Calculate correlation and investigate a radiation outlier

Plot $t$ horizontally and $I$ vertically, keeping all six pairs. In particular, $(0.4,3.5)$ sits above the main falling trend. The source directs you to the six-step plotting procedure in Example 1 and identifies time as explanatory and intensity as response. The first pair is $(0.2,3)$, not $(0.2,3.5)$. The source plot shows the endpoints 0.2 and 1.2 on the time axis and intensity extrema 0.2 and 3.5. This reconstruction labels a true zero origin. The sign follows the overall downward association; the visible blank explanatory answer is completed here from the supplied data.

Keep $t$ in L1 and $I$ in L2. Press STAT, choose CALC, then 2:2-Var Stats. Enter L1 as Xlist with 2nd 1, and L2 as Ylist with 2nd 2. Highlight Calculate, then ENTER; scroll up/down for all totals. Map calculator $x$ to $t$ and $y$ to $I$. The source screenshots also show $\bar t=0.7$, $S_t=0.3741657387$, $\sigma_t=0.3415650255$, and $\bar I=1.5$. The correlation formula uses the displayed sums, not a mixture of sample and population standard deviations.

Use the required formula rather than only reporting a calculator correlation:

$\begin{aligned}&r\\ & =\frac{3.9-(4.2)(9)/6}{\sqrt{[3.64-4.2^2/6][23.5-9^2/6]}}\\ & =\frac{-2.4}{\sqrt{(0.7)(10)}}\\ & =-0.907114\ldots.\end{aligned}$

The printed worked value is $-0.907$ to 3 significant figures. The source notes that a direct calculator method for $r$ will be illustrated in later regression examples; that does not replace the formula requested here.

The remaining scatter still falls as time increases but is less dispersed around a negative linear trend. This is the graphical explanation requested; a new numerical coefficient is unnecessary. The visible Student answer on PDF page 12 is blank, so this is an authored completion checked against the six exact pairs, not a supplied Teacher answer.