Fit crane profits and assess a sales target

Plot cranes sold horizontally and profit in thousand dollars vertically. Preserve all eight three-month observations. The reconstructed diagram includes the mean point and fitted line requested in the following parts; its input axis is cropped to 13–22 and its profit axis starts at zero.

The sums are $\sum x=136$ and $\sum y=2750$ across eight periods. Thus $\bar x=136/8=17$ and $\bar y=2750/8=343.75$. Both printed values are retained. The mean point is circled on the scatter and lies on the fitted line.

The independently calculated centred sums are $S_{xx}=72$ and $S_{xy}=1230$, giving

$b=\frac{1230}{72}=\frac{205}{12},\qquad a=343.75-17b=\frac{160}{3}.$

Hence $y=(205/12)x+160/3$, reported as the printed $y=17.1x+53.3$. The diagram uses the full-precision line rather than rounded coefficients.

Using $S_{yy}=22387.5$, calculate $r=1230/\sqrt{72(22387.5)}=0.9688043135\ldots$, matching the printed $0.969$. The scatter shows points relatively close to a rising line; neither perfect correlation nor causation is claimed.

Use the full coefficients at $x=20$:

$\begin{aligned}&y\\ & =\frac{205}{12}(20)+\frac{160}{3}\\ & =395.\end{aligned}$

Since $y$ is measured in thousands of dollars, the predicted profit is $395,000, agreeing with the printed answer. Using the rounded equation would give 395.3 thousand instead; retain precision until the final conversion. The target 20 lies inside the observed 13–22 sales range, though future conditions still need to remain comparable.

The observed quarterly sales range is 13–22 cranes, so 40 is well outside it. The fitted sales–profit relationship may change at that scale, for example if production costs or capacity change. These are possible reasons for caution, not extra data in the source. Strong correlation within the observed sample does not establish validity at 40.