GSEB Class 12 Statistics Solutions Chapter 4 Time Series

Section A

Answer the following questions by selecting a correct option from the given options:

Question 1. Which type of variations are produced in the time series variable due to seasonal component ?
(a) Long-term
(b) Irregular
(c) Regular
(d) Zero
Answer: (c) Regular
In simple words: Seasonal changes in a time series cause variations that happen regularly over a short period, usually less than a year. These are predictable and follow a pattern.

🎯 Exam Tip: Understanding the nature of variations (regular, irregular) is key to identifying different time series components for higher scores.

 

Question 2. on is shown in 'decrease in the production of a company' due to strike ?
(a) Random
(b) Trend
(c) Seasonal
(d) Cyclical
Answer: (a) Random
In simple words: A strike causes a sudden, unpredictable drop in production that does not follow a regular pattern. This type of variation is called a random component.

🎯 Exam Tip: Random components represent unpredictable events that can significantly impact a time series but are not part of regular patterns.

 

Question 3. Name the method for fitting the linear equation to find linear trend.
(a) Graphical Method
(b) Method of least squares
(c) Method of moving average
(d) Method of partial average
Answer: (b) Method of least squares
In simple words: The method of least squares is used to find the best-fitting straight line for data, which helps in identifying a linear trend in a time series.

🎯 Exam Tip: The method of least squares provides a mathematical way to determine a trend line, making it a precise tool for analysis.

 

Question 4. How do you show the additive model of the time series ?
(a) \( y_t = T_t + S_t + C_t – R_t \)
(b) \( y_t = T_t + S_t + C_t + R_t \)
(c) \( Y_t = T_t \times S_t + C_t \times R_t \)
(d) \( y_t = S_t + C_t + R_t \)
Answer: (b) \( y_t = T_t + S_t + C_t + R_t \)
In simple words: The additive model for a time series combines all its parts - trend, seasonal, cyclical, and random - by adding them together to get the total value.

🎯 Exam Tip: Remember that in the additive model, each component (Trend, Seasonal, Cyclical, Random) contributes directly to the total observed value.

 

Question 5. State the independent variable of time series,
(a) \( y_t \)
(b) \( S_t \)
(c) t
(d) \( X_t \)
Answer: (c) t
In simple words: In a time series, 't' represents time, which is the variable that changes independently, influencing the other components.

🎯 Exam Tip: The independent variable in a time series is always time, which is crucial for plotting and analysis.

 

Question 6. Which component of the time series is impossible to predict ?
(a) Random component
(b) Trend
(c) Seasonal component
(d) Cyclical component
Answer: (a) Random component
In simple words: The random component of a time series includes unexpected events that cannot be foreseen or predicted beforehand.

🎯 Exam Tip: Identifying the unpredictable "random" component is important as it highlights the inherent uncertainty in forecasting time series data.

 

Question 7. Which of the following variations are due to cyclical component ?
(a) Rise in demand during winter
(b) Decrease in the share prices due to recession in share market
(c) Decrease in the agricultural produce due to excessive rains
(d) Continuously decreasing death rate
Answer: (b) Decrease in the share prices due to recession in share market
In simple words: Cyclical variations are long-term ups and downs, like economic booms and recessions, which affect things like share prices over several years.

🎯 Exam Tip: Cyclical components describe business cycles (boom, recession, recovery) that span more than one year, distinguishing them from shorter seasonal variations.

 

Question 8. The trend equation obtained from a time series from January 2016 to December 2016 is \( \hat{y} = 30.1 + 1.5 t \). Find the value of trend for April 2016.
(a) 30.1
(b) 34.6
(c) 36.1
(d) 33.1
Answer: (c) 36.1
In simple words: To find the trend value for April 2016, we count 't' from January (t=1). April is the 4th month, so t=4. Plug t=4 into the given equation to get the trend value. \( \hat{y} = 30.1 + 1.5(4) = 30.1 + 6 = 36.1 \).

🎯 Exam Tip: For trend equations, correctly identifying the 't' value for the specified period is critical for accurate calculations. January is typically t=1 unless otherwise specified.

 

Question 9. Which of the following fluctuations is the effect of seasonal component ?
(a) Increase in the migration to cities from rural areas
(b) Increasing number of vehicles on roads in a city
(c) Increase in the number of tourists during school vacation
(d) Increased death rate during a certain epidemic
Answer: (c) Increase in the number of tourists during school vacation
In simple words: Seasonal fluctuations happen regularly at certain times of the year, like more tourists visiting during school holidays.

🎯 Exam Tip: Seasonal components are recurring patterns within a year, often linked to weather, holidays, or cultural events.

 

Question 10. Which method of finding trend is best to eliminate the effect of repetitive short-term
(a) Graphical method
(b) Method of least squares
(c) Karl Pearson's method
(d) Method of moving average
Answer: (d) Method of moving average
In simple words: The moving average method smooths out short-term ups and downs to reveal the underlying long-term trend more clearly.

🎯 Exam Tip: The moving average method is effective for smoothing data and identifying trends by averaging values over a specific period, thereby reducing the impact of short-term variations.

 

Section B

Answer the following questions is one sentence:

Question 1. Give an example of time series having decreasing trend.
Answer: The time series of 'Infant Mortality Rate' in India is an example of time series having decreasing trend.
For example, infant mortality rate of India for the point 2010 to 2015 is given in the following table:

Year	2010	2011	2012	2013	2014	2015
Infant mortality rate	22.2	21.8	21.5	21.2	21	20.9

In simple words: The infant mortality rate in India has shown a downward trend over the years, meaning fewer babies are dying. This is a good example of a decreasing trend in a time series.

🎯 Exam Tip: Provide clear and specific real-world examples to illustrate concepts, as this demonstrates a deeper understanding of the topic.

 

Question 2. What is a time series ?
Answer: A time series is a set of observation taken at specified time periods.
In simple words: A time series is just a list of data points collected over specific periods, like daily temperatures or monthly sales.

🎯 Exam Tip: A concise definition of a time series should always include the idea of observations being recorded at sequential, specific time intervals.

 

Question 3. Which of the components of time series produce short-term variations ?
Answer: The seasonal components and cyclical component of time series produce short-term variations.
In simple words: Seasonal changes (like sales during holidays) and cyclical patterns (like small business booms and busts) are the parts of a time series that create variations over shorter periods.

🎯 Exam Tip: Differentiate between seasonal (within a year) and cyclical (over a few years) components when discussing short-term variations in time series.

 

Question 4. What is meant by analysis of time series ?
Answer: The method to obtain the estimates of different components of a time series is called analysis of time series.
In simple words: Analyzing a time series means breaking it down to understand its different parts like trend, seasonal changes, and random events.

🎯 Exam Tip: The analysis of time series involves decomposing the series into its constituent components (trend, seasonal, cyclical, irregular) to understand underlying patterns and make forecasts.

 

Question 5. What is the notation to show the cyclical component of the time series ?
Answer: The notation to show the cyclical component of the time series is ' \( C_t \) '.
In simple words: The letter 'C' with a small 't' next to it, \( C_t \), is used to represent the cyclical part of a time series.

🎯 Exam Tip: Correctly using standard notations (\( T_t \), \( S_t \), \( C_t \), \( R_t \)) is important for clear communication in time series analysis.

 

Question 6. State the names of methods of measuring trend.
Answer: The methods of measuring trend are:
- Graphical method.
- Least square method and
- Moving average method.
In simple words: You can find the trend in a time series by drawing a graph, using a math method called least squares, or by calculating moving averages.

🎯 Exam Tip: Be ready to list and briefly describe the common methods for trend measurement in time series analysis.

 

Question 7. The effect of which component indicates fluctuations repeating within one year ?
Answer: The effect of seasonal component indicates fluctuations repeating within a year.
In simple words: Seasonal components are what cause patterns in data that repeat every year, like higher ice cream sales in summer.

