रूप में प्लॉट किया गया है और एक सीधी रेखा (ट्रेंड लाइन) को इन बिंदुओं के माध्यम से खींचा गया है ताकि डेटा के दीर्घकालिक रुझान को दर्शाया जा सके। ट्रेंड लाइन कुल मिलाकर एक ऊपर की ओर की प्रवृत्ति दिखाती है, जो उत्पादन में वृद्धि का संकेत देती है।
In simple words: To fit a trend line graphically, plot the given production data for each year as points on a graph and then draw a straight line or smooth curve that visually best represents the general upward or downward movement of these points over time.

🎯 Exam Tip: When fitting a trend line graphically, aim to draw a line that balances the points above and below it, effectively capturing the overall direction without being overly influenced by individual fluctuations.

Question 2. Fit a trend line to the data in Problem IV (1) by the method of least squares.
Answer:
Solution:

\(u = \frac{t-1978}{1}\)
\( \Sigma y = 57, \Sigma u = 0, \Sigma u^2 = 60, \Sigma uy = 38, n = 9 \)
Let the equation of the trend line be
\( Y = a + bu \)......(i)
\( \Sigma y = na + b\Sigma u \)......(ii)
\( \Sigma uy = a\Sigma u + b\Sigma u^2 \).........(iii)
Substituting the values of \( \Sigma u, n, \Sigma uy, \Sigma u^2 \) in (ii) & (iii) we get
\( 57 = 9a + 0 \)
\( \therefore a = 6.3333 \)
\( 38 = 0 + b \cdot 60 \)
\( \therefore b = 0.6333 \).
\( \therefore \) The equation of the trend line is
\( y = 6.3333 + 0.63333u \) where \( u = t - 1978 \)
In simple words: The least squares method involves calculating sums of the given variables (Y, u, u^2, uy) to derive two normal equations. Solving these equations determines the values for 'a' (the intercept) and 'b' (the slope), which define the best-fit linear trend line for the data.

🎯 Exam Tip: When using the least squares method, ensure accurate calculation of sums and products (like Σy, Σu, Σu^2, Σuy) as these are critical for correctly solving the normal equations and deriving the trend line equation.

Question 3. Obtain the trends values for the data on problem IV (1) using 5 yearly moving averages.
Answer:
Solution:

In simple words: To calculate 5-yearly moving averages, sum up five consecutive production values and then divide by 5 to get the average, centering this average on the middle year of the five-year period. Repeat this process by moving the window one year forward for each subsequent calculation.

🎯 Exam Tip: When computing moving averages, remember that you lose (window-1)/2 data points at the beginning and end of the series, meaning a 5-year moving average starts from the 3rd year and ends 2 years before the last data point.

Question 4. The following table shows the amount of sugar production (in lac tonnes) for the years 1971 to 1982.

Fit a trend line to the above data by graphical method.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र 1971 से 1982 तक चीनी उत्पादन (लाख टन में) के डेटा को दर्शाता है। प्रत्येक वर्ष के लिए उत्पादन मूल्यों को ग्राफ पर बिंदुओं के रूप में प्लॉट किया गया है, और एक ट्रेंड लाइन इन बिंदुओं के बीच से खींची गई है। ट्रेंड लाइन चीनी उत्पादन में दीर्घकालिक वृद्धि की प्रवृत्ति को दर्शाती है, हालांकि डेटा में वर्ष-दर-वर्ष उतार-चढ़ाव भी दिख रहा है।
In simple words: To fit a trend line graphically, plot each year's sugar production value as a point on a graph, and then draw a smooth, straight line that best captures the general upward or downward movement of the data over the entire period.

🎯 Exam Tip: When creating a graphical trend line, ensure your graph has properly labeled axes (Year on X-axis, Production on Y-axis) and a clear title, making the visual representation understandable and interpretable.

