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The table shows the reduction in costs (in hundreds) after a manager found ways each month to cut back in his store. Identify the best fit mathematical model with its corresponding 𝑅² value and tell whether it is a good model.

Month: 1 2 3 4 5
Profit loss: 86 82 72 45 15

A) quadratic model, 0.997. No, 0.997 is too high an 𝑅² value.
B) quadratic model, 0.997. Yes, 0.997 is very close to 1.
C) linear model, 0.902. No, 0.902 is too high an 𝑅² value.
D) linear model, 0.902. Yes, 0.902 is very close to 1.

1 answer

To find the best fit mathematical model for the given data and assess the appropriateness of the R² value, let's consider the provided options.

The options show two types of models: a quadratic model with an R² value of 0.997 and a linear model with an R² value of 0.902.

  1. Quadratic Model, 0.997:

    • An R² value of 0.997 implies that 99.7% of the variability in the profit loss data can be explained by the quadratic model. This is considered an excellent fit and indicates that the model closely resembles the actual data.
    • The statement "No, 0.997 is too high an R² value" is misleading, as a high R² value indicates a good fit.

  2. Linear Model, 0.902:

    • An R² value of 0.902 means that 90.2% of the variability in the profit loss data can be explained by the linear model, which is also a good fit, although not as strong as the quadratic model.
    • The assessment "Yes, 0.902 is very close to 1" is accurate, but the linear model is not the best fit since the quadratic model has a better R² value.

Given the above analysis:

  • The best option is B) quadratic model, 0.997. Yes, 0.997 is very close to 1. This indicates that the quadratic model fits the data very well, and the high R² value supports it being a good model.