Case study (4 Marks)

Question 14 Marks

Read the following text and answer the following questions on the basis of the same:
The following data shows the percentage of rural, urban and suburban Indians who have a high speed internet connection at home:

Year (t)	Rural	Urban	Suburban
2017	3	9	9
2018	6	18	17
2019	9	21	23
2020	16	29	29
2021	240	380	40

Q. 1. A straight line trend by the method of least square for the Rural Indians is:
(A) $y_t=11.6+5.2 x$
(B) $y_t=23+6.9 x$
(C) $y_t=23.6+7.4 x$
(D) $y_t=13.2+4.5 x$

Q. 2. A straight line trend by the method of least square for the Urban Indians is:
(A) $y_t=11.6+5.2 x$
(B) $y_t=23+6.9 x$
(C) $y_t=23.6+7.4 x$
(D) $y_t=13.2+4.5 x$

Q. 3. A straight line trend by the method of least square for the Suburban Indians is:
(A) $y_t=11.6+5.2 x$
(B) $y_t=23+6.9 x$
(C) $y_t=23.6+7.4 x$
(D) $y_t=13.2+4.5 x$

Q. 4. What is the forecast for the year 2022 for urban group using trend equation?
(A) 58.6% (B) 43.7% (C) 37.2% (D) 27.2%

Answer

(1) (A) $y_t=11.6+5.2 x$

Year (t)	Rural (y)	x = $t_i$-2019	$x^2$	xy
2017	3	-2	4	-6
2018	6	-1	1	-6
2019	9	0	0	0
2020	16	1	1	16
2021	24	2	4	48
	$\Sigma y=58$		$\Sigma x^2=10$	$\Sigma x y=52$

Here, n = 5 $a=\frac{\Sigma y}{n}$ and $b=\frac{\Sigma x y}{\Sigma x^2}$
So, $a=\frac{58}{5}=11.6$
and $b=\frac{52}{10}=5.2$
Thus, trend equation is given by $y_t=a+b x$
i.e $y_t=11.6+5.2 x$

(2) (B) $y_t=23+6.9 x$

Year (t)	Rural (y)	x = $t_i$-2019	$x^2$	xy
2017	9	-2	4	-18
2018	18	-1	1	-18
2019	21	0	0	0
2020	29	1	1	29
2021	38	2	4	76
	$\Sigma y=115$		$\Sigma x^2=10$	$\Sigma x y=69$

Here, n = 5 $a=\frac{\Sigma y}{n}$ and $b=\frac{\Sigma x y}{\Sigma x^2}$
So, $a=\frac{115}{5}=23$
and $b=\frac{69}{10}=6.9$
Thus, trend equation is given by $y_t=a+b x$
i.e. $y_t=23+6.9 x$

(3) (C) $y_t=23.6+7.4 x$

Year (t)	Rural (y)	x = $t_i$-2019	$x^2$	xy
2017	9	-2	4	-18
2018	17	-1	1	-17
2019	23	0	0	0
2020	29	1	1	29
2021	40	2	4	80
	$\Sigma y=118$		$\Sigma x^2=10$	$\Sigma x y=74$

Here, n = 5 $a=\frac{\Sigma y}{n}$ and $b=\frac{\Sigma x y}{\Sigma x^2}$
So, $a=\frac{118}{5}=23.6$
and $b=\frac{74}{10}=7.4$
Thus, trend equation is given by $y_t=a+b x$
i.e.$y_t=23.6+7.4 x$

(4) (B) 43.7%
Explanation: For year 2022, x = 3
The trend equation for urban is
$y_t=23+6.9 x$
So, at x = 3
we get $y_l=23+6.9(3)$
= 43.7%

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Question 24 Marks

Read the following text and answer the following questions on the basis of the same:
Today in class Mr. Agarwal is teaching the Method of Least Squares to measure the trend in time series. After explaining the method, he had taken an example which is as follows:
Given below are the data relating to the sales of a product in a district.
Fit a straight line trend by the method of least squares and tabulate the trend values.

Year	2015	2016	2017	2018	2019	2020	2021	2022
Sales	6.7	5.3	4.3	6.1	5.6	7.9	5.8	6.1

To solve the given example, he constructed the following table:
Computation of trend values by the method of least squares.

