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@ -223,7 +223,7 @@
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"cell_type": "code",
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"execution_count": 10,
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"execution_count": 2,
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"metadata": {},
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"outputs": [
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@ -441,6 +441,142 @@
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"所以学生的学习潜在学习能力与自学能力有较强的正相关。"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## 肯德尔和谐系数( Kendall)\n",
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"当多个(两个以上)变量值以等级次序排列或以等级次序表示，描述这几个变量之间的一性程度的量，称为肯徳尔和谐系数。\n",
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"<br>它常用来表示几个评定者对同组学生成绩用等级先后评定多次之间的一致性程度。\n",
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"<br>\n",
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"**同一评价者无相同等级评定时,计算公式：**\n",
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"$$\n",
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"W = \\frac{S}{\\frac{1}{12}K^2(N^3-N)}\n",
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"$$\n",
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"<ul>\n",
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" <li>N 被评的对象数；\n",
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" <li>K 评分者人数或评分所依据的标准数；\n",
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" <li>S 每个被评对象所评等级之和Ri与所有这些和的平均数的离差平方和\n",
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"$$\n",
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"S=\\sum^n_{i=1}(R_i-\\overline{R_i})^2\n",
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"= \\sum^n_{i=1}R^2_i-\\frac{1}{n}(\\sum^n_{i=1}R_i)^2\n",
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"$$\n",
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" 当评分者意见完成一致时，S取得最大值，和谐系数是实际求得的S与其最大可能取值的比值，故0≤W≤1."
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"**同一评价者有相同等级评定时，计算公式**\n",
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"<br>\n",
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"$$\n",
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"W = \\frac{S}{\\frac{1}{12}[K^2(N^3-N)-K\\sum^K_{i=1}T_i]}\n",
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"$$\n",
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"$$\n",
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"T_i = \\sum^{mi}_{i=1}(n^3_{ij}-n_{ij})\n",
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"$$\n",
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"<ul>\n",
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" <li>mi为第i个评价者的评定结果中有重复等级的个数。\n",
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" <li>nij为第i个评价者的评定结果中第j个重复等级的相同等级数。\n",
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" <li>对于评定结果无相同等级的评价者,T=0,因此只须对评定结果有相同等级的评价者计算Ti。"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"**实例1:同一评价者无相同等级评定时**\n",
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"某校开展学生小论文比赛，请6位教师对入选的6篇论文评定得奖等级，结果如下表所示，试计算6位教师评定结果的 kanda和谐系数\n",
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"\n",
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"|论文编号 | 论文1 | 论文2 | 论文3 | 论文4 | 论文5 | 论文6 ||\n",
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"| :----: | :----: | :----: | :----: | :----: | :----: | :----: |:----: |\n",
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"| 老师1 | 3 | 1 | 2 | 5 | 4 | 6 | |\n",
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"| 老师2 | 2 | 1 | 3 | 4 | 5 | 6 | |\n",
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"| 老师3 | 3 | 2 | 1 | 5 | 4 | 6 | |\n",
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"| 老师4 | 4 | 1 | 2 | 6 | 3 | 5 | |\n",
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"| 老师5 | 3 | 1 | 2 | 6 | 4 | 5 | |\n",
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"| 老师6 | 4 | 2 | 1 | 5 | 3 | 6 | |\n",
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"| R |19 | 8 | 11 | 31 | 23 |34| ∑K = 126 |\n",
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"| R^2 | 361 | 64| 121 |961|529|1156|R^2 = 3192|\n",
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"\n",
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"由于每个评分老师对6篇论文的评定都无相同的等级：\n",
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"$$\n",
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"S=\\sum^6_{i=1}-\\frac{1}{6}(\\sum^6_{i=1}R_i)^2\n",
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"= 3192- \\frac{1}{6} * 126^2 = 546\n",
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"$$\n",
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"$$\n",
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"W = \\frac{S}{\\frac{1}{12}K^2(N^3-N)} = \\frac{546}{\\frac{1}{12}6^2(6^3-6)}\n",
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"=\\frac{546}{630}=0.87\n",
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"$$\n",
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"由W=0.87表明6位老师的评定结果有较大的一致性"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"**实例1:同一评价者有相同等级评定时**\n",
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"3名专家对6篇心理学论文的评分经等级转换下表所示,试计算专家评定结果的肯德尔和谐系数。\n",
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"\n",
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"|论文编号 | A | B | C | D | E | F ||\n",
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"| :----: | :----: | :----: | :----: | :----: | :----: | :----: |:----: |\n",
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"| 甲 | 1 | 4 | 2.5 | 5 | 6 | 2.5 | |\n",
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"| 乙 | 2 | 3 | 1 | 5 | 6 | 4 | |\n",
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"| 丙 | 1.5 | 3 | 1.5 | 4 | 5.5 | 5.5 | |\n",
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"| R |4.5 | 10 | 5 | 14 | 17.5 |12| 63 |\n",
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"| R^2 | 20.25 | 100| 25 |196|306.25|144|791.5|\n",
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"\n",
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"由于甲、丙对6篇论文有相同等级的评定\n",
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"甲T = 23-2=6\n",
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"丙T = (23 - 2)+(23 - 2) = 12\n",
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"$$\n",
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"S=\\sum^6_{i=1}-\\frac{1}{6}(\\sum^6_{i=1}R_i)^2\n",
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"=791.5-\\frac{1}{6} * 63^2 = 130.00\n",
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"$$\n",
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"$$\n",
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"W = \\frac{S}{\\frac{1}{12}[K^2(N^3-N)-K\\sum^K_{i=1}T_i]}\n",
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"=\\frac{130}{\\frac{1}{12}[3^2(6^3-6)-3*(6+12)]}\n",
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"=\\frac{130}{153} = 0.849\n",
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"$$\n",
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"由W=0.849可看出专家评定结果有较大的一致性"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"**肯德尔和谐系数的显著性检验**\n",
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"评分者人数(k)在3-20之间，被评者(N)在3-7之间时，可查《肯德尔和谐系数(w)显著性临界值表》，检验W是否达到显著性水平。若实际计算的S值大于k、N相同的表内临界值，则W达到显著水平\n",
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"\n",
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"当K=6N=6，查表得检验水平分别为α=0.01，α=0.05的临界值各为S0.01=282.4,S0.05=221.4，均小于实算的S=546，故W达到显著水平，认为6位教师对6篇论文的评定相当一致\n",
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"\n",
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"当被评者n>7时，则可用如下的X2统计量对W是否达到显著水平作检验。"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 3,
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"tau 0.6\n",
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"p_value 0.23333333333333334\n"
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]
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}
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],
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"source": [
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"x1 = [10,9,8,7,6]\n",
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"x2 = [10,8,9,6,7]\n",
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"\n",
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"tau, p_value = stats.kendalltau(x1,x2)\n",
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"print('tau', tau)\n",
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"print('p_value', p_value)"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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benjas
4 years ago
