Statistics 简明教程
Statistics - F Test Table
F 检验以更为著名的分析师 R.A. 费希尔命名。F 检验用于测试总体变化的两个自主评估是否完全不同，或者是否可以将这两个样本视为从具有相同差异的典型总体中抽取。为了进行测试，我们计算 F 统计量，定义如下：
F-test is named after the more prominent analyst R.A. Fisher. F-test is utilized to test whether the two autonomous appraisals of populace change contrast altogether or whether the two examples may be viewed as drawn from the typical populace having the same difference. For doing the test, we calculate F-statistic is defined as:
Formula
Example
Problem Statement:
Problem Statement:
在 8 个观察值的样本中，事物从平均值的平方偏差总数为 94.5。在另一个由 10 个感知组成的样本中，观察到的值为 101.7。测试该差异在 5% 的水平上是否巨大。（我们给定在 5% 的中心性水平上，对于 ${v_1}$ = 7 和 ${v_2}$ = 9，${F_.05}$ 的基本估计值是 3.29）。
In a sample of 8 observations, the entirety of squared deviations of things from the mean was 94.5. In another specimen of 10 perceptions, the worth was observed to be 101.7 Test whether the distinction is huge at 5% level. (You are given that at 5% level of centrality, the basic estimation of ${F}$ for ${v_1}$ = 7 and ${v_2}$ = 9, ${F_.05}$ is 3.29).
Solution:
Solution:
让我们假设两个样本的方差的差异不显著，即 ${H_0: {\sigma_1}^2 = {\sigma_2}^2}$
Let us take the hypothesis that the difference in the variances of the two samples is not significant i.e. ${H_0: {\sigma_1}^2 = {\sigma_2}^2}$
我们得到以下内容：
We are given the following:
应用 F 检验
Applying F-Test
对于 ${v_1}$ = 8-1 = 7，${v_2}$ = 10-1 = 9，并且 ${F_.05}$ = 3.29。${F}$ 的计算值小于表值。因此，我们接受零假设，并得出结论：两个样本方差的差异在 5% 水平上并不显著。
For ${v_1}$ = 8-1 = 7, ${v_2}$ = 10-1 = 9 and ${F_.05}$ = 3.29. The Calculated value of ${F}$ is less than the table value. Hence, we accept the null hypothesis and conclude that the difference in the variances of two samples is not significant at 5% level.