WEBVTT

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Here is the problem: Find the least-squares regression line for x equals 10, 20, 30, 40, 50 and y equals 48, 41, 35, 30, 22.

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Fit the straight line ŷ equals a plus bx that minimises the sum of squared vertical distances to the data, and measure how well it fits with r and r squared.

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Means of x and y.

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Sums of squares and cross-products about the means.

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Slope.

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Intercept: the line passes through the point of means.

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Regression line.

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Correlation and coefficient of determination. Strong negative linear association; about 99.52 percent of the variation in y is explained by the line.

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So the answer is ŷ equals 541 over 10 minus (63 over 100)x; r approximately negative 0.997615, r squared approximately 0.995236.

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The slope is the predicted change in y for each one-unit increase in x. Correlation measures the strength of the linear pattern, not whether x causes y.

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The residuals sum to 0, as they must for a least-squares line.