The association between the total weight of produce picked from a vegetable garden and its width in non-linear. An x2 transformation is used to linearise the data. When a least squares line is fitted to the data, its y-intercept is 10 and its slope is 5. Assuming the weight is the dependent variable, the equation of this line is:

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In this problem, the dependent variable is weight and the independent variable has been transformed with a square to linearize the relationship. The least-squares line provided has an intercept of 10 and a slope of 5, which directly translate to the regression equation where y = intercept + slope × (transformed x). Since the transformation used is x^2 (width squared), the transformed predictor is width^2. Therefore, the linear model that fits the data is weight = 10 + 5 × (width^2). Option analysis follows: Option a states weight = 10 + 5 × width^2, which matches the intercept and slope exactly when the predictor is width^2; this is consistent with the stated transformation and the fitted line. If there were other options, we would evaluate them for mis-specifications, such as omitting the square on width, misplacing the intercept, or swapping the roles of x and y, but here only one option is provided and it aligns with the given intercept and slope. In short, applying the standard linear form y = intercept + slope × (transformed x) with transformed x = width^2 yields weight = 10 + 5 × width^2, which is the model described by the problem.

MCD1110 - T3 - 2024 10.3 Week 10 - Multi-choice quiz

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