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The results of the linear regression analysis are as follows:

- **Slope (m)**: 21773.036 (rounded to three decimal places)
- **Intercept (b)**: -70199.400 (rounded to three decimal places)
- **\( R^2 \) value**: 0.695, indicating that about 69.5% of the variance in sales is explained by the week variable.

### Model Equation
The equation of the linear model is:
\[
\text{Sales (y)} = 21773.036 \times \text{Week (x)} - 70199.400
\]

### Residual Plot Analysis
The residual plot shows some pattern rather than a random scatter, indicating that a linear model may not be the best fit for this data. A non-linear model might better capture the relationship between the weeks and sales values.

Would you like more details on any part of this analysis?

Here are some questions to explore further:
1. What does a residual plot tell us about the fit of a model?
2. How is the \( R^2 \) value interpreted in the context of model fit?
3. How could we improve the model if linear regression isn’t a good fit?
4. What other types of models might work better for this dataset?
5. How would you calculate the prediction for a specific week using this model?

**Tip:** When interpreting residuals, look for randomness; any pattern in residuals suggests that the model may be missing some aspect of the relationship between variables.

The data below represents an international corporation’s 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.

Explore linear regression modeling and residual analysis on an international corporation's sales estimates over time. Learn how to calculate slope, intercept, and R-squared values to determine model accuracy.

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