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extra-credit111

# extra-credit111 - D Y = 1969.09093 7.862146 EXCRATE...

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Extra Credit Assignment Prof. Clark A. A measure of the fit of the model is r2 = 0.0318, which means that approximately 3.8%  of the variation of Wheat(y) around it’s mean (ˉy) is explained by the regressor x. B. Ho: It is not a good fit Ha: It is a good fit  The value of F Critical one-tail is 1.330026.  We have 136-2 degrees of freedom for both Y (Shipment) and X (Excrate) because we  have 2 regressors for each (intercept and X-value). Afterwards I computed FDIST(4.372,134,134) =1.86093E-16 < 0.05. Also the very small  p-value =3.4055E-182 indicates that we should reject Ho. C. After computing the Residuals table in Excel, I calculated the standard deviation of the  residuals using the formula STDEV() and then divided the result by SQRT(135). The  result was 70.2576009.

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Unformatted text preview: D. Y = 1969.09093 + 7.862146 * EXCRATE . Intercept = 1969.09093 Slope = 7.862146 E. Ho: b2 = 0 Ha: b2 ≠ 0 There are 135 observations and 2 regressors. tb2= 1969.09093/ 413.6045925= 4.760805 TINV (0.1,133)= 1.656391 1.656 4.7608, therefore we reject H0 at 5% significance level F. Ho: b2 = 0 Ha: b2 ≠ 0 There are 135 observations and 2 regressors. tb2= 7.862146/ 3.759896783 = 2.091054 TINV (0.1,133)= 1.656391 1.656 < 2.09, therefore we reject H0 at 5% significance level G. The 95% C.I. for the slope is 7.8621 +- t.025 (df=133) * SE(excrate) = 7.8621 +- TINV(0.05,133) * 3.759896783. 95% C.I. 7.861 +- 1.977961 * 3.759896783 = ( 0.425171 , 15.29903 ) 1.86093E-16...
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extra-credit111 - D Y = 1969.09093 7.862146 EXCRATE...

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