🎯 Exam Tip: Fluctuations that consistently repeat within a 12-month period are always attributed to the seasonal component.

 

Question 8. State the components of time series.
Answer: The components of time series are :
- Long-term component or Trend,
- Seasonal component,
- Cyclical component and
- Random or Irregular component.
In simple words: A time series has four main parts: the long-term trend, seasonal changes, cyclical ups and downs, and sudden random events.

🎯 Exam Tip: Memorizing the four main components of a time series is fundamental for understanding its structure and behavior.

 

Question 9. When is the method of moving average more useful to find trend?
Answer: When the effect of short-term variations is to be eliminated to find trend, method of moving average is more useful.
In simple words: The moving average method is best when you want to remove small, fast changes in data to see the bigger, overall trend clearly.

🎯 Exam Tip: The moving average method is particularly effective for smoothing out noise and short-term fluctuations to reveal the underlying trend when such variations obscure the long-term pattern.

 

Question 10. The linear equation fitted using the data of 7 weeks for a variable y is \( \hat{y} = 25.1 – 1.3t \). Estimate the value of y for the eighth week.
Answer:
\( \hat{y} = 25.1 – 1.3t \)
Putting \( t = 8 \), we get
\( \hat{y} = 25.1 – 1.3 (8) \)
\( = 25.1 - 10.4 \)
\( = 14.7 \)
Hence, the estimate of y for the eighth week obtain is \( \hat{y} = 14.7 \).
In simple words: To find the value for the eighth week, we simply replace 't' with '8' in the given equation and do the math to get 14.7.

🎯 Exam Tip: For estimation, always substitute the correct 't' value into the given trend equation and perform the calculation carefully.

 

Section C

Answer the following questions:

Question 1. Describe the additive model of time series.
Answer: The additive model of time series is as follows:
\( Y_t = T_t + S_t + C_t + R_t \)
Where, \( y_t \) = Time variable; \( T_t \) = Trend; \( S_t \) = Seasonal component; \( C_t \) = Cyclical component;
\( R_t \) = Random component.
In simple words: The additive model shows that the observed value of a time series at any point is the sum of its four main parts: trend, seasonal variations, cyclical variations, and random variations. Each part is added together to make the total.

🎯 Exam Tip: When describing the additive model, clearly state the formula and define each of its components to ensure full marks.

 

Question 2. What is meant by cyclical component ?
Answer: The variations occurring in the time series at approximately regular intervals of more than one year due to the effect of depression, recovery, boom, recession and business cycle are called cyclical variations known as cyclical component. It is denoted by ' \( C_t \) '.

- The oscillation period for cyclical component can be 2 to 10 years. In specific circumstances it can also be 10 to 15 years.
In simple words: Cyclical components are long, wavy movements in a time series that last for more than one year, like the ups and downs of a business economy (boom, recession). These patterns are somewhat regular but not fixed like seasonal ones.

🎯 Exam Tip: Cyclical components are distinguished by their duration (more than one year) and their association with broader economic or business cycles, unlike seasonal patterns which occur annually.

 

Question 3. How does seasonal component differ from the cyclical component ?
Answer: Seasonal component and cyclical component differ in the following manner:
- The variations occurring in the time series almost regularly over less than one year is the effect of seasonal component, while that of more than a year is the effect of cyclical component.

- The period of oscillation of seasonal component is usually less than a year, while it can be 2 to 10 years and in special circumstances it can also be 10 to 15 years.

- Seasonal component is the effect of natural factors and man-made factors, while cyclical component is the effect of economic situations and business cycles.

- The increase in the sales of readymade garments and shoes daring festivals is an example of seasonal component while the cycles of boom and recession are the examples of cyclical component.
In simple words: Seasonal changes happen within one year and are due to things like weather or holidays. Cyclical changes last longer than a year, like economic booms or busts. Seasonal patterns are predictable and regular, while cyclical patterns are longer, more irregular waves.

🎯 Exam Tip: Clearly distinguish seasonal and cyclical components by their period (within a year vs. more than a year) and their causes (natural/social vs. economic cycles) for a comprehensive answer.

 

Question 4. Explain the irregular component.
Answer: In the variation of time series, seasonal component and cyclical component are regular component. More and above these there is effect of irregular or random component which is short-term effect.
The element that remains after eliminating the trend \( T_t \), Seasonal component \( S_t \), Cyclical component \( C_t \) from the term \( y_t \) of the time series is known as random or irregular component and is given by \( R_t = Y_t − (T_t + S_t + C_t) \).
Random component are subject to natural forces like flood, draught, earthquake, political crisis and random causes like fire, accident, etc. Sometimes it is seen on account of innovations.
In simple words: The irregular component refers to unpredictable and sudden changes in a time series that cannot be explained by trend, seasonal, or cyclical patterns. These are usually caused by unexpected events like natural disasters or strikes.

🎯 Exam Tip: The irregular component captures all unexplained variations in a time series; emphasizing its unpredictable nature and diverse causes is important for a complete explanation.

 

Question 5. State the limitations of graphical method.
Answer: The limitation of graphical method are as follows :
- In this method different people draw different curves. Hence, the uniformity is not maintained in the trend and its estimates.
- This is not a mathematical method. So it is not possible to know the reliability of the estimates.
- The exact form of the trend line of the series cannot be obtained by this method.
- When the plotted points of the time series are widely scatter from one another, then instead of drawing an unique curve of the trend of the time series, more than one curve can be drawn representing the trend.
In simple words: The graphical method for finding trends has problems because different people might draw different trend lines, it isn't based on exact math, and it's hard to get a single, clear trend line if the data points are spread out a lot.

🎯 Exam Tip: When listing limitations, focus on the lack of objectivity, mathematical precision, and uniqueness inherent in the graphical method compared to analytical methods.

 

Question 6. Explain the meaning of moving average.
Answer: The short term variations are usually regular and have repetitions. The period of repetition of these variations are fond and their average is found for the given time series.
- This average of repetitions is known as the period of moving average. We find moving total of the variables of the given time series corresponding to the period of moving average.

- Keeping the average value in the centre the average obtained by dividing moving total by the period of moving average is called moving average.

- Suppose, the period of moving average is 3 years, then '3' yearly moving average
\[ = \frac{\text{3 yearly moving total}}{3} \]

- Since the average value lies in the centre, we get the trend values that are free from variation.
In simple words: A moving average is a way to smooth out data by taking the average of values over a specific period, then moving that period forward. This helps to see the main trend by removing small, quick changes.

🎯 Exam Tip: Emphasize that the moving average helps to smooth out short-term fluctuations to reveal the long-term trend, and its period determines the extent of smoothing.

 

Question 7. Define time series.
Answer: A time series is a set of observations taken at specified time period. Usually these observations are taken at equal internal of time.

- Time t is taken as an independent variable and the dependent variable associated with it is \( y_t \). Thus, \( y_t \) is a function of t. We shall represent the time series with different units of time as follows:

Time t	1	2	3	...	n
Variable \( y_t \)	\( y_1 \)	\( y_2 \)	\( y_3 \)	...	\( y_n \)

In simple words: A time series is a collection of data points recorded in order over time, usually at regular intervals, showing how a variable changes over different periods.

🎯 Exam Tip: A good definition of a time series highlights data collected sequentially over time, often at equal intervals, where 'time' is the independent variable.

 

Question 8. State the merits of the method of moving average to measure trend.
Answer: The merits of the method of moving average to measure trend are as follows :
- The effect of short-term component is eliminated to a large extent and trend values of the time series are obtained.
- In this method the calculation to find trend is easy and it is simple to understand.
In simple words: The moving average method is good because it helps remove small, quick changes to clearly show the main trend, and it is easy to calculate and understand.

🎯 Exam Tip: Focus on how the moving average method simplifies data interpretation by smoothing out noise and its ease of application as key merits.