Question 5. Fit a trend line to data in problem 4 by the method of least squares.
Answer:
Solution:

\(u = \frac{t-1976.5}{\frac{1}{2}}\)
\( \Sigma y = 38, \Sigma u = 0, \Sigma u^2 = 572, \Sigma uy = 160, n = 12 \)
Let the equation of the trend line be
\( y = a + bu \).........(i)
\( u = 2t-3953 \)
\( \Sigma y = na + b\Sigma u \)..........(ii)
\( \Sigma uy = a\Sigma u + b\Sigma u^2 \).........(iii)
Substituting the values of \( \Sigma y, n, \Sigma u, \Sigma uy, \Sigma u^2 \) in (ii) & (iii) we get
\( 38 = 12a + 0 \)
\( \therefore a = 3.1667 \)
\( 160 = 0 + b \cdot 572 \)
\( \therefore b = 0.2797 \)
\( \therefore \) by (i), Equation of the trend line is
\( Y = 3.1667 + 0.2797u \) where \( u = 2t - 3953 \).
In simple words: The process involves transforming the time variable (t) into a simplified 'u' to make calculations easier, then setting up and solving two normal equations using the sums of Y, u, u^2, and uy to determine the intercept (a) and slope (b) of the linear trend line.

🎯 Exam Tip: When the number of years (n) is even, the origin for 'u' is set between the two middle years, and 'u' values are in half-year units to simplify calculations and ensure Σu = 0, which is crucial for easier determination of 'a'.

Question 6. Obtain trend values for data in Problem 4 using 4-yearly centered moving averages.
Answer:
Solution:

In simple words: To calculate 4-yearly centered moving averages, first compute 4-yearly moving totals and then their averages. Since these averages fall between years, sum two consecutive 4-yearly averages and divide by two to center the trend value on a specific year.

🎯 Exam Tip: Centered moving averages are used for even-period moving averages (like 4-yearly) to align the trend value with an actual time point, effectively smoothing the data while maintaining temporal consistency.

Question 7. The percentage of girls' enrollment in total enrollment for years 1960-2005 is shown in the following table.

Fit a trend line to the above data by graphical method.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र 1960 से 2005 तक कुल नामांकन में लड़कियों के नामांकन के प्रतिशत डेटा को दर्शाता है। प्रत्येक वर्ष के लिए प्रतिशत मूल्यों को ग्राफ पर बिंदुओं के रूप में प्लॉट किया गया है। इन बिंदुओं के माध्यम से एक ट्रेंड लाइन खींची गई है, जो लड़कियों के नामांकन प्रतिशत में समय के साथ एक स्पष्ट ऊपर की ओर की प्रवृत्ति दिखाती है, यह दर्शाता है कि यह अनुपात लगातार बढ़ रहा है।
In simple words: Plot the given data points of girls' enrollment percentage against their respective years on a graph, then draw a straight line that visually best represents the overall direction of the data, indicating a general increase, decrease, or stability.

🎯 Exam Tip: When graphically fitting a trend line, ensure that the line minimizes the visual distance to all data points, providing a balanced representation of the overall pattern in the time series.

Question 8. Fit a trend line to the data in Problem 7 by the method of least squares.
Answer:
Solution:

\(u = \frac{t-1980.5}{5}\)
\( \Sigma y = 51, \Sigma u = 0, \Sigma u^2 = 330, \Sigma uy = 157, n = 10 \)
Let the equation of the trend line be
\( Y = a + bu \).........(i)
\( \Sigma y = na + b\Sigma u \)..........(ii)
\( \Sigma uy = a\Sigma u + b\Sigma u^2 \).........(iii)
Substituting the values of \( \Sigma y, \Sigma u, n, \Sigma uy, \Sigma u^2 \) We get
\( 51 = 10a + 0 \)
\( \therefore a = 5.1 \)
and \( 157 = 0 + 6.330 \)
\( \therefore b = 0.4758 \)
by (i) equation of the trend line is
\( Y = 5.1 + 0.4758u \) where \( u = \frac{t-1980.5}{5} \)
In simple words: The least squares method involves preparing a table to compute sums for Y, u, u^2, and uy. These sums are then used in the normal equations to solve for the intercept 'a' and slope 'b', which define the linear trend line equation for forecasting.

