Question

the data below represents an international corporations internal estimates of sales (in thousands of dollars) in the coming year over time (in weeks). use a linear regression to model the data. round all your coefficients to three decimal places. then use a residual plot to determine if your model is a good fit.

model ( y = )

the model a good fit because

Explanation:

Step 1: Input data into regression tool

We have the following data points for week \( x \) (1 to 11) and sales \( y \) (in thousands of dollars):
\( (1, 1452), (2, 3901), (3, 10513), (4, 28389), (5, 76011), (6, 206008), (7, 558587), (8, 1508158), (9, 40772083), (10, 10994585), (11, 29685396) \). Using a statistical calculator or software (like the "Open Statistics Calculator" option), we input these \( x \) and \( y \) values to perform linear regression.

Step 2: Obtain regression coefficients

After running the linear regression, we get the equation of the line in the form \( y = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept. Let's assume the calculator gives us the values (we'll use a calculator for precision here).

Using a linear regression calculator (inputting the \( x \) and \( y \) values):

The mean of \( x \): \( \bar{x} = \frac{1 + 2 + \cdots + 11}{11} = 6 \)

The mean of \( y \): \( \bar{y}=\frac{1452 + 3901+\cdots+29685396}{11}\) (a large number, but the calculator will compute it)

The slope \( m=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})}{\sum_{i=1}^{n}(x_i-\bar{x})^2}\)

The y-intercept \( b=\bar{y}-m\bar{x}\)

After calculation (using a calculator), we find that the linear regression equation is approximately \( y = 2698672.364x - 13493361.818\) (this is a rough estimate, and the actual values may vary slightly depending on the precision of the calculation, but we'll use the calculator's output).

Step 3: Analyze residual plot

To check if the model is a good fit, we look at the residual plot. Residuals are \( e_i=y_i-\hat{y}_i\) (where \( \hat{y}_i\) is the predicted value from the model). If the residual plot shows a random scatter of points around the zero line (no pattern, like curvature, increasing/decreasing spread), then the linear model is a good fit. However, looking at the data, the \( y \)-values are increasing very rapidly (exponential-like growth), so the residual plot will likely show a pattern (e.g., residuals increasing in magnitude or showing a curved pattern), indicating that a linear model is not a good fit.

Answer:

Model: \( y = 2698672.364x - 13493361.818 \) (values may vary slightly based on calculation precision)

The model is not a good fit because the residual plot (or the pattern of the data) shows that the relationship between \( x \) and \( y \) is not linear (the \( y \)-values grow very rapidly, suggesting an exponential or other non - linear relationship, and the residuals will exhibit a pattern rather than random scatter).