the annual profits for a company are given in the following table, where x represents the number of years…

the annual profits for a company are given in the following table, where x represents the number of years since 2002, 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 2010, rounded to the nearest thousand dollars.\n| years since 2002 (x) | profits (y) (in thousands of dollars) |\n| ---- | ---- |\n| 0 | 135 |\n| 1 | 160 |\n| 2 | 157 |\n| 3 | 192 |\n| 4 | 192 |\n| 5 | 227 |\nanswer attempt 2 out of 3\nregression equation:\nfinal answer: thousand dollars

the annual profits for a company are given in the following table, where x represents the number of years since 2002, 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 2010, rounded to the nearest thousand dollars.\n| years since 2002 (x) | profits (y) (in thousands of dollars) |\n| ---- | ---- |\n| 0 | 135 |\n| 1 | 160 |\n| 2 | 157 |\n| 3 | 192 |\n| 4 | 192 |\n| 5 | 227 |\nanswer attempt 2 out of 3\nregression equation:\nfinal answer: thousand dollars

Answer

Explanation:

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=135 + 160+157+192+192+227 = 1063) (\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_iy_i=0\times135+1\times160 + 2\times157+3\times192+4\times192+5\times227) (=0 + 160+314+576+768+1135 = 2953)

Step2: Calculate slope (m)

The formula for the slope (m) of the linear - regression line (y=mx + b) 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 = 6), (\sum_{i = 1}^{n}x_i = 15), (\sum_{i = 1}^{n}y_i = 1063), (\sum_{i = 1}^{n}x_i^2 = 55), and (\sum_{i = 1}^{n}x_iy_i = 2953) into the formula: [ \begin{align*} m&=\frac{6\times2953-15\times1063}{6\times55 - 15^2}\ &=\frac{17718-15945}{330 - 225}\ &=\frac{1773}{105}\ &\approx16.89 \end{align*} ]

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\approx16.89), (n = 6), (\sum_{i = 1}^{n}x_i = 15), and (\sum_{i = 1}^{n}y_i = 1063) into the formula: [ \begin{align*} b&=\frac{1063-16.89\times15}{6}\ &=\frac{1063 - 253.35}{6}\ &=\frac{809.65}{6}\ &\approx134.94 \end{align*} ]

The linear - regression equation is (y = 16.89x+134.94)

Step4: Find (x) value for 2010

Since (x) represents the number of years since 2002, for 2010, (x = 2010 - 2002=8)

Step5: Predict profit

Substitute (x = 8) into the equation (y = 16.89x+134.94) (y=16.89\times8 + 134.94) (y = 135.12+134.94) (y = 270.06\approx270) (rounded to the nearest thousand dollars)

Answer:

Regression Equation: (y = 16.89x+134.94) Final Answer: 270 thousand dollars