Question 5
Determine the equation of the least-squares line where $r=0.926$, $\bar{x}=5.2$, $s_x=2.3$, $\bar{y}=68.6$ and $s_y=41.7$.
Not quite. The slope 16.79 is right, but the intercept has the means swapped. −1146.51 comes from $\bar{x}-m\bar{y}=5.2-16.78878\ldots\times68.6$. The intercept is $\bar{y}-m\bar{x}$, which makes the line pass through the mean point $(\bar{x},\bar{y})$.
That’s it. Step through the hints below to see why.
Not quite. 0.05 comes from turning the fraction upside down, $r\frac{s_x}{s_y}$. The 68.33 then follows correctly from that wrong slope. The slope is $r\frac{s_y}{s_x}$: the spread of $y$ over the spread of $x$.
Not quite. (D) makes both slips at once. The slope fraction is upside down, $r\frac{s_x}{s_y}\approx0.05$, and the intercept swaps the means, $5.2-0.0511\times68.6\approx1.70$.
Build the line yourself. Choose how to make the slope and the intercept, and two checks tell you whether the line behaves like a least-squares line.
The formula book gives the least-squares line $y=mx+c$ from the summary statistics:
$$m=r\frac{s_y}{s_x},\qquad c=\bar{y}-m\bar{x}$$
The 2020 formula sheet wrote the same line as $y=a+bx$ with $b=r\frac{s_y}{s_x}$ and $a=\bar{y}-b\bar{x}$. Same calculation, different letters.
Slope first. The $y$ spread goes on top, because the slope is change in $y$ per unit of $x$.
$$m=0.926\times\frac{41.7}{2.3}=16.78878\ldots$$
Sense check: $y$ spreads far more than $x$ (41.7 against 2.3), so the line must be steep. A slope of 0.05 is almost flat, which rules out (C) and (D).
Now the intercept, with the full slope still in your calculator.
$$c=68.6-16.78878\ldots\times5.2=-18.70166\ldots\approx-18.70$$
Round only at the end. If you round the slope to 16.79 first, you get , which isn’t quite any option. Store the slope, then use it.
$$\begin{aligned}m&=r\frac{s_y}{s_x}=0.926\times\frac{41.7}{2.3}=16.78878\ldots\approx16.79\\c&=\bar{y}-m\bar{x}=68.6-16.78878\ldots\times5.2=-18.70166\ldots\approx-18.70\\y&=16.79x-18.70\end{aligned}$$
Spread of $y$ over spread of $x$ for the slope, then $\bar{y}$ minus slope times $\bar{x}$ for the intercept, keeping the unrounded slope.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording is from the 2020 QCAA General Mathematics external assessment, © State of Queensland (QCAA) 2020, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
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