جد معامل الارتباط في الجدول الآتي:

$\begin{array}{r}\overline{x}=\frac{30}{5}=6\\ \overline{y}=\frac{15}{5}=3\end{array}$

$\begin{array}{rl}& {\mathrm{S}}_x=\sqrt{\frac{\sum {\mathrm{x}}^2}{\mathrm{n}}-{\overline{\mathrm{x}}}^2}=\sqrt{\frac{220}{5}-36}=\sqrt{8}\\ & {\mathrm{S}}_{\mathrm{y}}=\sqrt{\frac{\sum {\mathrm{x}}^2}{\mathrm{n}}-{\overline{\mathrm{y}}}^2}=\sqrt{\frac{55}{5}-9}=\sqrt{2}\\ & \mathrm{r}=\frac{\frac{1}{n}\sum (x-\overline{x})(y-\overline{y})}{s_x{s}_y}=\frac{\frac{1}{5}x20}{\sqrt{8}x\sqrt{2}}\\ & \therefore \mathrm{r}=\frac{4}{4}=1 \mathrm{تام} \mathrm{طردي} \mathrm{الارتباط}\end{array}$

تمارين (3-5)

(1)- جد معامل الارتباط للظاهرتين x,y من البيانات التالية:

$\begin{array}{rl}& {\mathrm{S}}_{\mathrm{x}}=\sqrt{\frac{\sum {\mathrm{x}}^2}{\mathrm{n}}-{\overline{\mathrm{x}}}^2}=\sqrt{\frac{14}{3}}-4=\sqrt{4.66-4}=\sqrt{0.66}\\ & {\mathrm{S}}_{\mathrm{y}}=\sqrt{\frac{\sum {\mathrm{x}}^2}{\mathrm{n}}-{\overline{\mathrm{y}}}^2}=\sqrt{\frac{56}{3}-16}=\sqrt{18.66-16}=\sqrt{2.66}\\ & \mathrm{r}=\frac{\frac{1}{n}\sum (x-\overline{x})(y-\overline{y})}{s_x{s}_y}=\frac{\frac{1}{3}x4}{\sqrt{0.66}x\sqrt{2.66}}=\frac{1.3}{1.3}=1 \mathrm{تام} \mathrm{طردي} \mathrm{الارتباط}\end{array}$

(2)- في الجدول المبين في السؤال الأول لو ضربت قيم الظاهرة x في 4 نحصل على جدول آخر وهو:

$$\begin{gathered} \begin{array}{rl}& \overline{x}=\frac{24}{3}=8\\ & \overline{x}=4\\ & S_x=\sqrt{\frac{\sum x^2}{n}-{\overline{x}}^2}=\sqrt{\frac{224}{3}-64}=3.265\\ & S_y=1.632 السابق السؤال من\end{array} \\ \begin{array}{r}r=\frac{\frac{1}{n}\sum (x-\overline{x})(y-\overline{y})}{s_x{s}_y}=\frac{\frac{1}{3}x16}{3.265x1.632}\\ \therefore r=\frac{5.33}{5.33}=1 \mathrm{تام} \mathrm{طردي} \mathrm{الارتباط}\end{array} \end{gathered}$$

(3)- جد معامل الارتباط في الجدول الآتي:

$\begin{array}{r}\overline{x}=\frac{30}{5}=6\\ \overline{y}=\frac{15}{5}=3\end{array}$

$\begin{array}{rl}& {\mathrm{S}}_x=\sqrt{\frac{\sum {\mathrm{x}}^2}{\mathrm{n}}-{\overline{\mathrm{x}}}^2}=\sqrt{\frac{220}{5}-36}=\sqrt{8}\\ & {\mathrm{S}}_{\mathrm{y}}=\sqrt{\frac{\sum {\mathrm{x}}^2}{\mathrm{n}}-{\overline{\mathrm{y}}}^2}=\sqrt{\frac{55}{5}-9}=\sqrt{2}\\ & \mathrm{r}=\frac{\frac{1}{n}\sum (x-\overline{x})(y-\overline{y})}{s_x{s}_y}=\frac{\frac{1}{5}x20}{\sqrt{8}x\sqrt{2}}\\ & \therefore \mathrm{r}=\frac{4}{4}=1 \mathrm{تام} \mathrm{طردي} \mathrm{الارتباط}\end{array}$