🎯 Exam Tip: For least squares with an even number of data points, it's crucial to correctly determine the origin for 'u' (midpoint between the two central years) and the unit of 'u' (usually half the interval between years) to simplify computations and ensure Σu = 0.

Question 9. Obtain trend values for data in Problem 7 using 4-yearly moving averages.
Answer:
Solution:

In simple words: To calculate 4-yearly moving averages, sum up four consecutive data points and divide by four, then center this average between the two middle years of that four-year span to get the trend value. Repeat this for the entire series.

🎯 Exam Tip: Remember that 4-yearly moving averages, like other even-period moving averages, require a second averaging step (centering) to align the trend values with specific time points rather than falling between them.

Question 10. The following data shows the number of boxes of cereal sold in the years 1977 to 1984.

Fit a trend line to the above data by graphical method.
Solution:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख 1977 से 1984 तक बेचे गए अनाज के बक्सों की संख्या ('000 में) को दर्शाता है। इसमें प्रत्येक वर्ष के डेटा पॉइंट्स को प्लॉट किया गया है और फिर इन पॉइंट्स के माध्यम से एक रुझान रेखा (ट्रेंड लाइन) खींची गई है, जो समय के साथ बिक्री के सामान्य पैटर्न को प्रदर्शित करती है। यह रेखा दिखाती है कि डेटा में समग्र रूप से कैसा उतार-चढ़ाव है और एक अनुमानित दीर्घकालिक प्रवृत्ति क्या है।
In simple words: This question asks us to visually represent the given cereal sales data over time and then draw a line that shows the general direction or trend of sales. This helps in understanding the pattern of sales movement without complex calculations.

🎯 Exam Tip: For graphical method questions, ensure your plot points are accurate and the trend line represents the central tendency of the data, extending appropriately for forecasting.

Question 11. Fit a trend line to data in Problem 10 by the method of least squares.
Solution:

\[ u = \frac{t-1980.5}{1/2} \] \( \Sigma y = 39, \Sigma u = 0, \Sigma u^2 = 168, \Sigma uy = 79, n = 8 \) Let the equation of the trend line by \( Y = a + bu \) .......(i) Where \( u = 2t - 3961 \) .......(i) \( \Sigma y = na + b\Sigma u \) .......(ii) \( \Sigma uy = a\Sigma u + b\Sigma u^2 \) ........(iii) Substituting the values of \( \Sigma y, n, \Sigma u, \Sigma uy, \Sigma u^2 \), in (ii) & (iii) \( 39 = 8a + 0 \)
\( \implies a = 4.875 \) \( 79 = 0 + b (168) \)
\( \implies b = 0.4702 \) by (i) the equation of the trend line is \( Y = 4.875 + 0.4702u \) Where \( u = 2t - 3961 \).
In simple words: This solution uses the least squares method to find a mathematical equation for the trend line. We calculate 'a' and 'b' coefficients from the given data to define the line \( Y = a + bu \), which best fits the past sales figures and can be used for future predictions.

🎯 Exam Tip: When applying the least squares method, correctly calculating \( \Sigma u \), \( \Sigma y \), \( \Sigma u^2 \), and \( \Sigma uy \) is crucial. Ensure 'u' is properly centered and scaled to simplify calculations, especially when \( \Sigma u = 0 \).

Question 12. Obtain trend values for data in Problem 10 using 3-yearly moving averages.
Solution:

In simple words: To find the 3-yearly moving averages, we sum up the sales for three consecutive years and then divide by three to get an average that smooths out short-term fluctuations, revealing the underlying trend. This process is repeated by shifting one year at a time.

