QUESTION IMAGE
Question
the annual profits for a company are given in the following table, where x represents the number of years since 2003, 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 2015, rounded to the nearest thousand dollars.
| years since 2003 (x) | profits (y) (in thousands of dollars) |
|---|---|
| 1 | 110 |
| 2 | 116 |
| 3 | 141 |
| 4 | 156 |
| 5 | 186 |
answer attempt 1 out of 3
regression equation:
final answer: thousand dollars
Step1: Calculate sums
Let \(n = 6\) (number of data - points).
\(\sum_{i = 1}^{n}x_{i}=0 + 1+2 + 3+4 + 5=15\)
\(\sum_{i = 1}^{n}y_{i}=99 + 110+116+141+156+186 = 808\)
\(\sum_{i = 1}^{n}x_{i}^{2}=0^{2}+1^{2}+2^{2}+3^{2}+4^{2}+5^{2}=0 + 1+4 + 9+16+25 = 55\)
\(\sum_{i = 1}^{n}x_{i}y_{i}=0\times99+1\times110 + 2\times116+3\times141+4\times156+5\times186\)
\(=0 + 110+232+423+624+930=2319\)
Step2: Calculate slope \(m\)
The formula for the slope \(m\) of the regression line is \(m=\frac{n\sum_{i = 1}^{n}x_{i}y_{i}-\sum_{i = 1}^{n}x_{i}\sum_{i = 1}^{n}y_{i}}{n\sum_{i = 1}^{n}x_{i}^{2}-(\sum_{i = 1}^{n}x_{i})^{2}}\)
Substitute \(n = 6\), \(\sum_{i = 1}^{n}x_{i}=15\), \(\sum_{i = 1}^{n}y_{i}=808\), \(\sum_{i = 1}^{n}x_{i}^{2}=55\), \(\sum_{i = 1}^{n}x_{i}y_{i}=2319\) into the formula:
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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 \(m\approx17.09\), \(n = 6\), \(\sum_{i = 1}^{n}x_{i}=15\), \(\sum_{i = 1}^{n}y_{i}=808\) into the formula:
\[
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The regression equation is \(y = 17.09x+91.94\)
Step4: Find \(x\) for 2015
Since \(x\) represents the number of years since 2003, for 2015, \(x=2015 - 2003=12\)
Step5: Predict the profit
Substitute \(x = 12\) into the regression equation \(y = 17.09\times12+91.94\)
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Regression Equation: \(y = 17.09x+91.94\)
Final Answer: 297 thousand dollars