Given below are the consumer price index numbers (CPI) of the industrial workers.
|
Year |
2014 |
2015 |
2016 |
2017 |
2018 |
2019 |
2020 |
|
Index Number |
145 |
140 |
150 |
190 |
200 |
220 |
230 |
Find the best fitted trend line by the method of least squares and tabulate the trend values.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $y_t=152.1+16.6x$
|
Year ($x_i$) |
Index number ($y$) |
$X = x_i – A = x_i – 2017$ |
$X^2$ |
$XY$ |
Trend value $y_t = a + bx$ |
|
2014 |
145 |
-3 |
9 |
-435 |
$152.1+ (-3)×16.6 = 102.3$ |
|
2015 |
140 |
-2 |
4 |
-280 |
$152.1+ (-2)×16.6 = 118.9$ |
|
2016 |
150 |
-1 |
1 |
-150 |
$152.1+ (-1)×16.6 = 135.5$ |
|
2017 |
190 |
0 |
0 |
0 |
$152.1+ (0)×16.6 = 152.1$ |
|
2018 |
200 |
1 |
1 |
200 |
$152.1+ (1)×16.6 = 168.7$ |
|
2019 |
220 |
2 |
4 |
440 |
$152.1+ (2)×16.6 = 185.3$ |
|
2020 |
230 |
3 |
9 |
690 |
$152.1+ (3)×16.6 = 201.9$ |
|
$n = 7$ |
$∑Y = 1065$ |
$∑X = 0$ |
$∑X^2 = 28$ |
$∑XY = 465$ |
$∑Y_t = 1064.7$ |
$a=\frac{∑y}{n}=\frac{1065}{7}=152.14$
and $b=\frac{∑xy}{∑x^2}=\frac{465}{28}=16.6$
Therefore, the required equation of the straight-line trend is given by
$y_t=a+bc⇒y_t=152.1+16.6x$