🌊Yuri🌊 · 2022年12月25日
NO.PZ2021071601000009
问题如下：
John, an analyst ran a simple regression explaining the variation in electricity consumption as a function of temperature. The total variation of the dependent variable was 132.08, and the explained variation was 60.38. She had 60 monthly observations.
Calculate the F-statistic to test the fit of the model.
选项：
A.49.6850.52
解释：
B is correct.
请问explained variation was 60.38是SSR吗？我理解字面的意思是x解释y的变化的值为60.38. 所以我以为R2等于60.38
NO.PZ2021071601000009 问题如下 John, analyst ra simple regression explaining the variation in electricity consumption a function of temperature. The totvariation of the pennt variable w132.08, anthe explainevariation w60.38. She h60 monthly observations. Calculate the F-statistic to test the fit of the mol. A.49.68 B.48.84 C.50.52 B is correct.F=60.38/1(132.08−60.38)/(60−2)=60.381.2362=48.84F = \frac{{60.38/1}}{{(132.08 - 60.38)/(60 - 2)}} = \frac{{60.38}}{{1.2362}} = 48.84F=(132.08−60.38)/(60−2)60.38/1=1.236260.38=48.84
NO.PZ2021071601000009问题如下 John, analyst ra simple regression explaining the variation in electricity consumption a function of temperature. The totvariation of the pennt variable w132.08, anthe explainevariation w60.38. She h60 monthly observations. Calculate the F-statistic to test the fit of the mol. A.49.68B.48.84C.50.52 B is correct.F=60.38/1(132.08−60.38)/(60−2)=60.381.2362=48.84F = \frac{{60.38/1}}{{(132.08 - 60.38)/(60 - 2)}} = \frac{{60.38}}{{1.2362}} = 48.84F=(132.08−60.38)/(60−2)60.38/1=1.236260.38=48.84 老师 可否再讲一下F检验的使用场景和意义？以及它跟Z检验 T检验的区别？还是不太理解
NO.PZ2021071601000009 如题
NO.PZ2021071601000009 这道题用的是哪个公式啊？
