The annual profits for a company are given in the following table, where x represents the number of years since 2006, 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 hundredth*. Using this equation, estimate the *calendar* year in which the profits would reach 482 thousand dollars.

Years since 2006 (x)Profits (y)
(in thousands of dollars)

00

147147

11

167167

22

209209

33

239239

44

2502505 267

Let's start by analyzing the data and calculating the linear regression equation. The data points are:

- $ (x_1, y_1) = (0, 147) $
- $ (x_2, y_2) = (1, 167) $
- $ (x_3, y_3) = (2, 209) $
- $ (x_4, y_4) = (3, 239) $
- $ (x_5, y_5) = (4, 250) $
- $ (x_6, y_6) = (5, 267) $

### Step 1: Calculate the Slope ($ m $)

The slope $ m $ is given by:

\[
m = \frac{n\sum{xy} - (\sum{x})(\sum{y})}{n\sum{x^2} - (\sum{x})^2}
\]

### Step 2: Calculate the Intercept ($ b $)

The intercept $ b $ is given by:

\[
b = \frac{\sum{y} - m\sum{x}}{n}
\]

### Calculations:

First, let's compute the sums:

- $ \sum{x} = 0 + 1 + 2 + 3 + 4 + 5 = 15 $
- $ \sum{y} = 147 + 167 + 209 + 239 + 250 + 267 = 1279 $
- $ \sum{xy} = (0)(147) + (1)(167) + (2)(209) + (3)(239) + (4)(250) + (5)(267) = 0 + 167 + 418 + 717 + 1000 + 1335 = 3637 $
- $ \sum{x^2} = 0^2 + 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 0 + 1 + 4 + 9 + 16 + 25 = 55 $

Now, using the slope formula:

\[
m = \frac{6(3637) - (15)(1279)}{6(55) - (15)^2} = \frac{21822 - 19185}{330 - 225} = \frac{2637}{105} \approx 25.11
\]

Next, calculate the intercept $ b $:

\[
b = \frac{1279 - 25.11 \times 15}{6} = \frac{1279 - 376.65}{6} \approx \frac{902.35}{6} \approx 150.39
\]

### The Linear Regression Equation:

The linear regression equation is:

\[
y = 25.11x + 150.39
\]

### Step 3: Estimate the Year in Which Profits Reach 482 Thousand Dollars

Set $ y = 482 $ and solve for $ x $:

\[
482 = 25.11x + 150.39
\]

\[
25.11x = 482 - 150.39
\]

\[
25.11x = 331.61
\]

\[
x = \frac{331.61}{25.11} \approx 13.21
\]

So, profits would reach 482 thousand dollars approximately **13.21 years** after 2006, which corresponds to the calendar year **2019**.

Would you like more details or have any questions? Here are some related problems:

1. What if we added more data points for years beyond 2011?
2. How does rounding coefficients impact the final prediction?
3. What is the profit estimate for the year 2025 using this model?
4. How would the equation change if there was a significant dip in profits in a certain year?
5. Can the data be better modeled using a logarithmic or exponential function?

**Tip:** Checking the correlation coefficient can help assess how well the linear model fits the data.

Learn how to calculate the linear regression equation for annual profits data using the slope-intercept form. Estimate the calendar year when profits would reach $482,000 based on the provided dataset.

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