If (t_1, y_1), (t_2, y_2),......,(t_n, y_n) denote the time series and y_t are the trend values of the variables y, then Option 1: _(i=1)^n (y_i - y_t) = 0 Option 2: _(i=1)^n (y_i - y_t) = 1 Option 3: _(i=1)^n (y_i - y_t) = ∞ Option 4: _(i=1)^n (y_i - y_t) 0

Answer: _(i=1)^n (y_i - y_t) = 0. Solution: If we fit a trend line to time series data using the Least Squares Method, the trend line passes through the data such that the deviations above and below it perfectly balance out. Here, y represents the actual observed values y_t represents the trend (fitted) values from the trend line y - y_t represents the deviation of the actual value from the trend value (also called a residual) The Least Squares Method finds the line that minimizes (y - y_t)^2 A fundamental property of this method is that the positive deviations (where actual is above the trend) and the negative deviations (where actual is below the trend) cancel each other out exactly. Think of it like a see-saw — if the deviations didn't cancel out, we could shift the line up or down to get a better fit, meaning it wasn't the best fit line in the first place. Therefore, (y - y_t) = 0 The sum of deviations of actual values from the trend values is always zero for a least squares trend line.

Statistics & Applications

Trends & Data

Medium

applied

If $(t_1, y_1)$ , $(t_2, y_2)$ ,......, $(t_n, y_n)$ denote the time series and $y_t$ are the trend values of the variables $y$ , then

$\sum_{i=1}^{n} (y_i - y_t) = 0$

$\sum_{i=1}^{n} (y_i - y_t) = 1$

$\sum_{i=1}^{n} (y_i - y_t) = \infty$

$\sum_{i=1}^{n} (y_i - y_t) \ne 0$

✅ Correct Option: 1

If we fit a trend line to time series data using the Least Squares Method, the trend line passes through the data such that the deviations above and below it perfectly balance out.

Here,

$y$ represents the actual observed values

$y_t$ represents the trend (fitted) values from the trend line

$y - y_t$ represents the deviation of the actual value from the trend value (also called a residual)

The Least Squares Method finds the line that minimizes $\sum(y - y_t)^2$

A fundamental property of this method is that the positive deviations (where actual is above the trend) and the negative deviations (where actual is below the trend) cancel each other out exactly.

Think of it like a see-saw — if the deviations didn't cancel out, we could shift the line up or down to get a better fit, meaning it wasn't the best fit line in the first place.

Therefore,

$\sum (y - y_t) = 0$

The sum of deviations of actual values from the trend values is always zero for a least squares trend line.

2025: 21 May Shift 1

The rise and fall of share market is an example of

2025: 30 May Shift 1

For the given values 8, 10, 12, 14, 16; the three-year moving averages are:

2025: 30 May Shift 2

A company is shut down due to unavailability of electricity due to non payment of electricity Bill because of some unavoidable circumstances. Under which component of time series does this situation fall?

2025: 15 May Shift 1

Year (t)	2010	2011	2012	2013	2014	2015	2016
Sales (in Rs. crores) (y)	6	8	9	11	13	17	20

In reference to the above data, which of the following statements are correct? (where $x = t - 2013$ )

(A) If the equation of the straight line trend is $y = 12 + 2.29 x$ , then the trend value for the year 2017 is 21.16.

(B) If the equation of the straight line trend is $y = 12 + 2.29 x$ , then the trend value for the year 2013 is 11.

(D) If $(t_1, y_2),(t_2, y_2),(t_3, y_3),......,(t_n, y_n)$ denote the time series and $y_t$ are the trend values of the variable y, then $\sum(y - y_t) = 0$ .

Choose the correct answer from the options given below:

2025: 3 June Shift 1

Based on the data available for the production ( $y_i$ in thousand tons) of a cloth factory for 7 years ( $x_i$ ) using the method of least squares, the straight line trend is given by $y - a + bx$ with $\sum y_i = 608, \sum x_i = 0, \sum x_iy_i = 116, \sum x_i^2 = 28$ . Then, the increase in production per year is:

2025: 22 May Shift 2

If three year moving averages for five data items given by 16, 18, 20, 10, 21 are $x$ , $y$ and $z$ respectively, then:

If (t1,y1)(t_1, y_1)(t1​,y1​), (t2,y2)(t_2, y_2)(t2​,y2​),......,(tn,yn)(t_n, y_n)(tn​,yn​) denote the time series and yty_tyt​ are the trend values of the variables yyy, then

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CUET Mathematics Past Year Questions (Topic-Wise)

If $(t_1, y_1)$ , $(t_2, y_2)$ ,......, $(t_n, y_n)$ denote the time series and $y_t$ are the trend values of the variables $y$ , then