🎯 Exam Tip: Remember that for an odd number of years (like 3-yearly), the moving average is placed against the middle year of the period. Be careful with the first and last few values as they won't have a full 3-year window.

Question 13. The following table shows the number of trade fatalities (in a state) resulting from drunken driving for the years 1975 to 1983.

Fit a trend line to the above data by graphical method.
Solution:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख 1975 से 1983 तक शराब पीकर गाड़ी चलाने से हुई मौतों की संख्या को दर्शाता है। प्रत्येक वर्ष के डेटा बिंदुओं को प्लॉट किया गया है, और फिर इन बिंदुओं से होकर एक प्रवृत्ति रेखा (ट्रेंड लाइन) खींची गई है। यह रेखा समय के साथ इन घटनाओं के दीर्घकालिक पैटर्न और समग्र दिशा को दर्शाती है, जो उतार-चढ़ाव के बावजूद एक सामान्य प्रवृत्ति प्रस्तुत करती है।
In simple words: This question asks for a visual plot of road fatalities due to drunken driving over several years, followed by drawing a trend line to show the general direction of these numbers. This graph helps to see if fatalities are increasing, decreasing, or staying stable over time.

🎯 Exam Tip: When drawing a trend line graphically, try to balance the points above and below the line to ensure it represents the overall direction of the data accurately.

Question 14. Fit a trend line to data in Problem 13 by the method of least squares.
Solution:

\[ u = \frac{t-1979}{1} \] \( \Sigma y = 47, \Sigma u = 0, \Sigma u^2 = 60, \Sigma uy = 40, n = 9 \) Let the equation of the trends line be \( Y = a + bu \) where \( u = t - 1979 \) .......(i) \( \Sigma y = na + b\Sigma u \) .........(ii) \( \Sigma uy = a\Sigma u + b\Sigma u^2 \) ........(iii) Substituting values of \( \Sigma y, n, \Sigma u, \Sigma uy, \Sigma u^2 \) in (ii) & (iii) We get \( 47 = 9a + 0 \)
\( \implies a = 5.2222 \) and \( 40 = 0 + b(60) \)
\( \implies b = 0.6667 \) \( \therefore \) by (i) the equation of the trend line is \( Y = 5.2222 + 0.6667u \) Where \( u = t - 1979 \).
In simple words: This solution calculates the specific trend line equation \( Y = a + bu \) for the fatalities data using the least squares method. By finding the values of 'a' and 'b', we create a mathematical model that describes the average change in fatalities over time.

🎯 Exam Tip: When the number of years 'n' is odd, centering 'u' around the middle year makes \( \Sigma u = 0 \), significantly simplifying the calculations for 'a' and 'b' in the least squares equations.

Question 15. Obtain trend values for data in Problem 13 using 4-yearly moving averages.
Solution:

In simple words: For 4-yearly moving averages, we sum four consecutive fatality values and then divide by four. Because it's an even number of years, these averages are centered between the second and third year of each group, providing a smoothed trend.

🎯 Exam Tip: For even-numbered moving averages (like 4-yearly), the average value is initially placed between the two central years. To properly center it, a 2-period moving average of these initial averages is then calculated and placed on the central year, yielding a centered moving average.

Question 16. The following table shows the all India infant mortality rates (per '000) for the years 1980 to 2000.

Fit a trend line to the above data by graphical method.
Solution:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख 1980 से 2010 तक अखिल भारतीय शिशु मृत्यु दर (प्रति 1000) को दर्शाता है। प्रत्येक दिए गए वर्ष के लिए IMR डेटा बिंदुओं को प्लॉट किया गया है। इसके बाद, इन बिंदुओं के माध्यम से एक रुझान रेखा खींची गई है, जो समय के साथ शिशु मृत्यु दर में गिरावट की दीर्घकालिक प्रवृत्ति को स्पष्ट रूप से इंगित करती है।
In simple words: This question asks us to plot the infant mortality rates over different years on a graph and then draw a line that visually represents the general pattern or trend of these rates. This helps to see if the IMR is increasing, decreasing, or stable over time.

