Respuesta :
To find the linear regression equation representing the given data, we need to find the equation of the line that best fits the data points. The linear regression equation has the form:
\[ y = mx + b \]
where:
- \( y \) is the dependent variable (profit in thousands of dollars)
- \( x \) is the independent variable (number of years since 1995)
- \( m \) is the slope of the line
- \( b \) is the y-intercept of the line
We can use the least squares method to find the values of \( m \) and \( b \) that minimize the sum of the squares of the differences between the observed and predicted values.
Using the given data points:
\[ (0, 96), \ (1, 122), \ (2, 147), \ (3, 147) \]
We first need to calculate the mean values of \( x \) and \( y \):
\[ \bar{x} = \frac{0 + 1 + 2 + 3}{4} = \frac{6}{4} = 1.5 \]
\[ \bar{y} = \frac{96 + 122 + 147 + 147}{4} = \frac{512}{4} = 128 \]
Now, calculate the slope (\( m \)) using the formula:
\[ m = \frac{\sum{(x_i - \bar{x})(y_i - \bar{y})}}{\sum{(x_i - \bar{x})^2}} \]
\[ m = \frac{(0 - 1.5)(96 - 128) + (1 - 1.5)(122 - 128) + (2 - 1.5)(147 - 128) + (3 - 1.5)(147 - 128)}{(0 - 1.5)^2 + (1 - 1.5)^2 + (2 - 1.5)^2 + (3 - 1.5)^2} \]
\[ m = \frac{(-1.5)(-32) + (-0.5)(-6) + (0.5)(19) + (1.5)(19)}{1.5^2 + 0.5^2 + 0.5^2 + 1.5^2} \]
\[ m = \frac{48 + 3 + 9 + 28.5}{2.25 + 0.25 + 0.25 + 2.25} \]
\[ m = \frac{88.5}{5.25} \]
\[ m ≈ 16.86 \]
Next, calculate the y-intercept (\( b \)):
\[ b = \bar{y} - m \cdot \bar{x} \]
\[ b = 128 - 16.86 \times 1.5 \]
\[ b ≈ 128 - 25.29 \]
\[ b ≈ 102.71 \]
So, the linear regression equation representing the given data is approximately:
\[ y ≈ 16.86x + 102.71 \]
To estimate the calendar year in which the profits would reach $244,000, we can set \( y \) to 244 and solve for \( x \):
\[ 244 = 16.86x + 102.71 \]
\[ 16.86x = 244 - 102.71 \]
\[ 16.86x = 141.29 \]
\[ x ≈ \frac{141.29}{16.86} \]
\[ x ≈ 8.38 \]
Since \( x \) represents the number of years since 1995, the estimated year would be \( 1995 + 8.38 = 2003.38 \).
So, the profits would reach $244,000 approximately in the year 2003.
\[ y = mx + b \]
where:
- \( y \) is the dependent variable (profit in thousands of dollars)
- \( x \) is the independent variable (number of years since 1995)
- \( m \) is the slope of the line
- \( b \) is the y-intercept of the line
We can use the least squares method to find the values of \( m \) and \( b \) that minimize the sum of the squares of the differences between the observed and predicted values.
Using the given data points:
\[ (0, 96), \ (1, 122), \ (2, 147), \ (3, 147) \]
We first need to calculate the mean values of \( x \) and \( y \):
\[ \bar{x} = \frac{0 + 1 + 2 + 3}{4} = \frac{6}{4} = 1.5 \]
\[ \bar{y} = \frac{96 + 122 + 147 + 147}{4} = \frac{512}{4} = 128 \]
Now, calculate the slope (\( m \)) using the formula:
\[ m = \frac{\sum{(x_i - \bar{x})(y_i - \bar{y})}}{\sum{(x_i - \bar{x})^2}} \]
\[ m = \frac{(0 - 1.5)(96 - 128) + (1 - 1.5)(122 - 128) + (2 - 1.5)(147 - 128) + (3 - 1.5)(147 - 128)}{(0 - 1.5)^2 + (1 - 1.5)^2 + (2 - 1.5)^2 + (3 - 1.5)^2} \]
\[ m = \frac{(-1.5)(-32) + (-0.5)(-6) + (0.5)(19) + (1.5)(19)}{1.5^2 + 0.5^2 + 0.5^2 + 1.5^2} \]
\[ m = \frac{48 + 3 + 9 + 28.5}{2.25 + 0.25 + 0.25 + 2.25} \]
\[ m = \frac{88.5}{5.25} \]
\[ m ≈ 16.86 \]
Next, calculate the y-intercept (\( b \)):
\[ b = \bar{y} - m \cdot \bar{x} \]
\[ b = 128 - 16.86 \times 1.5 \]
\[ b ≈ 128 - 25.29 \]
\[ b ≈ 102.71 \]
So, the linear regression equation representing the given data is approximately:
\[ y ≈ 16.86x + 102.71 \]
To estimate the calendar year in which the profits would reach $244,000, we can set \( y \) to 244 and solve for \( x \):
\[ 244 = 16.86x + 102.71 \]
\[ 16.86x = 244 - 102.71 \]
\[ 16.86x = 141.29 \]
\[ x ≈ \frac{141.29}{16.86} \]
\[ x ≈ 8.38 \]
Since \( x \) represents the number of years since 1995, the estimated year would be \( 1995 + 8.38 = 2003.38 \).
So, the profits would reach $244,000 approximately in the year 2003.
