Question

1. regression - 20 points

the owner of a company made a table of values to show the relationship between his employees annual salary and their number of years of experience. a linear function can be used to model the relationship.
which of the following functions best models the data?
regression equation:
correlation coefficient:
correlation strength: circle

• strong / weak
• positive / negative

Explanation:

Step1: Calculate means

Let \(x\) be years of experience and \(y\) be annual salary.
\(\bar{x}=\frac{1 + 3+4+6+8+10}{6}=\frac{32}{6}\approx5.33\)
\(\bar{y}=\frac{24000 + 27000+28500+30700+32000+34500}{6}=\frac{176700}{6}=29450\)

Step2: Calculate numerator and denominator for slope \(m\)

\[

$$\begin{align*} \sum_{i = 1}^{6}(x_i-\bar{x})(y_i - \bar{y})&=(1 - 5.33)(24000-29450)+(3 - 5.33)(27000 - 29450)+(4 - 5.33)(28500-29450)+(6 - 5.33)(30700-29450)+(8 - 5.33)(32000-29450)+(10 - 5.33)(34500-29450)\\ &=(- 4.33)(-5450)+(-2.33)(-2450)+(-1.33)(-950)+(0.67)(1250)+(2.67)(2550)+(4.67)(5050)\\ &=23698.5+5708.5 + 1263.5+837.5+6808.5+23583.5\\ &=61898 \end{align*}$$

\]
\[

$$\begin{align*} \sum_{i=1}^{6}(x_i-\bar{x})^2&=(1 - 5.33)^2+(3 - 5.33)^2+(4 - 5.33)^2+(6 - 5.33)^2+(8 - 5.33)^2+(10 - 5.33)^2\\ &=(-4.33)^2+(-2.33)^2+(-1.33)^2+(0.67)^2+(2.67)^2+(4.67)^2\\ &=18.7489+5.4289+1.7689+0.4489+7.1289+21.8089\\ &=55.3334 \end{align*}$$

\]
\(m=\frac{\sum_{i = 1}^{6}(x_i-\bar{x})(y_i - \bar{y})}{\sum_{i=1}^{6}(x_i-\bar{x})^2}=\frac{61898}{55.3334}\approx1120.48\)

Step3: Calculate \(y - intercept\) \(b\)

\(b=\bar{y}-m\bar{x}=29450-1120.48\times5.33=29450 - 5962.16=23487.84\)

The regression equation is \(y = 1120.48x+23487.84\)

Step4: Calculate correlation coefficient \(r\)

\[

$$\begin{align*} \sum_{i = 1}^{6}(y_i-\bar{y})^2&=(24000 - 29450)^2+(27000 - 29450)^2+(28500-29450)^2+(30700-29450)^2+(32000-29450)^2+(34500-29450)^2\\ &=(-5450)^2+(-2450)^2+(-950)^2+(1250)^2+(2550)^2+(5050)^2\\ &=29702500+6002500+902500+1562500+6502500+25502500\\ &=69175000 \end{align*}$$

\]
\(r=\frac{\sum_{i = 1}^{6}(x_i-\bar{x})(y_i - \bar{y})}{\sqrt{\sum_{i=1}^{6}(x_i-\bar{x})^2\sum_{i = 1}^{6}(y_i-\bar{y})^2}}=\frac{61898}{\sqrt{55.3334\times69175000}}=\frac{61898}{\sqrt{3826477500}}\approx\frac{61898}{61858.5}\approx0.997\)

Since \(r\approx0.997\) which is close to \(1\), the correlation is strong and positive.

Answer:

Regression Equation: \(y = 1120.48x+23487.84\)
Correlation Coefficient: \(r\approx0.997\)
Correlation Strength: strong, positive