🎯 Exam Tip: When interpreting a graphical trend line, look at its slope - an upward slope indicates an increasing trend, a downward slope indicates a decreasing trend, and a flat line suggests stability.

Question 17. Fit a trend line to data in Problem 16 by the method of least squares.
Solution:

\[ u = \frac{t-1995}{5} \] \( \Sigma y = 30, \Sigma u = 0, \Sigma u^2 = 70, \Sigma uy = -70, n = 7 \) Let the equation of the trend line be \( Y = a + bu \) Where \( u = \frac{t-1995}{5} \) .......(i) \( \Sigma y = na + b\Sigma u \) ......(ii) \( \Sigma uy = a\Sigma u + b\Sigma u^2 \) .......(iii) Substituting values of \( \Sigma y, n, \Sigma u, \Sigma uy \) & \( \Sigma u^2 \) in (ii) & (iii) we get \( 30 = 7a + 0 \)
\( \implies a = 4.2857 \) \( -70 = 0 + b(70) \)
\( \implies b = -1 \) \( \therefore \) by (i) the equation of the trend line is \( y = 4.2857 - 1(u) \) Where \( u = \frac{t-1995}{5} \).
In simple words: This solution finds a mathematical equation for the infant mortality rate trend using the least squares method. It establishes a linear relationship between year and IMR to predict future rates based on past data.

🎯 Exam Tip: When using the least squares method, remember that a negative 'b' coefficient in the trend equation \( Y = a + bu \) indicates a decreasing trend, while a positive 'b' indicates an increasing trend.

Question 18. Obtain trend values for data in Problem 16 using 3-yearly moving averages.
Solution:

In simple words: We calculate the 3-yearly moving average by summing up three consecutive IMR values and dividing by three. This helps to smooth out minor fluctuations in the data and reveal the general downward trend of infant mortality rates.

🎯 Exam Tip: Moving averages are effective for smoothing out short-term noise and making the underlying trend more apparent, especially for data with many fluctuations.

Question 19. The following table shows the wheat yield ('000 tonnes) in India for the years 1959 to 1968.

Fit a trend line to the above data by the method of least squares.
Solution:

\[ u = \frac{t-1963.5}{1/2} \] \( \Sigma y = 24, \Sigma u = 0, \Sigma u^2 = 330, \Sigma uy = 94, n = 10 \) Let the equation of the trend line be \( y = a + bu \) where \( u = \frac{t-1963.5}{1/2} \) ...(i) i.e. \( u = 2t - 3927 \) \( \Sigma y = na + b\Sigma u \) .......(ii) \( \Sigma uy = a\Sigma u + \Sigma u^2 \) .....(iii) Substituting values of \( \Sigma y, n, \Sigma u, \Sigma uy \) & \( \Sigma u^2 \) in (ii) & (iii) we get \( 24 = 10a + 0 \)
\( \implies a = 2.4 \) \( 94 = 0 + 6.330 \)
\( \implies b = 0.2848 \) \( \therefore \) Equation of the trend line is \( y = 2.4 + (0.2848)u \) where \( u = 2t - 3927 \).
In simple words: We are finding a linear equation for the wheat yield data using the least squares method. This equation helps us understand the average increase or decrease in yield over the years by fitting a straight line to the data.

🎯 Exam Tip: When 'n' is even, 'u' is typically centered between the two middle observations. This might involve using a fractional step in the 'u' transformation, so be precise with the calculations to maintain accuracy.

Question 20. Obtain trend values for data in problem 19 using 3-yearly moving averages.
Solution:

In simple words: This involves calculating the average of wheat yields over three consecutive years to smooth out short-term changes and highlight the underlying long-term trend in productivity.

🎯 Exam Tip: For moving averages, ensure the calculated average value is correctly positioned in the middle of the period it represents. For 3-yearly averages, the value is always associated with the second year of the three-year group.