Numa amostra de 30 pares de observações do tipo (xi , yi ), com i = 1, 2, ..., 30, a covariância obtida entre as variáveis X e Y foi -2. Os dados f...

Primeiramente, vamos calcular a média de X e Y: $\bar{x} = \frac{1}{n}\sum_{i=1}^{n}x_i$ $\bar{y} = \frac{1}{n}\sum_{i=1}^{n}y_i$ Agora, vamos calcular a covariância entre X e Y: $S_{xy} = \frac{1}{n-1}\sum_{i=1}^{n}(x_i - \bar{x})(y_i - \bar{y})$ Substituindo os valores, temos: $S_{xy} = \frac{1}{30-1}\sum_{i=1}^{30}(x_i - \bar{x})(y_i - \bar{y}) = -2$ Agora, vamos transformar os dados linearmente: $z_i = -3x_i + 1$ $w_i = 2y_i + 3$ Para calcular a covariância entre Z e W, vamos utilizar a seguinte fórmula: $S_{zw} = \frac{1}{n-1}\sum_{i=1}^{n}(z_i - \bar{z})(w_i - \bar{w})$ Substituindo os valores, temos: $S_{zw} = \frac{1}{30-1}\sum_{i=1}^{30}(z_i - \bar{z})(w_i - \bar{w})$ $S_{zw} = \frac{1}{29}\sum_{i=1}^{30}(-3x_i + 1 + 3)(2y_i + 3 - \frac{61}{15})$ $S_{zw} = \frac{1}{29}\sum_{i=1}^{30}(-6x_iy_i + 6x_i - 6y_i + 18)$ $S_{zw} = -\frac{6}{29}\sum_{i=1}^{30}(x_iy_i - x_i + y_i - 3)$ Agora, vamos utilizar a fórmula de covariância para X e Y: $S_{xy} = \frac{1}{29}\sum_{i=1}^{30}(x_i - \bar{x})(y_i - \bar{y})$ $S_{xy} = \frac{1}{29}\sum_{i=1}^{30}(x_iy_i - x_i\bar{y} - y_i\bar{x} + \bar{x}\bar{y})$ $S_{xy} = \frac{1}{29}\sum_{i=1}^{30}(x_iy_i - x_i\bar{y}) - \frac{1}{29}\sum_{i=1}^{30}(y_i\bar{x} - \bar{x}\bar{y})$ $S_{xy} = \frac{1}{29}\sum_{i=1}^{30}(x_iy_i - x_i\bar{y}) - \frac{1}{29}(\bar{x}\sum_{i=1}^{30}y_i - 30\bar{x}\bar{y})$ $S_{xy} = \frac{1}{29}\sum_{i=1}^{30}(x_iy_i - x_i\bar{y}) - \frac{1}{29}\bar{x}\sum_{i=1}^{30}y_i + \frac{30}{29}\bar{x}\bar{y}$ $S_{xy} = \frac{1}{29}\sum_{i=1}^{30}(x_iy_i - x_i\bar{y}) - \frac{1}{29}\bar{x}\sum_{i=1}^{30}y_i + \frac{30}{29}\bar{x}\bar{y}$ $S_{xy} = \frac{1}{29}\sum_{i=1}^{30}(x_iy_i - x_i\bar{y}) - \frac{1}{29}\bar{x}\cdot30\bar{y} + \frac{30}{29}\bar{x}\bar{y}$ $S_{xy} = \frac{1}{29}\sum_{i=1}^{30}(x_iy_i - x_i\bar{y}) - \frac{30}{29}\bar{x}\bar{y}$ Substituindo o valor de $S_{xy}$, temos: $-2 = \frac{1}{29}\sum_{i=1}^{30}(x_iy_i - x_i\bar{y}) - \frac{30}{29}\bar{x}\bar{y}$ $\sum_{i=1}^{30}(x_iy_i - x_i\bar{y}) = -58\bar{x}\bar{y}$ Agora, vamos substituir essa expressão em $S_{zw}$: $S_{zw} = -\frac{6}{29}\sum_{i=1}^{30}(x_iy_i - x_i + y_i - 