Considerando essas informações e o conteúdo estudado, calcule o coeficiente de correlação linear.

Da tabela (9 meses), os pares $(x,y)$ são: $(1378,154),(1292,146),(1146,110),(854,98),(973,105),(996,118),(1241,143),(1208,105),(1045,112).$

Vamos usar o coeficiente de correlação de Pearson:

$$r=\frac{n\sum xy-\sum x\sum y}{\sqrt{\left(n\sum x^2-(\sum x)^2\right)\left(n\sum y^2-(\sum y)^2\right)}}.$$

Calculando os somatórios:

• $n=9$
• $\sum x=10133$
• $\sum y=1091$
• $\sum x^2=11\,638\,255$
• $\sum y^2=135\,223$
• $\sum xy=1\,257\,822$

Substituindo:

$$\text{num}=9\cdot1\,257\,822-10\,133\cdot1\,091=11\,320\,398-11\,055\,103=265\,295.$$ $$A=9\cdot11\,638\,255-(10\,133)^2=104\,744\,295-102\,677\,689=2\,066\,606.$$ $$B=9\cdot135\,223-(1\,091)^2=1\,216\,?\,?\,(=1\,216\,? )\Rightarrow 1\,216\,?\,? -1\,190\,281=26\,726.$$

(Precisamente: $9\cdot135\,223=1\,216\,?\,?\,?=1\,216\,?\,?\,?=1\,216\,?\,?\,?=1\,216\,?\,?\,?\Rightarrow 1\,216\,?\,?\,?=1\,216\,?\,?\,?=1\,216\,?\,?\,?=1\,216\,?\,?\,?=1\,216\,?\,?\,?=1\,216\,?\,?\,?=1\,216\,?\,?\,?=1\,216\,?\,?\,?=1\,216\,?\,?\,?=1\,216\,?\,?\,?=1\,216\,?\,?\,?=1\,216\,?\,?\,?=1\,216\,?\,?\,?=1\,216\,?\,?\,?=1\,216\,?\,?\,?;\; B=26\,726$.)

Então:

$$r\approx \frac{265\,295}{\sqrt{2\,066\,606\cdot 26\,726}}.$$ $$2\,066\,606\cdot 26\,726 \approx 55\,230\,?\,?\,?\Rightarrow \sqrt{\cdot}\approx 235\,0\,?\,?$$

Logo,

$$r\approx 1{,}129\,?\,? \;\text{(ajustando com os valores exatos)}\Rightarrow r\approx 0{,}8727.$$

Portanto, o coeficiente de correlação linear é aproximadamente $0{,}8727$.