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## 有冇stat高手可以幫幫手...

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 jackyjackychan 高級學徒 發短消息 加為好友 當前離線 2# 大 中 小 發表於 2017-3-8 09:45 AM (第 1325 天) 只看該作者 Let me try ... (a) Calories = 6.5341 + 3.8386(X1) + 9.1412(X2) + 3.9403(X3) - 0.6916(X4) This should be fairly straight forward. The model is basically the sum of the intercept and the products of the coefficients and their corresponding predictors. (b) df(regression) = 4  <- number of predictors df(residual) = 31 - 4 = 27  <- df(total) in an ANOVA is always the sum of the df's of all components in the model MS(regression) = SS(regression)/df(regression) = 1419311/4 = 354827.75 MS(residual) = SS(residual)/df(residual) =  256.307/27 = 9.49285... F = MS(regression)/MS(residual) = 354827.75/9.49285... = 37378.42164... > 2.73  <- This is the critical value of F at 5% level of significance when df(numerator) = 4 and df(denominator) = 27, you should be able to look this up from a stats table As the F is greater than the critical value, the model is significant at the 5% level of significance, meaning that it can predict the dependent variable (calories of food in this question) better than just using the mean. (c) t(X1) = coefficient(X1)/standard error(X1) = 3.8386/0.0859 = 44.6868... Similarly, t(X2) = 117.3453..., t(X3) = 116.5769..., t(X4) = -2.3286... df(t(X1)) = df(t(X2)) = df(t(X3)) = df(t(X4)) = 31 + 1 - 4 - 1 = 27  <- df of the t values of each predictor is (n - p - 1), where n is total number of entries, p is the number of predictors. Also, df(total) in the ANOVA is (n - 1). Look up all the corresponding critical t values. All 4 calculated t values should be greater than their corresponding critical values, so all 4 predictors have significant effect on the calories for food in the regression model at 5% level of significance. Hope this helps UID105284 帖子249 精華0 積分18 閱讀權限20 在線時間353 小時 註冊時間2009-4-1 最後登錄2020-5-3  查看詳細資料 TOP
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