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Test results from the table 4.5 showed that the calculated Durbin-Watson values is equal to 1.46.And Durbin-Watson from the table are as follows: dU is equal to 1.3 dL equals 1.1 (N = 32 & K ' = 2)When comparing the value calculated with the Durbin-Watson Durbin-Watson values from the table reveals that the Durbin-Watson is less than the calculated 4-dL and older dU therefore will accept H0.Conclude that independent tolerance value, which is in accordance with the terms of the regression analysis.
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