Answer to Question #143272 in Statistics and Probability for Auguste Nkole

Question

QUESTION 4 (10 MARKS)



The asset turnovers, excluding cash and short-term investments, for the Blue Waters Fishing Company from 2010 to 2019 are listed below (in $mil):







YEAR

2010

2011

2012

2013

2014

2015

2016

2017

2018

2019

TURNOVERSS

3.0

4.2

4.8

3.7

3.4

4.3

5.6

4.4

3.8

4.1





4.1 Determine the least squares trend line equation, using the sequential coding method with X=1 in 2010 (6)



4.2 Use the trend line equation to estimate turnovers for 2008 and 2022. (2 x 2 = 4)

Answer 1

4.1

The independent variable is $x,$ and the dependent variable is $y.$

In order to compute the regression coefficients, the following table needs to be used:

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} & x & y &xy & x^2 & y^2 \\ \hline & 1& 3.0 & 3 & 1 & 9 \\ & 2 & 4.2 & 8.4 & 4 & 17.64 \\ & 3 & 4.8 & 14.4 & 9 & 23.04 \\ & 4 & 3.7 & 14.8 & 16 & 13.69 \\ & 5 & 3.4 & 17 & 25 & 11.56 \\ & 6 & 4.3 & 25.8 & 36 & 18.49 \\ & 7 & 5.6 & 39.2 & 49 & 31.36 \\ & 8 & 4.4 & 35.2 & 64 & 19.36 \\ & 9 & 3.8 & 34.2 & 81 & 14.44 \\ & 10 & 4.1 &41 & 100 & 16.81 \\ \hdashline Sum= & 55 & 41.3 & 233 & 385 & 175.39 \end{array}$$

Based on the above table, the following is calculated:

$$\bar{x}=\dfrac{1}{n}\displaystyle\sum_{i=1}^nx_i=\dfrac{55}{10}=5.5$$

$$\bar{y}=\dfrac{1}{n}\displaystyle\sum_{i=1}^ny_i=\dfrac{41.3}{10}=4.13$$

$$SS_{xx}=\displaystyle\sum_{i=1}^nx_i^2-\dfrac{1}{n}\big(\displaystyle\sum_{i=1}^nx_i\big)^2=385-\dfrac{55^2}{10}=82.5$$

$$SS_{yy}=\displaystyle\sum_{i=1}^ny_i^2-\dfrac{1}{n}\big(\displaystyle\sum_{i=1}^ny_i\big)^2=175.39-\dfrac{41.3^2}{10}=4.821$$

$$SS_{xy}=\displaystyle\sum_{i=1}^nx_iy_i-\dfrac{1}{n}\big(\displaystyle\sum_{i=1}^nx_i\big)\big(\displaystyle\sum_{i=1}^ny_i\big)=$$

$$=233-\dfrac{55\cdot41.3}{10}=5.85$$

The regression coefficients (the slope $m,$ and the y-intercept $n$) are obtained as follows:

$$m=\dfrac{SS_{xy}}{SS_{xx}}=\dfrac{5.85}{82.5}=0.0709$$

$$n=\bar{y}-\bar{x}\cdot m=4.13-5.5\cdot\dfrac{5.85}{82.5}=3.74$$

We find that the regression equation is:

$$y=3.74+0.0709x$$

4.2

$$y=3.74+0.0709x$$

2008: $x=-1, y=3.74+0.0709(-1)=3.6691$

3.6691 $mil in 2008.

2022: $x=12, y=3.74+0.0709(12)=4.5908$

4.5908 $mil in 2022.