Tag: time series

Questions Related to time series

What would be the estimated sale on the advertisement expenditure of Rs $15$ lakhs,on the basis of following data obtained from the company?.The coefficient of correlation is $0.8$.

Advertising expenditure(in Rs.lakhs) $x$ Sale (in Rs lakhs) $y$
Mean $20$ $90$
standard Deviation $5$ $12$
statistics time series moving average and variation simple moving average

  1. Rs $105$ lakhs

  2. Rs$106$ lakhs

  3. Rs $110$ lakhs

  4. Rs $120$ lakhs

Show answer
Correct Option: B
Explanation:

$\bar x=20$, $\bar y=90$, $\sigma_x=5$, $\sigma_y=12$ and $r=0.8$
Regression line of y on x will be $y-\bar y=r\dfrac{\sigma_y}{\sigma_x}(x-\bar x)$
Subsitute all the abpve value in an equation of regression line, we get
$y-90=\dfrac{0.8\times 12}{5}(x-20)$
$y-90=1.92(x-20)$
$y=1.92x+90-38.4$
Now, Subsitute the value of x=15, we get
$y=1.92\times 15+51.6=80.4$

Find the equation of $y$ on $x$ for the following data

$x$ $8$ $6$ $4$ $7$ $5$
$y$ $9$ $8$ $5$ $6$ $2$
statistics time series moving average and variation simple moving average

  1. $y=2x -1.2$

  2. $y=1.2x +1.2$

  3. $y=1.2x -1.2$

  4. $y=1.2x -2$

Show answer
Correct Option: C
Explanation:
 $x$  $y$  $x^2$  $x\times y$
 $8$  $9$  $64$  $72$
 $6$  $8$  $36$  $48$
 $4$  $5$  $16$  $20$
 $7$  $6$  $49$  $42$
 $5$  $2$  $25$  $10$

$\sum x=30$
$\sum y=30$
$\sum x^2=190$
$\sum x\times y=192$
So, $\bar x=\dfrac{\sum x}{n}=\dfrac{30}{5}=6$
$\bar y=\dfrac{\sum y}{n}=\dfrac{30}{5}=6$
$b_{yx}=\dfrac{\sum xy-\dfrac{\sum x \times \sum y}{n}}{\sum x^2-\dfrac{(\sum
x)^2}{n}}=\dfrac{192-180}{190-180}=\dfrac{12}{10}=1.2$
Regression line of y on x will be $y-\bar y=b_{yx}(x-\bar x)$
$y-6=1.2(x-6)$
$y=1.2x-7.2+6$
$y=1.2x-1.2$

For the variables $x$ and $y$, the regression equations are given as $7x-3y-18=0$ and $4x-y-11=0$. Find the arithmetic means of $x$ and $y$ respectively.

statistics time series moving average and variation simple moving average

  1. $3$ and $1$

  2. $1$ and $3$

  3. $2$ and $4$

  4. $4$ and $2$

Show answer
Correct Option: A
Explanation:

Since the two lines of regression interest at the point $(\bar{X}, \bar{Y} )$


Replace $x$ and $y$ by $\bar{X}$ and $\bar{Y}$ respectively in the given regression equations.

We get,
$7\bar{X}- 3\bar{Y} - 18 = 0$ and $4\bar{X}-\bar{Y} - 11 = 0$ 


Solving these equations, we get $\bar{X} = 3$ and $\bar{Y} = 1$ 

Thus the arithmetic mean of $x$ and $y$ is given by $3$ and $1$ respectively.

The two lines of regression are $x+2y-5=0$ and $x+3y-8=0$. The coefficient of correlation between $x$ and $y$ is 

statistics time series moving average and variation simple moving average

  1. $-0.72$

  2. $0.72$

  3. $-0.82$

  4. $0.82$

Show answer
Correct Option: C
Explanation:

Given two lines $x+2y-5=0, x+3y-8=0$.

Consider $x+2y-5=0$
$\Rightarrow x=-2y+5$
$\Rightarrow r_1=-2$
Consider $x+3y-8=0$
$\Rightarrow y=-\dfrac{1}{3}x+\dfrac{8}{3}$
$\Rightarrow r_2=-\dfrac{1}{3}$
We know that $r^2=r_1 \times r_2$
$\Rightarrow r^2=-2 \times -\dfrac{1}{3}$
$\Rightarrow r^2=\dfrac{2}{3}$
$\Rightarrow r=\pm \sqrt{\dfrac{2}{3}}$
We know that, If both regression coefficients are negative, $r$ would be negative.
$\Rightarrow r=-\sqrt{\dfrac{2}{3}}=-0.82$