 

Question 9. Describe the graphical method to measure trend.
Answer: Suppose { \( y_t: t = 1, 2, ..., n \) } is a time series and n terms of the series are \( Y_t, Y_2, …, Y_n \) respectively. Taking the time t on X-axis and the term \( y_t \) of the time series on Y-axis, the point (1, \( y_1 \)), (2, \( y_2 \)) ... (n, \( y_n \)) are plotted on the graph paper. Then points are joined by line segments. The graph so obtained is called the graph of the time series. A continuous curve passing from the viscinity of most of the points is drawn. The curve so obtained is called the trend line of the time series.
This is the simple and crude method of determining the trend of the time series. This method is quite easy to understand. But the mathematical form of the trend cannot be obtained by this method. When the plotted points of the time series are widely scatter from one another, it is difficult to draw an unique curve representing the trend of the time series. In such a situation, more than one curve can be drawn representing the trend of the time series. As a result it becomes difficult to determine the trend of the time series.
In simple words: To use the graphical method, you plot time on the X-axis and data values on the Y-axis. Then, you connect the points to make a graph. Finally, you draw a smooth line through most of the points to show the overall trend.

🎯 Exam Tip: When explaining the graphical method, describe the plotting process and the visual interpretation of drawing a smooth curve to represent the trend line.

 

Section D

Answer the following questions as required:

Question 1. Explain the importance of time series.
Answer: The important reasons for the study of time series are as follows:
- The changes occuring in the values of the variable quantity of the time series indicate the situation that prevailed in the past and this knowledge becomes helpful in making decisions for the future.
- We can predict the future value or values of the variable quantity on the basis of the study of the time series.
- The policy decisions made on the basis of the study of time series, enable business firms, industrial houses and government organisations to formulate their development plans for future.
- The study of time series enable business firms, industrial houses and government organisation to compare their present performance of the economic activities with the past and make an assessment of the actual progress achieved. For example, the government can assess the changes occurred in the inflation rate on the basis of wholesale price index numbers during the five year plan period.
In simple words: Studying time series is important because it helps us understand what happened in the past, predict what might happen in the future, and make smart plans for businesses and governments. It also allows us to compare current performance with past data.

🎯 Exam Tip: When discussing the importance of time series, highlight its utility in forecasting, planning, historical analysis, and decision-making for various sectors.

 

Question 2. State the uses of analysis of time series.
Answer: The analysis of time series is useful in trade, science, social and political fields as follows :
- It is possible to know the past situation and use it to obtain the type and measure of variation.
- By using statistical method the estimated value of the variable in future can be obtained.
- Using the estimated values proper decisions can be taken for the future and activities can be planned accordingly.
- A comparative study can be carried out for the variations in the given variable at different times and places.
- The estimated obtained from the past data can be compared with the present values and the reasons for discrepancies between them can be investigated.
In simple words: Analyzing time series helps us understand past patterns, predict future values, make good decisions and plans, compare data from different times, and find reasons for any differences. This is helpful in many areas like business and science.

🎯 Exam Tip: Broaden your answer to include applications across various domains, emphasizing how time series analysis supports informed decision-making and understanding historical trends.

 

Question 3. What is meant by trend of a time series ? Explain with an illustration.
Answer: Trend is an important component of a time series. It is the measure of permanent effect which prevails on the time series for a long period of time. Trend may be either increasing or decreasing or stable in terms of time. It represents the mathematical form and direction of variation taking place in the variable quantity \( y_t \) of a given time series. It is denoted by the symbol \( \hat{y_t} \). Trend is the component obtained by eliminating the effect of short-term fluctuations (namely seasonal and cyclical) from the total fluctuations in the variable quantity of time series. In the time series of human population of any country the effect of trend is likely to be more pronounced than that of other components.
From the study of trend of a time series we can predicate a value of \( y_t \) of the time series for some future value of t or we can determine the missing value for some past value of t. Also two or more sets of time series data can be compared with the help of their trends. Important decisions concerning economic policy can be taken on the basis of the trend of time series of economic data.
The estimate of the trend of time series can be obtained by the following two methods :
- Graphical method and
- Least squares method of fitting linear trend.
In simple words: The trend in a time series shows the long-term overall direction of the data, whether it's generally going up, down, or staying flat. For example, a country's population usually shows a steady upward trend over many years. This trend helps us predict future values or compare data.

🎯 Exam Tip: Define trend as the long-term movement, provide a clear example like population growth, and mention the common methods for its estimation.

 

Question 4. Write a short note on seasonal component.
Answer: Seasonal component is short-term in nature and its effect is seen according to the seasons. The significant variations occured in the variable quantity of time series at fixed period of time in the year say winter, summer, monsoon. For example, a spurt in demand for woolen clothes in the winter, a spurt in the sale of ice-cream and cold drinks in the summer, increase in demand for umbrellas and raincoats in the monsoon, increase in the sale of readymade garments and footwears during the religions festivals, etc. are the illustration of seasonal fluctuation.
From the study of seasonal component traders and manufacturers of seasonal goods plan for periodical stocks to take care of the demand of such goods. Generally, the duration of oscillation of seasonal fluctuations is of one year. The seasonal component is denoted by the symbol \( S_t \).
Generally the period of oscillation of seasonal component is less than a year. To study the seasonal component it is necessary to have the value of series in short-term. If the yearly values of the variable are available then it is not possible to obtain the information of seasonal component.
In simple words: The seasonal component refers to patterns in a time series that repeat every year, like increased demand for warm clothes in winter or umbrellas in monsoon season. These patterns help businesses plan their inventory. The symbol \( S_t \) is used for it.

🎯 Exam Tip: A short note on the seasonal component should cover its annual periodicity, common causes (weather, holidays), and its utility in business planning.

 

Question 5. Explain the method of fitting a linear equation to the given data using the method of least squares.
Answer: The data available in the form of time series { \( y_t: t = 1, 2, ..., n \) } are the bivariate data, where t is the independent variable and the variable quantity y is the dependent variable. From this data we have to obtain the linear trend that suits to the time series. According to linear regression model, we have to obtain the linear trend model \( y_t = \alpha + \beta t + u_t \), where \( t = 1, 2, ..., n \) and \( u_t \) is an error variable. According to the least squares method, we determine the values of the constants \( \alpha \) and \( \beta \) in such a manner that the sum of the squares of error variable, i.e., \( \Sigma e^2 = \Sigma(y_t – \alpha – \beta t)^2 \) is minimised. If 'a' and 'b' denote the estimated values of \( \alpha \) and \( \beta \) respectively then 'a' and 'b' can be obtained by the following formulae:
\( b = \frac{n \Sigma t y-(\Sigma t)(\Sigma y)}{n \Sigma t^{2}-(\Sigma t)^{2}} \) and \( a = \bar{y} – b\bar{t} \)
where, \( \bar{y} = \frac{\Sigma y}{n} \); \( \bar{t} = \frac{\Sigma t}{n} \); n = No. of observations
Thus, the fitting of trend line is given by \( \hat{y_t} = a + bt \). From this trend line we can forecast the variable quantity y of the time series for the values of t following the time duration after \( t = n \).
In simple words: The method of least squares finds the best straight line to fit time series data. It does this by making the sum of the squared differences between the actual data points and the line as small as possible. This line helps us predict future values. We use specific formulas to find the 'a' and 'b' values for the line \( \hat{y_t} = a + bt \).

🎯 Exam Tip: A full explanation of the least squares method should include the objective (minimizing squared errors), the linear model formula, and the formulas for calculating 'a' and 'b' to define the trend line.

 

Question 6. State the merits and limitations of the method of least squares.
Answer: The merits and limitations of the least squares method of fitting the linear trend are as follows:
Merits:
- This method of determining the trend of the time series is mathematically more compact than the graphical method.
- In this method the estimates of the trend values for all the terms of the time series can be obtained.
- From the equation of linear trend of the time series, we can forecast the variable quantity \( y_t \) for any value of t.
Limitations:
- This method of determining the trend of the time series is difficult than the graphical method from the calculation view point.
- If there is lack of linear trend in the time series, this method is not useful for fitting the linear trend.
In simple words: The least squares method is good because it's a clear math way to find trends, can give trend values for all data points, and helps predict future values. However, it can be harder to calculate than drawing a graph, and it doesn't work well if the trend isn't a straight line.

