Question

the annual profits for a company are given in the following table, where x represents the number of years since 2011, 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, find the projected profit (in thousands of dollars) for 2022, rounded to the nearest thousand dollars.

years since 2011 (x)	profits (y) (in thousands of dollars)
1	141
2	151
3	145

answer attempt 1 out of 2

regression equation:

final answer: thousand dollars

Explanation:

Step1: Calculate sums

Let \(n = 4\) (number of data - points).
\(\sum_{i = 1}^{n}x_i=0 + 1+2 + 3=6\)
\(\sum_{i = 1}^{n}y_i=88 + 141+151+145 = 525\)
\(\sum_{i = 1}^{n}x_i^2=0^2 + 1^2+2^2 + 3^2=0 + 1+4 + 9 = 14\)
\(\sum_{i = 1}^{n}x_iy_i=(0\times88)+(1\times141)+(2\times151)+(3\times145)=0 + 141+302+435 = 878\)

Step2: Calculate slope \(m\)

The formula for the slope \(m\) of the regression line is \(m=\frac{n\sum_{i = 1}^{n}x_iy_i-\sum_{i = 1}^{n}x_i\sum_{i = 1}^{n}y_i}{n\sum_{i = 1}^{n}x_i^2-(\sum_{i = 1}^{n}x_i)^2}\)
Substitute \(n = 4\), \(\sum_{i = 1}^{n}x_i = 6\), \(\sum_{i = 1}^{n}y_i = 525\), \(\sum_{i = 1}^{n}x_i^2 = 14\), and \(\sum_{i = 1}^{n}x_iy_i = 878\) into the formula:

$$
LATEXBLOCK0
$$

Step3: Calculate y - intercept \(b\)

The formula for the y - intercept \(b\) is \(b=\frac{\sum_{i = 1}^{n}y_i-m\sum_{i = 1}^{n}x_i}{n}\)
Substitute \(n = 4\), \(m = 18.10\), \(\sum_{i = 1}^{n}x_i = 6\), and \(\sum_{i = 1}^{n}y_i = 525\) into the formula:

$$
LATEXBLOCK1
$$

The linear regression equation is \(y = 18.10x+104.10\)

Step4: Find \(x\) for 2022

Since \(x\) represents the number of years since 2011, for 2022, \(x=2022 - 2011=11\)

Step5: Predict profit

Substitute \(x = 11\) into the regression equation \(y = 18.10x+104.10\)

$$
LATEXBLOCK2
$$

Answer:

Regression Equation: \(y = 18.10x+104.10\)
Final Answer: 303 thousand dollars