Years (t)	Sales (y)	x = $\frac{(t-2018.5)}{0.5}$	XY	$x^2$
2015	6.7	-7	-46.9	49
2016	5.3	-5	-26.5	25
2017	4.3	-3	-12.9	9
2018	6.1	-1	-6.1	1
2019	5.6	1	5.6	1
2020	7.9	3	23.7	9
2021	5.8	5	29.0	25
2022	6.1	7	42.7	49

Here, n = 8 (even)
So, origin is mean of two middle years
i.e. $\frac{2018+2019}{2}=2018.5$

Q. 1. If the straight line is $y_t=a+b x$, then value of ' $a$ ' is:
(A) 5.615 (B) 5.759 (C) 5.829 (D) 5.975

Q. 2. If the straight line is $y_t=a+b x$, then value of 'b' is:
(A) 0.051 (B) 0.011 (C) 0.411 (D) None of these

Q. 3. The trend equation is:
(A) $y_t=5.975+0.051 x$
(B) $y_t=0.051+5.975 x$
(C) $y_t=5.975-0.051 x$
(D) None of the above

Q. 4. The trend value for year 2015 is:
(A) 5.691 (B) 5.618 (C) 5.987 (D) 5.719

Answer

(1) (D) 5.975
$\Sigma y=6.7+5.3+4.3+6.1$$+5.6+7.9+5.8+6.1$
$a=\frac{\Sigma y}{n}$
Here, $n=8$ and $\Sigma y=47.8$
So, $a=\frac{47.8}{8}$
= 5.975

(2) (A) 0.051
Explanation :
$\Sigma x y=-46.9-26.5-12.9-6.1$$+5.6+23.7+29.0$
= 8.6
$b=\frac{\Sigma x y}{\Sigma x^2}$
Here, $n=7$ and $\Sigma x y=8.6$ and $\Sigma x^2=168$
So, $b=\frac{8.6}{168}$
= 0.051

(3) (A) $y_t=5.975+0.051 x$
Since, a = 5.975
and b = 0.051
Thus, trend equation is given by
$y_1=5.975+0.051 x$

(4) (B) 5.618
Explanation:
For year 2015,
x = - 7
$\therefore \quad y_t=5.975+0.051(-7)$
= 5.618

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Question 34 Marks

Read the following text and answer the following questions on the basis of the same:

Year (t)	Profit (y)	x = $t_i-$2018	$x^2$	xy
2015	114	-3	9	-342
2016	130	-2	4	-260
2017	126	-1	1	-126
2018	144	0	0	0
2019	138	1	1	138
2020	156	2	4	312
2021	164	3	9	492

The trend equation can be considered as $y_t=a+b x$.

Q. 1. The value of 'a' in the trend equation is:
(A) 138.86 (B) 7.64 (C) 183.86 (D) 7.46

Q. 2. The value of 'b' in the trend equation is:
(A) 138.86 (B) 7.64 (C) 183.86 (D) 7.46

Q. 3. The trend equation is:
(A) $y_t=138.86+7.64 x$
(B) $y_1=7.64+138.86 x$
(C) $y_t=138.86-7.64 x$
(D) None of these

Q. 4. The trend value for year 2015 is:
(A) 161.78 (B) 154.14 (C) 115.94 (D) 138,86

Answer

(1) (A) 138.86
Explanation: $a=\frac{\Sigma y}{n}$
Here, $n=7$ and $\sum y=972$
So, $a=\frac{972}{7}$
= 138.86

(2) (B) 7.64
Explanation:
$b=\frac{\Sigma x y}{\Sigma x^2}$
Here, $n=7, \sum x y=214$
and $\sum x^2=28$
So, $b=\frac{214}{28}$
= 7.64

(3) (A) $y_t=138.86+7.64 x$
Explanation: Since, a = 138.86 and b = 7.64
Thus, trend equation is given by $y_t=138.86+7.64 x$

(4) (C) 115.94
Explanation: For year $2015, x=-3$
$\begin{aligned} y_t & =138.86+7.64(-3) \\ & =115.94\end{aligned}$

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Applied Maths Case study (4 Marks)

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