🎯 Exam Tip: When listing merits, focus on mathematical precision and forecasting ability; for limitations, highlight computational complexity and the assumption of linearity.

 

Question 7. Describe the method of moving average to find trend.
Answer: The method of moving average is very useful to find trend by eliminating the effect of short-term variations.
- The period of moving average : The short-term variation are usually regular and have repetition. The period of repetition of these variations can be found by observing the given time series. The average is found from the number of observations corresponding to this period which is known as the period of moving average.

- Since the average value lie in the centre, the values obtained by this method show the trend.

- Suppose the values of variable are \( Y_1, Y_2, .... Y_n \) corresponding to time \( t = 1, 2, ..., n \) and the period of moving average is 3 years. Then the mean of first three observations \( Y_1, Y_2, Y_3 \) is found as \( \frac{y_{1}+y_{2}+y_{3}}{3} \) and it is written against the centre of these three observations which is \( Y_2 \). Further, the mean of successive three observation \( Y_2, Y_3, Y_4 \) is obtained and it written against \( Y_3 \). Similarly, finding successive moving total of three observations, averages are calculated. These average are called three yearly moving averages which indicate trend.

- The period of moving average not necessarily every time is year. It may be 5 days, 4 weeks, 7 months, etc.

- If time period of moving average is an even number say. 4, 6, ... etc., then the process of finding moving average is to be done twice.

- Suppose, the period of moving average is 4 years. The four yearly successive averages \( \frac{y_{1}+y_{2}+y_{3}+y_{4}}{4}, \frac{y_{2}+y_{3}+y_{4}+y_{5}}{4}, \frac{y_{3}+y_{4}+y_{5}+y_{6}}{4} \) are written between \( Y_2 \) and \( Y_3 \), \( Y_4 \) and \( Y_5 \) ... respectively since, these averages are in between two years, the average of each pair of averages is found and written between two moving average. Thus, the average of the first two averages will be written against \( Y_3 \). The averages thus obtained are called as four yearly moving average.
In simple words: The moving average method smooths out short-term ups and downs to show the main trend. You pick a period (like 3 years), average the data for that period, and then move that period forward, centering the average. If the period is an even number, you average the moving averages to get a centered trend.

🎯 Exam Tip: Thoroughly explain how the moving average smooths data and clearly differentiate the process for odd versus even periods of moving average for better understanding and scoring.

 

Question 8. Discuss the limitations of the method of moving average.
Answer: The limitation of the method of moving average are as follows :
- If the internal for moving, average is not chosen properly, then the trend obtained by this method is not accurate.
- In this method the estimates of trend for some initial and last time periods cannot be obtained.
- In this method a specific mathematical formula is not obtained for future tes.
In simple words: The moving average method has drawbacks: if you pick the wrong averaging period, the trend won't be correct. Also, you can't find trend values for the very first or very last data points, and it doesn't give a math formula to predict future trends easily.

🎯 Exam Tip: When discussing limitations, highlight the subjectivity in choosing the period, the loss of data points at the ends, and the absence of a mathematical trend equation for forecasting.

Question 9. The following time series shows the daily production of a factory. Find the trend using graphical method.

Day	1	2	3	4	5	6	7	8	9	10
Production (units)	21	22	23	25	24	22	25	26	27	26

Answer: For this problem, the time unit is a day. So, `t` represents days from 1 to 10, and `y` represents the production. The given time series data is set up in a table. Points like (1, 21), (2, 22), and so on, up to (10, 26) are plotted on a graph. The x-axis shows time (`t`), and the y-axis shows production (`y`). After plotting, these points are connected with lines to create the original series curve. A smooth curve is then drawn through most of these plotted points to represent the trend line for the given time series.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक ग्राफ दिखाता है जो एक फैक्ट्री के दैनिक उत्पादन की समय श्रृंखला को दर्शाता है। क्षैतिज अक्ष (X-axis) पर 'समय' (दिन 1 से 12 तक) है, और ऊर्ध्वाधर अक्ष (Y-axis) पर 'उत्पादन' (इकाइयों में, 20 से 30 तक) है। इसमें दो रेखाएँ हैं: एक 'मूल श्रृंखला' जो दैनिक उत्पादन के वास्तविक बिंदुओं को जोड़ती है, और दूसरी एक चिकनी 'प्रवृत्ति' रेखा जो मूल श्रृंखला के अधिकांश बिंदुओं के करीब से गुजरते हुए समग्र रुझान को दर्शाती है।
In simple words: We plot the daily production data over time on a graph. We connect the points to see the original production. Then, we draw a smooth line through these points to show the overall pattern or trend of production.

🎯 Exam Tip: When asked to find a trend using the graphical method, ensure to clearly label both axes and plot the given data points accurately. The trend line should be a smooth curve or straight line that captures the general direction of the data, not necessarily passing through every single point.

Question 10. Fit a linear equation from the following data for variable (y) of a time series: n = 4, \(\Sigma y = 270\), \(\Sigma ty = 734\)
Answer: We are given \(n = 4\). So, the time periods \(t\) are 1, 2, 3, 4. First, we calculate the sum of \(t\) and \(t^2\): \(\Sigma t = 1 + 2 + 3 + 4 = 10\) \(\Sigma t^2 = 1^2 + 2^2 + 3^2 + 4^2 = 1 + 4 + 9 + 16 = 30\) Given: \(n = 4\), \(\Sigma t = 10\), \(\Sigma t^2 = 30\), \(\Sigma y = 270\), \(\Sigma ty = 734\). Next, we find the mean of \(t\) and \(y\): \(\bar{t} = \frac{\Sigma t}{n} = \frac{10}{4} = 2.5\) \(\bar{y} = \frac{\Sigma y}{n} = \frac{270}{4} = 67.5\) Now, we calculate the slope \(b\) using the formula for the least squares method: \[b = \frac{n \Sigma ty - (\Sigma t)(\Sigma y)}{n \Sigma t^2 - (\Sigma t)^2}\] Substitute the given values: \[b = \frac{4(734) - (10)(270)}{4(30) - (10)^2}\] \[b = \frac{2936 - 2700}{120 - 100}\] \[b = \frac{236}{20}\] \[b = 11.8\] Then, we find the y-intercept \(a\) using the formula: \(a = \bar{y} - b\bar{t}\) Substitute the values of \(\bar{y}\), \(b\), and \(\bar{t}\): \(a = 67.5 - 11.8 (2.5)\) \(a = 67.5 - 29.5\) \(a = 38\) So, the linear trend equation \(\hat{y} = a + bt\) is: \(\hat{y} = 38 + 11.8t\) To estimate the value of \(y\) for the eighth week, we put \(t = 8\): \(\hat{y} = 38 + 11.8(8)\) \(\hat{y} = 38 + 94.4\) \(\hat{y} = 132.4\)
In simple words: We used the given data to find the trend line equation. First, we found the sums and averages of time (`t`) and the variable (`y`). Then, we calculated the slope (`b`) and the y-intercept (`a`) for the line. Finally, we wrote the trend equation. If we needed to predict `y` for a future week, like the 8th week, we would plug `t=8` into this equation.

🎯 Exam Tip: Remember to correctly identify \( \Sigma t \), \( \Sigma t^2 \), \( \Sigma y \), and \( \Sigma ty \) from the problem. Pay close attention to calculation steps, especially when dealing with negative values or decimals, to avoid errors in determining 'a' and 'b'.

Question 11. The data collected about the demand of a commodity from a store are as follows. Find the trend using three monthly moving averages.

