Model for Sales $y$

The linear regression model for sales, $y$, based on the given weeks $x$ is:

$y = 560.018x + 6267.255$

where:

• Slope ($m$) = 560.018 (representing the weekly increase in sales in thousands of dollars),
• Intercept ($b$) = 6267.255 (indicating the starting sales amount when $x = 0$).

Goodness of Fit

The coefficient of determination (R-squared value) is $R^2 = 0.959$, which suggests that approximately 95.9% of the variation in sales is explained by the linear model. This high R-squared value indicates that the model fits the data well.

Residual Analysis

The residual plot above shows the differences between observed and predicted sales values. Although there is some variance, the residuals do not exhibit a clear pattern, suggesting that the linear model is appropriate for this data.

Conclusion

The model is a good fit because:

1. The R-squared value is high, indicating strong explanatory power.
2. The residuals appear randomly scattered without any systematic pattern, supporting the appropriateness of the linear model.

Would you like further details or have any questions?

Here are some related questions to consider:

1. How would the model change if there were additional weekly sales data?
2. What would the model predict for sales in the 15th week?
3. How might seasonal fluctuations affect the accuracy of this linear model?
4. Could a non-linear model provide a better fit for this data set?
5. What are the implications if R-squared were significantly lower?

Tip: Residual plots are useful in detecting non-linearity; if you see a pattern, consider a non-linear model.