To solve this problem, we will:

1. Determine the linear regression equation in the form $y = mx + b$, where $m$ is the slope and $b$ is the y-intercept.
2. Estimate the calendar year when the profit reaches $y = 283$ thousand dollars.

Step 1: Find the linear regression equation.

We are given the data points:

$(0, 162), (1, 170), (2, 162), (3, 208), (4, 214), (5, 226).$

1.1 Calculate the slope $m$.

The slope formula is:

$m = \frac{\text{Change in } y}{\text{Change in } x} = \frac{y_2 - y_1}{x_2 - x_1}.$

We will approximate using two data points: $(0, 162)$ and $(5, 226)$.

$m = \frac{226 - 162}{5 - 0} = \frac{64}{5} = 12.8.$

1.2 Determine the y-intercept $b$.

Using the slope $m = 12.8$ and the point $(0, 162)$:

$y = mx + b.$ At $x = 0$, $y = 162$. Therefore:

$b = 162.$

Thus, the linear regression equation is:

$y = 12.8x + 162.$

Step 2: Estimate when $y = 283$.

We substitute $y = 283$ into the equation $y = 12.8x + 162$:

$283 = 12.8x + 162.$

Solve for $x$:

$12.8x = 283 - 162,$ $12.8x = 121.$ $x = \frac{121}{12.8} \approx 9.45.$

Step 3: Find the calendar year.

The value $x$ represents the number of years since 2013. Therefore, the estimated year is:

$2013 + 9.45 \approx 2022.45.$

Rounding to the nearest year, the profits will reach $283$ thousand dollars in 2022.

Final Answer:

1. Linear Regression Equation: $y = 12.8x + 162$
2. Estimated Calendar Year: 2022

Let me know if you'd like a detailed verification or explanation for any step!

Related Questions:

1. How is the slope $m$ affected if we choose different pairs of points?
2. Can you calculate the residuals for each data point based on the regression line?
3. How would you calculate profits for any year beyond $x = 5$?
4. What is the correlation coefficient $r$ for this data set?
5. How can you determine if a linear regression model is the best fit for this data?

Tip:

When estimating values using a regression line, always check the accuracy by calculating residuals (differences between observed and predicted values).