The Printout Below Shows a Regression Analysis for a Time $\mathrm { C } 2 = 1.20 + 0.0362 \mathrm { C } 1$

The printout below shows a regression analysis for a time series that included 20 observations.

Regression Analysis: C2 versus C1

The regression equation is
$\mathrm { C } 2 = 1.20 + 0.0362 \mathrm { C } 1$

$\begin{array} { l r r r r } \text { Predictor } & \text { Coef } & \text { SE Coef } & \mathrm { T } & \mathrm { P } \\ \text { Constant } & 1.19947 & 0.08471 & 14.16 & 0.000 \\ \mathrm { C } 1 & 0.036241 & 0.007072 & 5.12 & 0.000 \\\end{array}$

$\begin{array} { lll}\mathrm {~S} = 0.182361 & \mathrm { R } - \mathrm { Sq } = 59.3 \% \quad \mathrm { R } - \mathrm { Sq } ( \text { adj } ) = 57.1 \% \end{array}$

Analysis of Variance
$\begin{array}{lcrrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\\text { Regression } & 1 & 0.87340 & 0.87340 & 26.26 & 0.000 \\\text { Residual Error } & 18 & 0.59860 & 0.03326 & & \\\text { Total } & 19 & 1.47200 & & &\end{array}$

Locate the Durbin-Watson d-statistic and test the null hypothesis that there is no autocorrelation of residuals. Use $\alpha = 0.10$ .