Least Squares: Calculate the Fit and Interpret Its Coefficients — Worked Solutions

L has squared sum 1+4+1=6; M has 0+9+0=9. L better satisfies the criterion among these two candidates because its sum of squared residuals is smaller. Least squares minimizes that sum, not the signed sum or the number of exact hits. An LSRL with an intercept passes through (4,9), the sample mean point. This comparison alone does not establish that L is the minimizer among every possible line.

The LSRL passes through (4,9), so 9=a+1.5(4), giving a=3. The equation is y-hat=3+1.5x. At the sample mean of 4 hours the predicted score equals the sample mean of 9 points.

r squared=(−0.80)²=0.64, the coefficient of determination. About 64% of variation in target scores is explained by the linear relationship with practice hours for these data. Squaring removes the sign; r squared is not a negative quantity and does not describe the fraction of individual scores that are “correct.”

The slope is b=−0.4 minutes per kilometer and the intercept is a=7 minutes; they are the fitted line’s coefficients. For each one-kilometer increase in distance within the studied interval, predicted delivery duration decreases by 0.4 minutes under this model. This is a prediction from the fitted association, not an automatic causal claim.

The intercept predicts −12 centimeters at age 0 weeks. Zero weeks is outside the observed 4–10 week interval, so this is extrapolation. In addition, a plant height cannot be negative, so the predicted intercept value has no logical physical interpretation in this setting. These are distinct limitations even though both apply here.

With x=[1,2,3,4,5,6] and y=[3,4,6,8,9,12], a linear regression with an intercept gives a=0.800000…, b=1.771428571…, r=0.990258676…, and r²=0.980612245…. The model is $\hat y=0.8+1.771428571x$. A valid calculator or software report identifies the paired lists/table and the linear-with-intercept request. Software may label coefficients differently; their equation roles determine a and b.

x-bar=3.5 hours and y-bar=7 points. At x=3.5, 0.8+1.771428571(3.5)=7, so the line passes through (3.5,7). For x=1,2,3,4,5,6, predictions are approximately 2.571429,4.342857,6.114286,7.885714,9.657143,11.428571 points. Matched observed-minus-predicted residuals are 0.428571,−0.342857,−0.114286,0.114286,−0.657143,0.571429 points. Their squared sum is 38/35≈1.085714 square points. The line is called the least-squares regression line because, among straight lines with an intercept for these paired observations, it minimizes the sum of squared residuals. Minimizing that squared-error sum is different from minimizing the sum of signed residuals, whose positive and negative values can cancel.

The slope coefficient b≈1.771429 points per hour and intercept coefficient a=0.8 points come from these six sampled paired values. Under the fitted model, a one-hour increase in practice is associated with an increase of about 1.771 points in predicted target score. Individual scores can differ; the chosen practice times do not establish the effect of forcing more practice.

The intercept predicts a target score of 0.8 points at zero practice hours. Zero lies outside the observed 1–6 hours, so this is extrapolation and may lack a reasonable interpretation for the studied group. A negative-height intercept has an additional logical problem because height cannot be negative; an outside-range input and an impossible response are distinct limitations.

About 0.980612, or 98.06%, of the variation in target scores among these observations is explained by the linear relationship with practice hours. It is a proportion of variation across responses, not a fraction of each member’s score. The observational relationship does not establish that practice causes that percentage of score.