Question at position 28 In a statistics course, a linear regression equation was computed to predict the final-exam score from the score on the first test. The equation was y^=10+0.9x where y is the final-exam score and x is the score on the first test. Carla scored 95 on the first test. What is the predicted value of her score on the final exam?9595.585.590None of these (answer not listed)

Question

BlackTom AI · Accepted Answer

We are told to predict the final-exam score using the regression equation ŷ = 10 + 0.9x, where x is the score on the first test.

First, evaluate each option by plugging x = 95 into the equation.
Option A: 95
- If we substitute x = 95 into ŷ = 10 + 0.9x, we get ŷ = 10 + 0.9(95) = 10 + 85.5 = 95.5, not 95. The value 95 would correspond to a different regression result or rounding issue, but it does not match the calculated prediction from the given model.

Option B: 95.5
- Calculation: ŷ = 10 + 0.9(95) = 10 + 85.5 = 95.5. This matches precisely the model’s prediction for Carla’s 95 on the first test.

Option C: 85.5
- This equals 0.9 × 95, which would be the slope component alone, forgetting the intercept of 10. The full prediction must include both the intercept and the slope term, so 85.5 is incomplete.

Option D: 90
- This value is not obtained by the model with x = 95. It would require a different intercept or slope, or perhaps a different x value, but it does not align with ŷ = 10 + 0.9x when x = 95.

Option E: None of these (answer not listed)
- Since one of the provided options exactly equals the model’s computed prediction (95.5), this choice is incorrect.

Summary of the reasoning: the regression equation adds a fixed starting point (intercept) of 10 to 0.9 times the first-test score. Substituting x = 95 yields 95.5, which uniquely matches Option B. The other options either omit the intercept, misapply the slope, or disagree with the direct substitution result, so they do not reflect the model’s prediction.

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