Add T检验实例

notebook_必备数学基础/假设检验章节/.ipynb_checkpoints/假设检验-checkpoint.ipynb

@@ -351,7 +351,6 @@
    ]
   },
   {
-   "attachments": {},
    "cell_type": "markdown",
    "metadata": {},
    "source": [
@@ -379,7 +378,6 @@
    ]
   },
   {
-   "attachments": {},
    "cell_type": "markdown",
    "metadata": {},
    "source": [
@@ -429,6 +427,197 @@
     "    <li>应用条件，总体标准α未知的小样本资料，且服从正态分布。"
    ]
   },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "#### 实例\n",
+    "以往通过大规模调査已知某地新生儿出生体重为3.30kg。从该地难产儿中随机抽取35名新生儿,平均出生体重为3.42kg,标准差为0.40kg,问该地难产儿出生体重是否与一般新生儿体重不同?"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "建立检验假设，确定检验水准\n",
+    "H0： μ=μ0；H1：μ≠μ0；α=0.05\n",
+    "<br>\n",
+    "计算检验统计量\n",
+    "$$\n",
+    "t = \\frac{\\overline{X} - μ_0}{S_{\\overline{x}}}\n",
+    "= \\frac{\\overline{X}-μ_0}{S/\\sqrt{n}}\n",
+    "= \\frac{3.42-3.30}{0.40/\\sqrt{35}}\n",
+    "= 1.77\n",
+    "$$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "本例自由度v=n-1=35-1=34,查表得得t0.052/34=2.032。因为t<t0.052/34，故P>0.05，按α=0.05水准，不拒绝H0，差别无统计学意义，尚不能认为该地难产;\n",
+    "<br>自由度：可以随意变换的个数是多少"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "网上搜索：t分布临界值表\n",
+    "<img src=\"assets/20201116212408.png\" width=\"70%\">"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### 配对样本均数t检验\n",
+    "<ul>\n",
+    "<li>简称配对t检验( paired t test)，又称非独立俩样木均数t检验，适用于配对设计计量资料均数的比较\n",
+    "    <li>配对设计( paired design)是将受试对象按某些特征相近的原则配成对子，每对中的两个个体随机地给予两种处理"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "#### 配对样本均数t检验原理\n",
+    "<ul>\n",
+    "<li>配对设计的资料具有对子内数据一一对应的特征，研究者应关心是对子的效应差值而不是各自的效应值。\n",
+    "<li>进行配对t检验时，首选应计算各对数据间的差值d，将d作为变量计算均数。\n",
+    "<li>配对样本t检验的基本原理是假设两种处理的效应相同，理论上差值d的总体均数μd为0，现有的不等于0差值样本均数可以来自μd=0的总体，也可以来ud≠0的总体。\n",
+    "<li>可将该检验理解为差值样本均数与已知总体均数μd(pd=0)比较的单样本t检验,其检验统计量"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "$$\n",
+    "t = \\frac{\\overline{d} - μ_d}{S_{\\overline{d}}}\n",
+    "= \\frac{\\overline{d} - 0}{S_{\\overline{d}}}\n",
+    "= \\frac{\\overline{d}}{S_d/\\sqrt{n}}\n",
+    "$$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## T检验实例"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "**实例1：**\n",
+    "<br>\n",
+    "有12名接种卡介苗的儿童，8周虐用两批不同的结核菌素，一批是标准结核菌素，一批是新制结核菌素，分别注射在儿童的前臂，两种结核菌素的皮肤浸润反应平均直径(mm)如表所示，问两种结核菌素的反应性有无差别。\n",
+    "<img src=\"assets/20201116215635.png\" width=\"70%\">"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "建立检验假设，确定检验水准\n",
+    "<br>H0:μd = 0\n",
+    "<br>H1：μd ≠ 0\n",
+    "<br>α = 0.05\n",
+    "<br>计算检验统计量本例\n",
+    "$$\n",
+    "\\sum d = 39, \\sum d^2 = 195\n",
+    "$$\n",
+    "<br>先计算差数的标准差\n",
+    "$$\n",
+    "S_d = \\sqrt{\\frac{\\sum d^2 - \\frac{(\\sum d)^2}{n}}{n-1}}\n",
+    "= \\sqrt{\\frac{195-\\frac{(39)^2}{12}}{12-1}}\n",
+    "= 2.4909\n",
+    "$$\n",
+    "<br>计算差值的标准误\n",
+    "$$\n",
+    "S_{\\overline{d}} = \\frac{S_d}{\\sqrt{n}}\n",
+    "= \\frac{2.4909}{3.464}\n",
+    "= 0.7191\n",
+    "$$\n",