3)$ $S_{zw} = -\frac{6}{29}(\sum_{i=1}^{30}x_iy_i - \sum_{i=1}^{30}x_i + \sum_{i=1}^{30}y_i - 90)$ $S_{zw} = -\frac{6}{29}(\sum_{i=1}^{30}x_iy_i - 30\bar{x} + 30\bar{y} - 90)$ $S_{zw} = -\frac{6}{29}(\sum_{i=1}^{30}x_iy_i - 30\bar{x} + 30\bar{y} - 30 - 60)$ $S_{zw} = -\frac{6}{29}(\sum_{i=1}^{30}x_iy_i - 30\bar{x} - 30\bar{y} - 90)$ $S_{zw} = -\frac{6}{29}(\sum_{i=1}^{30}x_iy_i - 58\bar{x}\bar{y} - 90)$ Substituindo o valor de $S_{xy}$, temos: $S_{zw} = -\frac{6}{29}(\sum_{i=1}^{30}x_iy_i + 170)$ Agora, vamos calcular $\sum_{i=1}^{30}x_iy_i$: $\sum_{i=1}^{30}x_iy_i = \frac{1}{2}\sum_{i=1}^{30}(x_iy_i + x_iy_i)$ $\sum_{i=1}^{30}x_iy_i = \frac{1}{2}\sum_{i=1}^{30}x_iy_i + \frac{1}{2}\sum_{i=1}^{30}x_iy_i$ $\sum_{i=1}^{30}x_iy_i = \frac{1}{2}\sum_{i=1}^{30}x_iy_i + \frac{1}{2}\sum_{i=1}^{30}y_ix_i$ $\sum_{i=1}^{30}x_iy_i = \frac{1}{2}\sum_{i=1}^{30}(x_iy_i + y_ix_i)$ $\sum_{i=1}^{30}x_iy_i = \frac{1}{2}\sum_{i=1}^{30}(x_iy_i + x_iy_i + y_ix_i - x_iy_i)$ $\sum_{i=1}^{30}x_iy_i = \frac{1}{2}\sum_{i=1}^{30}(x_iy_i + y_ix_i) - \frac{1}{2}\sum_{i=1}^{30}x_iy_i$ $\sum_{i=1}^{30}x_iy_i = \frac{1}{2}\sum_{i=1}^{30}(x_iy_i + y_ix_i) - \frac{1}{2}(-2\cdot29)$ $\sum_{i=1}^{30}x_iy_i = \frac{1}{2}\sum_{i=1}^{30}(x_iy_i + y_ix_i) + 29$ Agora, vamos calcular $\sum_{i=1}^{30}(x_iy_i + y_ix_i)$: $\sum_{i=1}^{30}(x_iy_i + y_ix_i) = \sum_{i=1}^{30}x_iy_i + \sum_{i=1}^{30}y_ix_i$ $\sum_{i=1}^{30}(x_iy_i + y_ix_i) = \frac{1}{2}\sum_{i=1}^{30}(x_iy_i + y_ix_i) + \frac{1}{2}\sum_{i=1}^{30}(x_iy_i + y_ix_i)$ $\sum_{i=1}^{30}(x_iy_i + y_ix_i) = \frac{1}{2}\sum_{i=1}^{30}(x_iy_i + y_ix_i + x_iy_i + y_ix_i)$ $\sum_{i=1}^{30}(x_iy_i + y_ix_i) = \frac{1}{2}\sum_{i=1}^{30}(x_iy_i + y_ix_i + 2x_iy_i - x_iy_i - y_ix_i)$ $\sum_{i=1}^{30}(x_iy_i + y_ix_i) = \frac{1}{2}\sum_{i=1}^{30}(3x_iy_i)$ $\sum_{i=1}^{30}(x_iy_i + y_ix_i) = \frac{3}{2}\sum_{i=1}^{30}(x_iy_i)$ Agora, vamos substituir esses valores em $S_{zw}$: $S_{zw} = -\frac{6}{29}(\frac{3}{2}\sum_{i=1}^{30}(x_iy_i) + 170)$ $S_{zw} = -\frac{9}{29}\sum_{i=1}^{30}(x_iy_i) - \frac{1020}{29}$ Substituindo o valor de $\sum_{i=1}^{30}(x_iy_i)$, temos: $S_{zw} = -\frac{9}{29}(-58\bar{x}\bar{y}) - \frac{1020}{29}$ $S_{zw} = \frac{522}{29} - \frac{1020}{29}$ $S_{zw} = -\frac{498}{29}$ Portanto, a alternativa correta é a letra (C) -7.