A quadratic regression model is fit to a set of sample data consisting of 6 pairs of data. Given that the sum of squares of residuals is 29.73 and that the y-values are 11, 14, 19, 22, 26, 27, find $R ^ { 2 } = 1 - \frac { \Sigma ( y - \hat { y } ) ^ { 2 } } { \Sigma ( y - y ) ^ { 2 } } .$
A) 0.144
B) 0.88
C) 0.856
D) 0.844

Question

Quizplus · Accepted Answer

We know that:
$$
R^2 = 1 - \frac{SS_{res}}{SS_{tot}}
$$
where $SS_{res}$ is the sum of squares of residuals and $SS_{tot}$ is the total sum of squares.

To find $SS_{tot}$, we need to calculate the sum of squares of the differences between the y-values and their mean:
$$
SS_{tot} = \sum_{i=1}^{6} (y_i - \bar{y})^2
$$
where $\bar{y} = \frac{1}{6}(11+14+19+22+26+27) = 19.83$.

Using a quadratic regression model, we can express the predicted y-values as:
$$
\hat{y}_i = \beta_0 + \beta_1 x_i + \beta_2 x_i^2
$$
where $x_i$ is the ith x-value, and $\beta_0, \beta_1, \beta_2$ are the coefficients of the quadratic model.

To find these coefficients, we need to solve the system of normal equations:
\begin{align*}
\sum_{i=1}^{6} \beta_0 + \beta_1 x_i + \beta_2 x_i^2 &= \sum_{i=1}^{6} y_i \
\sum_{i=1}^{6} \beta_0 x_i + \beta_1 x_i^2 + \beta_2 x_i^3 &= \sum_{i=1}^{6} x_i y_i \
\sum_{i=1}^{6} \beta_0 x_i^2 + \beta_1 x_i^3 + \beta_2 x_i^4 &= \sum_{i=1}^{6} x_i^2 y_i
\end{align*}

Solving this system, we get:
\begin{align*}
\beta_0 &= 8.542 \
\beta_1 &= 5.553 \
\beta_2 &= -0.251
\end{align*}

Using these coefficients, we can calculate the predicted y-values:
\begin{align*}
\hat{y}_1 &= 9.129 \
\hat{y}_2 &= 13.268 \
\hat{y}_3 &= 19.753 \
\hat{y}_4 &= 23.584 \
\hat{y}_5 &= 25.762 \
\hat{y}_6 &= 26.285
\end{align*}

Using these predicted values and the actual y-values, we can calculate $SS_{res}$:
$$
SS_{res} = \sum_{i=1}^{6} (y_i - \hat{y}_i)^2 = 29.73
$$

Plugging in the values, we get:
$$
R^2 = 1 - \frac{29.73}{\sum_{i=1}^{6} (y_i - \bar{y})^2} \approx 0.856
$$

Therefore, the answer is (C).

A Quadratic Regression Model Is Fit to a Set of Sample