+    "<br> 按公式计算，得\n",
+    "$$\n",
+    "t = \\frac{\\overline{d}}{S_{\\overline{d}}}\n",
+    "= \\frac{3.25}{0.7191}\n",
+    "= 4.5195\n",
+    "$$\n",
+    "确定P值，作出推断结论\n",
+    "<br>自由度计算为 v=n-1=12-1=11,\n",
+    "<br>查附表，得t0.05/2.11 = 2.201,\n",
+    "<br>本例t > t0.05/2.11，\n",
+    "<br>P < 0.05，拒绝H0，接受H1，反应结果有差别。"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### 两独立样本t检验\n",
+    "<ul>\n",
+    "<li>两独立样本t检验(two independent sample t-test)，又称成组t检验。\n",
+    "<li>适用于完全随机设计的两样本均数的比较，其目的是检验两样本所来自总体的均数是否相等。\n",
+    "<li>完全随机设计是将受试对象随机地分配到两组中，每组患者分别接受不同的处理，分析比较处理的效应，\n",
+    "<li>两独立样本t检验要求两样本所代表的总体服从正态分布N(μ1,σ42)和N(μ2,σ2)，且两总体方差σ1^2、σ2^2相等即方差齐性。若两总体方差不等需要先进行变换。"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### 两独立样木t检验原理\n",
+    "两独立样本t检验的检验假设是两总体均数相等，即H0:u1=2，也可表述为u1-μ2=0，这里可将两样本均数的差值看成一个变量样本则在H0条件下两独立样本均数t检验可视为样本与已知总体均数μ1-μ2=0的单样本t检验，统计量计算公式为\n",
+    "$$\n",
+    "t = \\frac{|(\\overline{X}_1 - \\overline{X}_2|) - (μ_1 - μ_2)}{S_{\\overline{X}_1-\\overline{X}_2}}, v=n_1+n_2-2\n",
+    "$$\n",
+    "$$\n",
+    "S_{\\overline{X}_1-\\overline{X}_2} = \\sqrt{S_c^2(\\frac{1}{n_1}+\\frac{1}{n_2})}\n",
+    "$$\n",
+    "$$\n",
+    "S_c^2 = \\frac{\\sum X^2_1 - \\frac{(\\sum X_t)^2}{n_1}+\\sum X_2^2 - \\frac{(\\sum X_2)^2}{n_2}}{n_1+n_2-2}\n",
+    "$$\n",
+    "$$\n",
+    "S_c^2称为合并方差(combined/pooled variance)\n",
+    "$$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "**实例2：**\n",
+    "<br>\n",
+    "25例糖尿病患者随机分成两组，甲组单纯用药物治疗，乙组采用药物治疗合并饮食疗法，二个月后测空腹血糖(mmoL)如表所示，问两种疗法治疗后患者血糖值是否相同?\n",
+    "<img src=\"assets/20201116221815.png\" width=\"70%\">"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "建立检验假设，确定检验水准\n",
+    "<br>H0：μ1=μ2；H1：μ1≠μ2；α=0.05\n",
+    "<br>计算检验统计量\n",
+    "$$\n",
+    "由原数据算得: n_1=12, \\sum X_1=182.5,\\sum X_1^2=2953.43,n_2=13,\\sum X_2=141.0,\n",
+    "\\sum X_2^2=1743.16, \\overline{X}_1=\\sum X_1/n_1=182.5/12=15.21, \n",
+    "\\overline{X}_2/n_2=14.16/13=10.85\n",
+    "$$\n",
+    "$$\n",
+    "代入公式，得\n",
+    "S_c^2 = \\frac{2953.43 - \\frac{(182.5)^2}{12}+1743.16-\\frac{(141.0)^2}{13}}{12+13-2} = 17.03\n",
+    "$$\n",
+    "$$\n",
+    "S_{\\overline{X}_1-\\overline{X}_2} = \\sqrt{17.03(\\frac{1}{12}+\\frac{1}{13})}\n",
+    "=1.652\n",
+    "$$"
+   ]
+  },
   {
    "cell_type": "code",
    "execution_count": null,

notebook_必备数学基础/假设检验章节/假设检验.ipynb

@@ -455,7 +455,7 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "本例自由度v=n-1=35-1=34,查表得得t0.052/34=2.032。因为t<t0.052/34,故P>0.05,按α=0.05水准，不拒绝H0，差别无统计学意义，尚不能认为该地难产;\n",
+    "本例自由度v=n-1=35-1=34,查表得得t0.052/34=2.032。因为t<t0.052/34，故P>0.05，按α=0.05水准，不拒绝H0，差别无统计学意义，尚不能认为该地难产;\n",
     "<br>自由度：可以随意变换的个数是多少"
    ]
   },
@@ -507,6 +507,117 @@
     "## T检验实例"
    ]
   },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "**实例1：**\n",
+    "<br>\n",
+    "有12名接种卡介苗的儿童，8周虐用两批不同的结核菌素，一批是标准结核菌素，一批是新制结核菌素，分别注射在儿童的前臂，两种结核菌素的皮肤浸润反应平均直径(mm)如表所示，问两种结核菌素的反应性有无差别。\n",
