the annual profits for a company are given in the following table, where x represents the number of years…

Explanation:

Step1: Calculate the means of x and y

Let $x_i$ be the values of years since 2006 and $y_i$ be the profit - values. $n = 6$ $\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}=\frac{0 + 1+2+3+4+5}{6}=\frac{15}{6}=2.5$ $\bar{y}=\frac{\sum_{i = 1}^{n}y_i}{n}=\frac{147 + 167+209+239+250+267}{6}=\frac{1289}{6}\approx214.83$

Step2: Calculate the slope (m)

$m=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})}{\sum_{i = 1}^{n}(x_i-\bar{x})^2}$ $(x_1-\bar{x})(y_1 - \bar{y})=(0 - 2.5)(147-214.83)=(-2.5)(-67.83)=169.575$ $(x_2-\bar{x})(y_2 - \bar{y})=(1 - 2.5)(167 - 214.83)=(-1.5)(-47.83)=71.745$ $(x_3-\bar{x})(y_3 - \bar{y})=(2 - 2.5)(209 - 214.83)=(-0.5)(-5.83)=2.915$ $(x_4-\bar{x})(y_4 - \bar{y})=(3 - 2.5)(239 - 214.83)=(0.5)(24.17)=12.085$ $(x_5-\bar{x})(y_5 - \bar{y})=(4 - 2.5)(250 - 214.83)=(1.5)(35.17)=52.755$ $(x_6-\bar{x})(y_6 - \bar{y})=(5 - 2.5)(267 - 214.83)=(2.5)(52.17)=130.425$ $\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})=169.575+71.745 + 2.915+12.085+52.755+130.425=439.5$

$(x_1-\bar{x})^2=(0 - 2.5)^2 = 6.25$ $(x_2-\bar{x})^2=(1 - 2.5)^2=2.25$ $(x_3-\bar{x})^2=(2 - 2.5)^2 = 0.25$ $(x_4-\bar{x})^2=(3 - 2.5)^2=0.25$ $(x_5-\bar{x})^2=(4 - 2.5)^2 = 2.25$ $(x_6-\bar{x})^2=(5 - 2.5)^2=6.25$ $\sum_{i = 1}^{n}(x_i-\bar{x})^2=6.25+2.25 + 0.25+0.25+2.25+6.25 = 17.5$

$m=\frac{439.5}{17.5}\approx25.11$

Step3: Calculate the y - intercept (b)

$b=\bar{y}-m\bar{x}$ $b = 214.83-25.11\times2.5$ $b=214.83 - 62.78$ $b = 152.05$

The linear regression equation is $y = 25.11x+152.05$

Step4: Estimate the year when $y = 482$

Set $y = 482$ in the equation $482=25.11x+152.05$ $25.11x=482 - 152.05$ $25.11x=329.95$ $x=\frac{329.95}{25.11}\approx13.14$

The year is $2006 + 13.14\approx2019$ (rounding up to the next - whole year)

Answer:

The linear regression equation is $y = 25.11x+152.05$, and the profit would reach 482 thousand dollars in the year 2019.