In a raw score (unstandardized) regression equation, when r=0.5, what value would you predict for Y with an X value of 6?

Question

BlackTom AI · Accepted Answer

Consider the information given and what's missing in order to make a prediction.

- In a raw score (unstandardized) regression equation, the prediction for Y requires the regression equation Y = a + bX, where a is the intercept and b is the slope.
- The slope b for an unstandardized regression can be computed (in part) from the correlation r, specifically b = r * (Sy / Sx), but this also requires the standard deviations of Y and X (Sy and Sx). Without those standard deviations (and without the actual data or the regression intercept a), you cannot determine the numeric value of b.
- Even knowing r = 0.5 by itself does not specify a or b; r tells you the strength and direction of the linear relationship, not the scale or location parameters needed to predict a specific Y for X = 6.
- Therefore, with only r = 0.5 and an X value of 6, and without the regression equation coefficients (a and b) or the data dispersion measures, you cannot compute a numerical predicted Y. The prediction is indeterminable from the information provided.

In short, the missing pieces—namely the regression intercept a and the slope b (which itself depends on Sy and Sx)—prevent any numeric prediction of Y at X = 6. If an option states that you cannot determine Y with the available information, that would align with the logical limitation described above.

In a raw score (unstandardized) regression equation, when r=0.5, what value would you predict for Y with an X value of 6?单项选择题

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