+    "<img src=\"assets/20201116215635.png\" width=\"70%\">"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "建立检验假设，确定检验水准\n",
+    "<br>H0:μd = 0\n",
+    "<br>H1：μd ≠ 0\n",
+    "<br>α = 0.05\n",
+    "<br>计算检验统计量本例\n",
+    "$$\n",
+    "\\sum d = 39, \\sum d^2 = 195\n",
+    "$$\n",
+    "<br>先计算差数的标准差\n",
+    "$$\n",
+    "S_d = \\sqrt{\\frac{\\sum d^2 - \\frac{(\\sum d)^2}{n}}{n-1}}\n",
+    "= \\sqrt{\\frac{195-\\frac{(39)^2}{12}}{12-1}}\n",
+    "= 2.4909\n",
+    "$$\n",
+    "<br>计算差值的标准误\n",
+    "$$\n",
+    "S_{\\overline{d}} = \\frac{S_d}{\\sqrt{n}}\n",
+    "= \\frac{2.4909}{3.464}\n",
+    "= 0.7191\n",
+    "$$\n",
+    "<br> 按公式计算，得\n",
+    "$$\n",
+    "t = \\frac{\\overline{d}}{S_{\\overline{d}}}\n",
+    "= \\frac{3.25}{0.7191}\n",
+    "= 4.5195\n",
+    "$$\n",
+    "确定P值，作出推断结论\n",
+    "<br>自由度计算为 v=n-1=12-1=11,\n",
+    "<br>查附表，得t0.05/2.11 = 2.201,\n",
+    "<br>本例t > t0.05/2.11，\n",
+    "<br>P < 0.05，拒绝H0，接受H1，反应结果有差别。"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### 两独立样本t检验\n",
+    "<ul>\n",
+    "<li>两独立样本t检验(two independent sample t-test)，又称成组t检验。\n",
+    "<li>适用于完全随机设计的两样本均数的比较，其目的是检验两样本所来自总体的均数是否相等。\n",
+    "<li>完全随机设计是将受试对象随机地分配到两组中，每组患者分别接受不同的处理，分析比较处理的效应，\n",
+    "<li>两独立样本t检验要求两样本所代表的总体服从正态分布N(μ1,σ42)和N(μ2,σ2)，且两总体方差σ1^2、σ2^2相等即方差齐性。若两总体方差不等需要先进行变换。"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### 两独立样木t检验原理\n",
+    "两独立样本t检验的检验假设是两总体均数相等，即H0:u1=2，也可表述为u1-μ2=0，这里可将两样本均数的差值看成一个变量样本则在H0条件下两独立样本均数t检验可视为样本与已知总体均数μ1-μ2=0的单样本t检验，统计量计算公式为\n",
+    "$$\n",
+    "t = \\frac{|(\\overline{X}_1 - \\overline{X}_2|) - (μ_1 - μ_2)}{S_{\\overline{X}_1-\\overline{X}_2}}, v=n_1+n_2-2\n",
+    "$$\n",
+    "$$\n",
+    "S_{\\overline{X}_1-\\overline{X}_2} = \\sqrt{S_c^2(\\frac{1}{n_1}+\\frac{1}{n_2})}\n",
+    "$$\n",
+    "$$\n",
+    "S_c^2 = \\frac{\\sum X^2_1 - \\frac{(\\sum X_t)^2}{n_1}+\\sum X_2^2 - \\frac{(\\sum X_2)^2}{n_2}}{n_1+n_2-2}\n",
+    "$$\n",
+    "$$\n",
+    "S_c^2称为合并方差(combined/pooled variance)\n",
+    "$$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "**实例2：**\n",
+    "<br>\n",
+    "25例糖尿病患者随机分成两组，甲组单纯用药物治疗，乙组采用药物治疗合并饮食疗法，二个月后测空腹血糖(mmoL)如表所示，问两种疗法治疗后患者血糖值是否相同?\n",
+    "<img src=\"assets/20201116221815.png\" width=\"70%\">"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "建立检验假设，确定检验水准\n",
+    "<br>H0：μ1=μ2；H1：μ1≠μ2；α=0.05\n",
+    "<br>计算检验统计量\n",
+    "$$\n",
+    "由原数据算得: n_1=12, \\sum X_1=182.5,\\sum X_1^2=2953.43,n_2=13,\\sum X_2=141.0,\n",
+    "\\sum X_2^2=1743.16, \\overline{X}_1=\\sum X_1/n_1=182.5/12=15.21, \n",
+    "\\overline{X}_2/n_2=14.16/13=10.85\n",
+    "$$\n",
+    "$$\n",
+    "代入公式，得\n",
+    "S_c^2 = \\frac{2953.43 - \\frac{(182.5)^2}{12}+1743.16-\\frac{(141.0)^2}{13}}{12+13-2} = 17.03\n",
+    "$$\n",
+    "$$\n",
+    "S_{\\overline{X}_1-\\overline{X}_2} = \\sqrt{17.03(\\frac{1}{12}+\\frac{1}{13})}\n",
+    "=1.652\n",
+    "$$"
+   ]
+  },
   {
    "cell_type": "code",
    "execution_count": null,