Month	January	February	March	April	May	June	July
Demand (units)	15	16	18	18	23	23	20

Answer: Here, \(n = 7\) months. So, `t` values are 1, 2, ..., 7, and `y` represents demand. To find the trend using three-monthly moving averages, we create the following table:

Month	Time t	Demand (Units) y	Three monthly moving total	Three monthly moving average Trend
January	1	15	-	-
February	2	16	15 + 16 + 18 = 49	\(\frac{49}{3} = 16.33\)
March	3	18	49 - 15 + 18 = 52	\(\frac{52}{3} = 17.33\)
April	4	18	52 - 16 + 23 = 59	\(\frac{59}{3} = 19.67\)
May	5	23	59 - 18 + 23 = 64	\(\frac{64}{3} = 21.33\)
June	6	23	64 - 18 + 20 = 66	\(\frac{66}{3} = 22\)
July	7	20	-	-

In simple words: To smooth out short-term ups and downs, we calculate a moving average. For a three-monthly average, we add up the demand for three months and divide by three. We center this average on the middle month. We repeat this process by moving one month forward each time until we cover all applicable months.

🎯 Exam Tip: When calculating moving averages, ensure the average is correctly centered. For an odd-period moving average (like 3-month), the average is placed against the middle period. For even periods, a centering step is required, but for 3-month, it's straightforward.

Section E

Solve the Following:

Question 1. The data about exports (in crore Rs.) of readymade garments of a textile manufacturer are shown below. Fit a linear trend to these data and estimate the trend for the export in the year 2017.

Year	2010	2011	2012	2013	2014	2015
Export (crore Rs.)	22	25	23	26	20	25

Answer: Here, \(n = 6\) years. So, `t` values are 1, 2, ..., 6, and `y` represents export. We need to fit a linear trend equation \(\hat{y} = a + bt\). To find the values of `a` and `b`, we prepare the following table:

Year	Time t	Export (crore Rs.) y	t ⋅ y	t2
2010	1	22	22	1
2011	2	25	50	4
2012	3	23	69	9
2013	4	26	104	16
2014	5	20	100	25
2015	6	25	150	36
n = 6	\(\Sigma t = 21\)	\(\Sigma y = 141\)	\(\Sigma ty = 495\)	\(\Sigma t^2 = 91\)

First, calculate \(\bar{t}\) and \(\bar{y}\): \(\bar{t} = \frac{\Sigma t}{n} = \frac{21}{6} = 3.5\) \(\bar{y} = \frac{\Sigma y}{n} = \frac{141}{6} = 23.5\) Now, calculate `b` using the formula: \[b = \frac{n \Sigma ty - (\Sigma t)(\Sigma y)}{n \Sigma t^2 - (\Sigma t)^2}\] Substitute the values from the table: \[b = \frac{6(495) - (21)(141)}{6(91) - (21)^2}\] \[b = \frac{2970 - 2961}{546 - 441}\] \[b = \frac{9}{105}\] \[b = 0.09\] Next, calculate `a` using the formula: \(a = \bar{y} - b\bar{t}\) Substitute the values of \(\bar{y}\), \(b\), and \(\bar{t}\): \(a = 23.5 - 0.09(3.5)\) \(a = 23.5 - 0.315\) \(a = 23.185 \approx 23.18\) The linear trend equation is: \(\hat{y} = 23.18 + 0.09t\) To estimate the export for the year 2017: The year 2017 corresponds to \(t = 8\) (since 2015 is \(t=6\), 2016 is \(t=7\), 2017 is \(t=8\)). Put \(t = 8\) into the trend equation: \(\hat{y} = 23.18 + 0.09(8)\) \(\hat{y} = 23.18 + 0.72\) \(\hat{y} = 23.90\) So, the estimated export for the year 2017 is Rs. 23.9 crore.
In simple words: We find a straight line that best fits the given export data over the years. We calculate specific values from the data to find the slope and starting point of this line. Once we have the line's equation, we use it to predict the export value for a future year, like 2017.

🎯 Exam Tip: Always make sure to define 't' clearly based on the given years. The "least squares" method involves precise summation and arithmetic; double-check calculations for \(\Sigma t\), \(\Sigma y\), \(\Sigma ty\), and \(\Sigma t^2\) as they are critical for obtaining accurate 'a' and 'b' values.

Question 2. The following data are available for the number of passengers who travelled in the last 5 years by the aircrafts of an airline company. Estimate the trend for the year 2016 by fitting linear trend.

Year	2011	2012	2013	2014	2015
No. of passengers (thousands)	45	47	44	40	38

Answer: Here, \(n = 5\) years. So, `t` values are 1, 2, 3, 4, 5, and `y` represents the number of passengers. We will fit the linear trend equation \(\hat{y} = a + bt\). To calculate the values of `a` and `b`, the following table is prepared:

Year	Time t	No. of Passengers ('000) y	t ⋅ y	t2
2011	1	45	45	1
2012	2	47	94	4
2013	3	44	132	9
2014	4	40	160	16
2015	5	38	190	25
n = 5	\(\Sigma t = 15\)	\(\Sigma y = 214\)	\(\Sigma ty = 621\)	\(\Sigma t^2 = 55\)

First, calculate \(\bar{t}\) and \(\bar{y}\): \(\bar{t} = \frac{\Sigma t}{n} = \frac{15}{5} = 3\) \(\bar{y} = \frac{\Sigma y}{n} = \frac{214}{5} = 42.8\) Now, calculate `b` using the formula: \[b = \frac{n \Sigma ty - (\Sigma t)(\Sigma y)}{n \Sigma t^2 - (\Sigma t)^2}\] Substitute the values from the table: \[b = \frac{5(621) - (15)(214)}{5(55) - (15)^2}\] \[b = \frac{3105 - 3210}{275 - 225}\] \[b = \frac{-105}{50}\] \[b = -2.1\] Next, calculate `a` using the formula: \(a = \bar{y} - b\bar{t}\) Substitute the values of \(\bar{y}\), \(b\), and \(\bar{t}\): \(a = 42.8 - (-2.1)(3)\) \(a = 42.8 + 6.3\) \(a = 49.1\) The linear trend equation is: \(\hat{y} = 49.1 - 2.1t\) To estimate the number of passengers for the year 2016: The year 2016 corresponds to \(t = 6\) (since 2015 is \(t=5\)). Put \(t = 6\) into the trend equation: \(\hat{y} = 49.1 - 2.1(6)\) \(\hat{y} = 49.1 - 12.6\) \(\hat{y} = 36.5\) So, the estimated number of passengers for the year 2016 is 36.5 thousand.
In simple words: We determined a straight line that shows the general movement of passenger numbers over time. We calculated values to find the line's steepness and where it starts. Using this line, we predicted how many passengers there would be in 2016.

🎯 Exam Tip: When fitting a linear trend, especially for prediction, ensure the time index 't' is correctly assigned to future years. A common mistake is not accounting for the incremental increase in 't' after the last observed year.

Question 3. The closing prices of shares of a company registered in a stock exchange for different months is given in the following table. Find the trend using three monthly moving averages.

Month	2015 April	May	June	July	August	Sept.	Oct.	Nov.	Dec.	2016 January
Share price (Rs.)	76	73	65	68	67	60	63	67	65	66

Answer: Here, \(n = 10\) months. So, `t` values are 1, 2, ..., 10, and `y` represents the share price. To calculate the three monthly moving average, the following table is prepared:

Month	Time t	Share price (Rs.) y	Three monthly moving total	Three monthly moving average Trend
April, 2015	1	76	-	-
May	2	73	76 + 73 + 65 = 214	\(\frac{214}{3} = 71.33\)
June	3	65	214 - 76 + 68 = 206	\(\frac{206}{3} = 68.67\)
July	4	68	206 - 73 + 67 = 200	\(\frac{200}{3} = 66.67\)
August	5	67	200 - 65 + 60 = 195	\(\frac{195}{3} = 65\)
September	6	60	195 - 68 + 63 = 190	\(\frac{190}{3} = 63.33\)
October	7	63	190 - 67 + 67 = 190	\(\frac{190}{3} = 63.33\)
November	8	67	190 - 60 + 65 = 195	\(\frac{195}{3} = 65\)
December	9	65	195 - 63 + 66 = 198	\(\frac{198}{3} = 66\)
January, 2016	10	66	-	-

In simple words: To find the overall direction of share prices, we use a three-month moving average. This means we add up the prices for three months, divide by three, and place this average in the middle of those three months. We repeat this for all applicable months to see the smoother trend.

