the annual profits for a company are given in the following table, where x represents the number of years since 1999, and y represents the profit in thousands of dollars. write the linear regression equation that represents this set of data, rounding all coefficients to the nearest tenth. using this equation, estimate the calendar year in which the profits would reach 330 thousand dollars.

Question

Accepted Answer

# Explanation:
## Step1: Calculate the means of $x$ and $y$
$\bar{x}=\frac{0 + 1+2 + 3}{4}=\frac{6}{4}=1.5$
$\bar{y}=\frac{76+71 + 113+129}{4}=\frac{389}{4}=97.25$

## Step2: Calculate the numerator and denominator for the slope $b$
$$
\begin{align*}
\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})&=(0 - 1.5)(76-97.25)+(1 - 1.5)(71 - 97.25)+(2-1.5)(113 - 97.25)+(3 - 1.5)(129-97.25)\
&=(-1.5)(-21.25)+(-0.5)(-26.25)+(0.5)(15.75)+(1.5)(31.75)\
&=31.875 + 13.125+7.875+47.625\
&=100.5
\end{align*}
$$
$$
\begin{align*}
\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}&=(0 - 1.5)^{2}+(1 - 1.5)^{2}+(2 - 1.5)^{2}+(3 - 1.5)^{2}\
&=2.25+0.25 + 0.25+2.25\
&=5
\end{align*}
$$
$b=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})}{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}=\frac{100.5}{5}=20.1$

## Step3: Calculate the intercept $a$
$a=\bar{y}-b\bar{x}=97.25-20.1\times1.5=97.25 - 30.15=67.1$

The linear regression equation is $y = 20.1x+67.1$

## Step4: Estimate the year when $y = 330$
Set $y = 330$ in the equation $330=20.1x+67.1$
$20.1x=330 - 67.1=262.9$
$x=\frac{262.9}{20.1}\approx13.1$

Since $x$ represents the number of years since 1999, the year is $1999+13.1\approx2012$

# Answer:
The linear regression equation is $y = 20.1x + 67.1$. The profits would reach 330 thousand dollars around the year 2012.

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QUESTION IMAGE

Question

Explanation:

Step1: Calculate the means of \(x\) and \(y\)

Step2: Calculate the numerator and denominator for the slope \(b\)

Step3: Calculate the intercept \(a\)

Step4: Estimate the year when \(y = 330\)

Answer: