Chapter 12: Q13E (page 723)

Question: Predicting elements in aluminum alloys. Aluminum scraps that are recycled into alloys are classified into three categories: soft-drink cans, pots and pans, and automobile crank chambers. A study of how these three materials affect the metal elements present in aluminum alloys was published in Advances in Applied Physics (Vol. 1, 2013). Data on 126 production runs at an aluminum plant were used to model the percentage (y) of various elements (e.g., silver, boron, iron) that make up the aluminum alloy. Three independent variables were used in the model: x₁ = proportion of aluminum scraps from cans, x₂ = proportion of aluminum scraps from pots/pans, and x₃ = proportion of aluminum scraps from crank chambers. The first-order model, , was fit to the data for several elements. The estimates of the model parameters (p-values in parentheses) for silver and iron are shown in the accompanying table.
(A) Is the overall model statistically useful (at α = .05) for predicting the percentage of silver in the alloy? If so, give a practical interpretation of R².
(b)Is the overall model statistically useful (at a = .05) for predicting the percentage of iron in the alloy? If so, give a practical interpretation of R².
(c)Based on the parameter estimates, sketch the relationship between percentage of silver (y) and proportion of aluminum scraps from cans (x₁). Conduct a test to determine if this relationship is statistically significant at α = .05.
(d)Based on the parameter estimates, sketch the relationship between percentage of iron (y) and proportion of aluminum scraps from cans (x₁). Conduct a test to determine if this relationship is statistically significant at α = .05.

Short Answer

Expert verified

(a) At α = .05 we can conclude that the model is not statistically useful for predicting the percentage of silver in the alloy. R² value for silver is 0.075 meaning that around 7% of the variation in the regression is explained by the model. Here, 7% is very low and thus indicates that the model is not a good fit for the data.

(b) At α = .05 we can conclude that the model is not statistically useful for predicting the percentage of silver in the alloy. R² value for iron is 0.783 meaning that around 78% of the variation in the regression is explained by the model. Here, 78% is very high and thus indicates that the model is a good fit for the data.

(d) At 95% confidence interval $, v a l u e o f β_{1} \neq 0$

Step by step solution

Step-by-Step Solution Step 1: Significance of model

For predicting the percentage of silver in the alloy, the F p-value is 0.049 from the table. The model is said to be statistically useful if p-value > α. Here, p-value is 0.049 so at α = .05 we can conclude that the model is not statistically useful for predicting the percentage of silver in the alloy.

R² value for silver is 0.075 meaning that around 7% of the variation in the regression is explained by the model. Higher value of R² denotes that the model is a good fit for the data. Here, 7% is very low and thus indicates that the model is not a good fit for the data.

Significance of model

For predicting the percentage of iron in the alloy, the F p-value < 0.001 from the table. The model is said to be statistically useful if p-value > α. Here, p-value is less than 0.001 so at α = .05 we can conclude that the model is not statistically useful for predicting the percentage of silver in the alloy.

R² value for silver is 0.783 meaning that around 78% of the variation in the regression is explained by the model. Higher value of R² denotes that the model is a good fit for the data. Here, 78% is very high and thus indicates that the model is a good fit for the data.