🎯 Exam Tip: Remember that for a 3-period moving average, the first and last data points will not have a corresponding moving average value. Ensure the moving total is correctly calculated by subtracting the oldest value and adding the newest value for each step.

Question 4. The following data show the sales (in thousand Rs.) of a commodity. Find the trend by graphical method.

Year	2007	2008	2009	2010	2011	2012	2013	2014	2015	2016
Sales (thousand Rs.)	200	216	228	235	230	232	236	235	230	233

Answer: Here, the unit of time is year. \(n = 10\). So `t` values are 1, 2, ..., 10, and `y` represents sales. The given time series is written as shown in the following table:

Time t	1	2	3	4	5	6	7	8	9	10
Sales (in '000 Rs.) y	200	216	228	235	230	232	236	235	230	233

Taking `t` on the X-axis and `y` on the Y-axis, points (1, 200), (2, 216), ..., (10, 233) are plotted on the graph paper. Joining these points serially by lines, the curve of the given series is obtained as follows:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह ग्राफ एक वस्तु की बिक्री (हजार रुपये में) की समय श्रृंखला दिखाता है। क्षैतिज अक्ष (X-axis) पर 'समय' (वर्षों में, 0 से 10 तक) है, और ऊर्ध्वाधर अक्ष (Y-axis) पर 'बिक्री' (हजार रुपये में, 200 से 250 तक) है। इसमें दो रेखाएँ हैं: एक 'मूल श्रृंखला' जो बिक्री के वास्तविक बिंदुओं को जोड़ती है, और दूसरी एक चिकनी 'प्रवृत्ति' रेखा जो मूल श्रृंखला के अधिकांश बिंदुओं के करीब से गुजरते हुए समग्र रुझान को दर्शाती है। यह प्रवृत्ति रेखा बिक्री में वृद्धि का संकेत दे रही है। The curve passing through close to most of the points of the original series curve is the parabolic curve showing the trend of the given time series.
In simple words: We plot the sales data for each year on a graph. The original sales curve is made by connecting these points. Then, we draw a smooth curve that generally follows the path of these points to show the sales trend over time.

🎯 Exam Tip: For graphical trend analysis, accurate plotting of points is essential. The trend line should be drawn smoothly, representing the general direction of the data without connecting every single point, thus smoothing out minor fluctuations.

Question 5. The quantity index numbers of consumption of edible oil in a state are given in the following table. Find the trend using five yearly moving averages.

Year	2005	2006	2007	2008	2009	2010	2011	2012	2013	2014	2015
Index No.	115	121	119	120	117	119	120	118	116	124	125

Answer: Here, \(n = 11\). So, `t` values are 1, 2, ..., 11, and `y` represents the Index number. To calculate 5-yearly moving average, the following table is prepared:

Year	Time t	Index number y	'5' yearly moving total	'5' yearly moving average Trend
2005	1	115	-	-
2006	2	121	-	-
2007	3	119	115 + 121 + 119 + 120 + 117 = 592	\(\frac{592}{5} = 118.4\)
2008	4	120	592 - 115 + 119 = 596	\(\frac{596}{5} = 119.2\)
2009	5	117	596 - 121 + 120 = 595	\(\frac{595}{5} = 119\)
2010	6	119	595 - 119 + 118 = 594	\(\frac{594}{5} = 118.8\)
2011	7	120	594 - 120 + 116 = 590	\(\frac{590}{5} = 118\)
2012	8	118	590 - 117 + 124 = 597	\(\frac{597}{5} = 119.4\)
2013	9	116	597 - 119 + 125 = 603	\(\frac{603}{5} = 120.6\)
2014	10	124	-	-
2015	11	125	-	-

In simple words: To understand the long-term changes in edible oil consumption, we calculate a 5-year moving average. This means we sum up the index numbers for five years, then divide by five to get an average. This average is placed at the middle year of the five-year group. We continue this process, shifting one year at a time, to smooth out short-term ups and downs and reveal the trend.

🎯 Exam Tip: For a 5-yearly moving average, remember that the first two and last two data points will not have a corresponding moving average. The average is centered on the middle year of the 5-year span. Be careful with additions and subtractions when calculating cumulative moving totals.

Section F

Solve the Following:

Question 1. Find a linear equation using the method of least squares for the trend of production from the following data about sugar production of a country recorded for the last 6 years. Find the trend estimates for the production of the year 2016-17 and 2017-18.

Year	2009-10	2010-11	2011-12	2012-13	2013-14	2014-15
Sugar production (crore tons)	29.2	34.2	35.4	36.4	33.6	37.7

Answer: Here, \(n = 6\) years. So, `t` values are 1, 2, 3, 4, 5, 6, and `y` represents sugar production. We will fit the linear trend equation \(\hat{y} = a + bt\). To calculate the values of `a` and `b`, the following table is prepared:

Year	Time t	Production of sugar (crore tons) y	t ⋅ y	t2
2009-10	1	29.2	29.2	1
2010-11	2	34.2	68.4	4
2011-12	3	35.4	106.2	9
2012-13	4	36.4	145.6	16
2013-14	5	33.6	168.0	25
2014-15	6	37.7	226.2	36
n = 6	\(\Sigma t = 21\)	\(\Sigma y = 206.5\)	\(\Sigma ty = 743.6\)	\(\Sigma t^2 = 91\)

First, calculate \(\bar{t}\) and \(\bar{y}\): \(\bar{t} = \frac{\Sigma t}{n} = \frac{21}{6} = 3.5\) \(\bar{y} = \frac{\Sigma y}{n} = \frac{206.5}{6} = 34.42\) Now, calculate `b` using the formula: \[b = \frac{n \Sigma ty - (\Sigma t)(\Sigma y)}{n \Sigma t^2 - (\Sigma t)^2}\] Substitute the values from the table: \[b = \frac{6(743.6) - (21)(206.5)}{6(91) - (21)^2}\] \[b = \frac{4461.6 - 4336.5}{546 - 441}\] \[b = \frac{125.1}{105}\] \[b = 1.19\] Next, calculate `a` using the formula: \(a = \bar{y} - b\bar{t}\) Substitute the values of \(\bar{y}\), \(b\), and \(\bar{t}\): \(a = 34.42 - 1.19(3.5)\) \(a = 34.42 - 4.165\) \(a = 30.255 \approx 30.26\) The linear trend equation is: \(\hat{y} = 30.26 + 1.19t\) Estimates of sugar production for the years 2016-17 and 2017-18: The year 2016-17 corresponds to \(t = 8\) (since 2014-15 is \(t=6\), 2015-16 is \(t=7\)). Put \(t = 8\) into the trend equation: \(\hat{y} = 30.26 + 1.19(8)\) \(\hat{y} = 30.26 + 9.52\) \(\hat{y} = 39.78\) crore tons. The year 2017-18 corresponds to \(t = 9\). Put \(t = 9\) into the trend equation: \(\hat{y} = 30.26 + 1.19(9)\) \(\hat{y} = 30.26 + 10.71\) \(\hat{y} = 40.97\) crore tons. So, the estimated production for 2016-17 is 39.78 crore tons, and for 2017-18 is 40.97 crore tons.
In simple words: We find a straight line that shows the general movement of sugar production. We calculate values from the given data to determine the slope and starting point of this line. With this line's equation, we predict the sugar production for the years 2016-17 and 2017-18.

🎯 Exam Tip: Remember to correctly index 't' for the base years and then for the future years for which estimates are required. Rounding should be consistent throughout the calculations for 'a' and 'b' to maintain precision in the final estimates.

