For the following data set, compute the coefficient of determination.
$\begin{array} { l | l l l l l l }
x & 8 & 6 & 3 & 2 & 10 & 4 \
\hline y & 21 & 24 & 20 & 33 & 27 & 27
\end{array}$

A) 0.272
B) 0.728
C) 0.074
D) 0.926

Question

For the following data set, compute the coefficient of determination. 
$\begin{array} { l | l l l l l l } 
x & 8 & 6 & 3 & 2 & 10 & 4 \
\hline y & 21 & 24 & 20 & 33 & 27 & 27
\end{array}$

A) 0.272
B) 0.728
C) 0.074
D) 0.926

Quizplus · Accepted Answer

The coefficient of determination, denoted as $R^2$, measures the proportion of the variance in the dependent variable that is predictable from the independent variable. To compute $R^2$, we first need to calculate the correlation coefficient ($r$) between $x$ and $y$, and then square it ($R^2 = r^2$).Given the data set, we calculate the mean of $x$ and $y$, the deviations from the mean for each $x$ and $y$, the product of these deviations, the squared deviations for $x$ and $y$, and finally use these to calculate $r$.Mean of $x = \frac{8 + 6 + 3 + 2 + 10 + 4}{6} = \frac{33}{6} = 5.5$Mean of $y = \frac{21 + 24 + 20 + 33 + 27 + 27}{6} = \frac{152}{6} = 25.33$Using the formula for $r$:$$r = \frac{n(\sum xy) - (\sum x)(\sum y)}{\sqrt{[n\sum x^2 - (\sum x)^2][n\sum y^2 - (\sum y)^2]}}$$After calculating $r$, we find $r \approx -0.272$. Squaring this to get $R^2$, we obtain $R^2 \approx 0.074$.Therefore, the coefficient of determination is approximately 0.074, which indicates that only about 7.4% of the variance in $y$ can be predicted from $x$.

For the Following Data Set, Compute the Coefficient of Determination x8632104y212420332727\begin{array} { l | l l l l l l } x & 8 & 6 & 3 & 2 & 10 & 4 \\\hline y & 21 & 24 & 20 & 33 & 27 & 27\end{array}xy​821​624​320​233​1027​427​​

For the Following Data Set, Compute the Coefficient of Determination $\begin{array} { l | l l l l l l } x & 8 & 6 & 3 & 2 & 10 & 4 \\\hline y & 21 & 24 & 20 & 33 & 27 & 27\end{array}$