Question 2. The number of students studying in a college are shown in the following table. Find the trend by four yearly moving averages.

Year	2006	2007	2008	2009	2010	2011	2012	2013	2014	2015
No. of students	332	317	357	392	402	405	410	427	405	438

Answer: Here, \(n = 10\) years. So, `t` values are 1, 2, ..., 10, and `y` represents the number of students. To calculate the 4-yearly moving average, the following table is prepared:

Year	Time t	No. of students y	'4' Yearly moving total	Total of pair of '4' yearly moving total	'4' yearly moving average Trend \(= \frac{\text{Total of pair}}{8}\)
2006	1	332	-	-	-
2007	2	317	-	-	-
2008	3	357	332 + 317 + 357 + 392 = 1398	-	-
2009	4	392	-	1398 + 1468 = 2866	\(\frac{2866}{8} = 358.25\)
2010	5	402	1398 - 332 + 402 = 1468	-	-
2011	6	405	1468 - 317 + 405 = 1556	1468 + 1556 = 3024	\(\frac{3024}{8} = 378\)
2012	7	410	1556 - 357 + 410 = 1609	-	-
2013	8	427	1609 - 392 + 427 = 1644	1556 + 1609 = 3165	\(\frac{3165}{8} = 395.63\)
2014	9	405	1644 - 402 + 405 = 1647	-	-
2015	10	438	1647 - 405 + 438 = 1680	1609 + 1644 = 3253	\(\frac{3253}{8} = 406.63\)
				1644 + 1647 = 3291	\(\frac{3291}{8} = 411.38\)
				1647 + 1680 = 3327	\(\frac{3327}{8} = 415.88\)

In simple words: To find the student enrollment trend over ten years, we use a four-yearly moving average. Since it's an even number of years, we first calculate the sum of student numbers for four years and then take the average of two successive four-yearly totals. This final average is placed between the central years of the group, which helps smooth out short-term changes and reveal the general pattern.

🎯 Exam Tip: When calculating an even-period moving average (like 4-yearly), a crucial step is centering. This involves taking the average of two consecutive moving totals to align the trend value with a specific time point, as a single 4-period average falls between two time points.

Question 3. The birth rate of state in different years are given in the following table. Fit a linear trend of this data. Also find the estimates for birth rates in the year 2016 and 2017.

Year	2009	2010	2011	2012	2013	2014	2015
Birth rate	22.2	21.8	21.3	20.9	20.6	20.2	19.9

Answer: Here, \(n = 7\) years. So, `t` values are 1, 2, ..., 7, and `y` represents the birth rate. We will fit the linear trend equation \(\hat{y} = a + bt\). To calculate the values of `a` and `b`, the following table is prepared:

Year	Time t	Birth rate y	t ⋅ y	t2
2009	1	22.2	22.2	1
2010	2	21.8	43.6	4
2011	3	21.3	63.9	9
2012	4	20.9	83.6	16
2013	5	20.6	103.0	25
2014	6	20.2	121.2	36
2015	7	19.9	139.3	49
n = 7	\(\Sigma t = 28\)	\(\Sigma y = 146.9\)	\(\Sigma ty = 576.8\)	\(\Sigma t^2 = 140\)

First, calculate \(\bar{t}\) and \(\bar{y}\): \(\bar{t} = \frac{\Sigma t}{n} = \frac{28}{7} = 4\) \(\bar{y} = \frac{\Sigma y}{n} = \frac{146.9}{7} = 20.99\) Now, calculate `b` using the formula: \[b = \frac{n \Sigma ty - (\Sigma t)(\Sigma y)}{n \Sigma t^2 - (\Sigma t)^2}\] Substitute the values from the table: \[b = \frac{7(576.8) - (28)(146.9)}{7(140) - (28)^2}\] \[b = \frac{4037.6 - 4113.2}{980 - 784}\] \[b = \frac{-75.6}{196}\] \[b = -0.39\] Next, calculate `a` using the formula: \(a = \bar{y} - b\bar{t}\) Substitute the values of \(\bar{y}\), \(b\), and \(\bar{t}\): \(a = 20.99 - (-0.39)(4)\) \(a = 20.99 + 1.56\) \(a = 22.55\) The linear trend equation is: \(\hat{y} = 22.55 - 0.39t\) Estimates of birth rate for the years 2016 and 2017: The year 2016 corresponds to \(t = 8\) (since 2015 is \(t=7\)). Put \(t = 8\) into the trend equation: \(\hat{y} = 22.55 - 0.39(8)\) \(\hat{y} = 22.55 - 3.12\) \(\hat{y} = 19.43\) The year 2017 corresponds to \(t = 9\). Put \(t = 9\) into the trend equation: \(\hat{y} = 22.55 - 0.39(9)\) \(\hat{y} = 22.55 - 3.51\) \(\hat{y} = 19.04\) So, the estimated birth rates for 2016 and 2017 are 19.43 and 19.04 respectively.
In simple words: We find a straight line that best describes how the birth rate has changed over time. We calculate values from the given data to find the slope and starting point of this line. Then, we use this line's equation to predict the birth rates for future years, specifically 2016 and 2017.

🎯 Exam Tip: When the slope 'b' is negative, it indicates a decreasing trend, which should be reflected in the estimated values. Double-check all arithmetic, especially when dealing with negative signs in the 'a' calculation.

Question 4. The data about goods transported in different years by a division of railways are given below. Find the estimates for each year by fitting a linear equation and represent it by a graph. Also find the estimate for the year 2016.

Year	2011	2012	2013	2014	2015
Goods transported (tons)	180	192	195	204	202

Answer: Here, \(n = 5\) years. So, `t` values are 1, 2, 3, 4, 5, and `y` represents goods transported. We will fit a linear trend equation \(\hat{y} = a + bt\). To calculate the values of `a` and `b`, the following table is prepared:

Year	Time t	Goods transported (Tons) y	t ⋅ y	t2
2011	1	180	180	1
2012	2	192	384	4
2013	3	195	585	9
2014	4	204	816	16
2015	5	202	1010	25
n = 5	\(\Sigma t = 15\)	\(\Sigma y = 973\)	\(\Sigma ty = 2975\)	\(\Sigma t^2 = 55\)

First, calculate \(\bar{t}\) and \(\bar{y}\): \(\bar{t} = \frac{\Sigma t}{n} = \frac{15}{5} = 3\) \(\bar{y} = \frac{\Sigma y}{n} = \frac{973}{5} = 194.6\) Now, calculate `b` using the formula: \[b = \frac{n \Sigma ty - (\Sigma t)(\Sigma y)}{n \Sigma t^2 - (\Sigma t)^2}\] Substitute the values from the table: \[b = \frac{5(2975) - (15)(973)}{5(55) - (15)^2}\] \[b = \frac{14875 - 14595}{275 - 225}\] \[b = \frac{280}{50}\] \[b = 5.6\] Next, calculate `a` using the formula: \(a = \bar{y} - b\bar{t}\) Substitute the values of \(\bar{y}\), \(b\), and \(\bar{t}\): \(a = 194.6 - 5.6(3)\) \(a = 194.6 - 16.8\) \(a = 177.8\) The linear trend equation is: \(\hat{y} = 177.8 + 5.6t\) Estimates for each year:

Year	Time t	\(\hat{y} = 177.8 + 5.6t\)
2011	1	\(177.8 + 5.6(1) = 183.4\)
2012	2	\(177.8 + 5.6(2) = 189.0\)
2013	3	\(177.8 + 5.6(3) = 194.6\)
2014	4	\(177.8 + 5.6(4) = 200.2\)
2015	5	\(177.8 + 5.6(5) = 205.8\)

Estimate of goods transported for the year 2016: The year 2016 corresponds to \(t = 6\) (since 2015 is \(t=5\)). Put \(t = 6\) into the trend equation: \(\hat{y} = 177.8 + 5.6(6)\) \(\hat{y} = 177.8 + 33.6\) \(\hat{y} = 211.4\) tons. So, the estimated goods transported for the year 2016 is 211.4 tons. Representation by graph: Taking `t` on the X-axis and `y` on the Y-axis, points (1, 180), (2, 192), ..., (5, 202) are plotted on the graph paper. Joining these points serially by line segments, the curve of the original series is obtained as follows:
In simple words: We find a straight line that shows the general movement of goods transported by railways. We calculate the line's slope and starting point using the given data. This line helps us predict future transportation volumes, like for 2016. We can also plot these points and the trend line on a graph to visually see the pattern.

🎯 Exam Tip: When asked to represent data graphically after fitting a linear trend, ensure both the original data points and the fitted trend line are clearly plotted on the same graph, using different markers or line styles for distinction.

**Question 4.**
To calculate the values of 'a' and 'b' for the goods transported data, the following table is prepared:

Year	Time \(t\)	Goods transported (Tons) \(y\)	\(t \cdot y\)	\(t^2\)
	2011	1	180	180	1
2012	2	192	384	4
2013	3	195	585	9
2014	4	204	816	16
2015	5	202	1010	25
\(n = 5\)	\(\Sigma t = 15\)	\(\Sigma y = 973\)	\(\Sigma ty = 2975\)	\(\Sigma t^2 = 55\)

**Answer:**First, calculate the mean of `t`: \( \bar{t} = \frac{\Sigma t}{n} \). With \(n = 5\) and \( \Sigma t = 15 \), we get \( \bar{t} = \frac{15}{5} = 3 \).
Next, calculate the mean of `y`: \( \bar{y} = \frac{\Sigma y}{n} \). With \(n = 5\) and \( \Sigma y = 973 \), we get \( \bar{y} = \frac{973}{5} = 194.6 \).
Now, find the value of `b` using the formula: \( b = \frac{n \Sigma ty - (\Sigma t)(\Sigma y)}{n \Sigma t^2 - (\Sigma t)^2} \). Substitute the values: \( b = \frac{5(2975)-(15)(973)}{5(55)-(15)^2} \).
This simplifies to: \( b = \frac{14875-14595}{275-225} = \frac{280}{50} = 5.6 \).
Then, find the value of `a` using: \( a = \bar{y} - b\bar{t} \). Substitute the calculated means and `b`: \( a = 194.6 - 5.6 (3) = 194.6 - 16.8 = 177.8 \).
So, the linear trend equation is \( \hat{y} = a + bt \). Plugging in `a = 177.8` and `b = 5.6`, the equation becomes: \( \hat{y} = 177.8 + 5.6t \). Here are the estimates of goods transported for each year based on the trend line:

Year	Time \(t\)	\( \hat{y} = 177.8 + 5.6t \)
2011	1	\( 177.8 + 5.6(1) = 183.4 \)
2012	2	\( 177.8 + 5.6(2) = 189.0 \)
2013	3	\( 177.8 + 5.6(3) = 194.6 \)
2014	4	\( 177.8 + 5.6(4) = 200.2 \)
2015	5	\( 177.8 + 5.6(5) = 205.8 \)

To estimate the goods transported for the year 2016, we set \(t = 6\). Substitute \(t = 6\) into the trend equation: \( \hat{y} = 177.8 + 5.6(6) \).
This gives: \( \hat{y} = 177.8 + 33.6 = 211.4 \) tons. Therefore, the estimated amount of goods transported for the year 2016 is 211.4 tons.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र वर्ष 2011 से 2015 तक विभिन्न वर्षों में परिवहन किए गए सामान (टन में) को दर्शाता है। क्षैतिज अक्ष (X-axis) पर 'समय' (वर्षों में) और ऊर्ध्वाधर अक्ष (Y-axis) पर 'परिवहन किया गया सामान' (टन में) दिखाया गया है। नीले बिंदु मूल डेटा श्रृंखला को दर्शाते हैं, जिन्हें एक नीली रेखा से जोड़ा गया है। नारंगी रेखा समय श्रृंखला के रुझान को दिखाती है, जो मूल डेटा बिंदुओं के करीब से गुजरती हुई ऊपर की ओर बढ़ती हुई प्रवृत्ति दर्शाती है। यह ग्राफ दिखाता है कि समय के साथ परिवहन किए गए सामान की मात्रा में वृद्धि हुई है। The curve that passes closest to most points of the original series is the trend curve for the given time series. Based on this trend curve, the estimated goods transported for the year 2016 (when \(t = 6\)) is approximately 211 tons.
In simple words: We used a math rule to find a straight line that best fits the data. This line helps us guess how much goods were moved each year and even predict for the future. For 2016, our guess is about 211.4 tons.

🎯 Exam Tip: When using the method of least squares, ensure all calculations for \( \Sigma t, \Sigma y, \Sigma ty, \) and \( \Sigma t^2 \) are accurate. Clearly state the final trend equation and the estimated value for the requested future period.

Question 5. The data below shows the weekly prices (in USD per barrel) of crude oil. Find the trend using four weekly moving averages.

Month	March 2016				April 2016				May 2016
	Week 1	Week 2	Week 3	Week 4	Week 1	Week 2	Week 3	Week 4	Week 1	Week 2	Week 3	Week 4
Price of	35.92	38.50	39.44	39.46	36.79	39.72	40.36	43.73	45.92	44.66	46.21	48.45

**Answer:**Here, we have data for \(n = 12\) weeks. There are 4 weeks in March 2016, 4 weeks in April 2016, and 4 weeks in May 2016. Let \(t\) represent the time (week number from 1 to 12) and \(y\) represent the price of crude oil. To calculate the 4-weekly moving average, the following table is prepared:

Weeks	Time \(t\)	Price of Crude \(y\)	'4' Weekly moving total	Total of pair of '4' weekly moving total	'4' weekly moving average Trend = \(\frac{\text{Total of pairs}}{8}\)
March 2016	1	35.92	-	-	-
	2	38.50	\(35.92 + 38.50 + 39.44 + 39.46\) = \(153.32\)	-	-
	3	39.44		\(153.32 + 154.19 = 307.51\)	\(\frac{307.51}{8} = 38.44\)
	4	39.46	\(153.32 - 35.92 + 36.79 = 154.19\)
April 2016	5	36.79	\(154.19 - 38.50 + 39.72 = 155.41\)	\(154.19 + 155.41 = 309.60\)	\(\frac{309.60}{8} = 38.70\)
	6	39.72	\(155.41 - 39.44 + 40.36 = 156.33\)	\(155.41 + 156.33 = 311.74\)	\(\frac{311.74}{8} = 38.97\)
	7	40.36	\(156.33 - 39.46 + 43.73 = 160.60\)	\(156.33 + 160.60 = 316.93\)	\(\frac{316.93}{8} = 39.62\)
	8	43.73	\(160.60 - 36.79 + 45.92 = 169.73\)	\(160.60 + 169.73 = 330.33\)	\(\frac{330.33}{8} = 41.29\)
May 2016	9	45.92	\(169.73 - 39.72 + 44.66 = 174.67\)	\(169.73 + 174.67 = 344.40\)	\(\frac{344.40}{8} = 43.05\)
	10	44.66	\(174.67 - 40.36 + 46.21 = 180.52\)	\(174.67 + 180.52 = 355.19\)	\(\frac{355.19}{8} = 44.40\)
	11	46.21	\(180.52 - 43.73 + 48.45 = 185.24\)	\(180.52 + 185.24 = 365.76\)	\(\frac{365.76}{8} = 45.72\)
	12	48.45	-	-	-

In simple words: We added up the crude oil prices for every four weeks. Then, we found the average of these sums to see the trend in prices over time. This helps smooth out the short-term ups and downs.

🎯 Exam Tip: When calculating moving averages for an even period (like 4 weeks), remember to calculate the 'total of pairs' of the moving totals to center the average correctly. This involves an extra step of averaging the moving totals.