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@ -268,6 +268,137 @@
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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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"## Z检验实例\n",
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"### Z检验实例1:"
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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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"研究正常人与高血压患者胆固醇含量(mg%)的资料如下,试比较两组血清胆固醇含量有无差别。\n",
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"<br>\n",
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"正常人组 \n",
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"$$\n",
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"n_1 = 506,\\overline{X}_1 = 180.6,S_1 = 34.2\n",
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"$$\n",
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"样本数506,均值1800.6,标准差34.2\n",
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"<br>\n",
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"<br>\n",
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"高血压组\n",
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"$$\n",
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"n_2 = 142,\\overline{X}_2 = 223.6,S_2 = 45.8\n",
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"$$"
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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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"建立检验假设,确定检验水平\n",
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"<ul>\n",
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"<li>H0: μ1 = μ2;认为没有差别\n",
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"<li>H1: μ1 ≠μ2;认为有区别\n",
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"<li>α = 0.05 ;有5%的置信空间,即误差在这个范围内是允许的\n",
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"</li>\n",
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"</ul>\n",
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"计算统计量Z\n",
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"<ul>\n",
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" <li>将已知数据代入公式</li>\n",
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"$$\n",
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"Z = \\frac{|180.6-223.6|}{\\sqrt{34.2^2/506+45.8^2/142}} = 10.40\n",
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"$$"
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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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"这里是双侧,即二分之一α,1-α/2=0.975,查表\n",
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"<br>网上搜索:统计分布临界值\n",
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"<img src=\"assets/20201115111943.png\" width=\"70%\">\n",
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"<br><br>\n",
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"1.9+0.6=1.96,统计量为10.4比1.96大,意味着面积肯定小于1.96临界值(α/2)的面积\n",
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"<img src=\"assets/20201115112520.png\" width=\"70%\">\n",
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"根据双侧检验,若p值 < α/2,拒绝H0"
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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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"确认P值,作为判断结论;本例Z=10.40>1.96(查表得0.975对应值),故P<0.05,按α=0.05水准拒绝H0,接受H1,可以认为正常人与高血压患者的血清胆固醇含量有差别,高血压患者高于正常人。\n",
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"<br>\n",
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"**注意:我们的第一反应可能是不应该越小表示差异也越小吗,其实是越小于α/2,表示两者值的越偏离,因为我们已经假定了对比值A在中间(H0值),那么对比A(H0)的B(H1)应该越趋向中间(越高),才表示相似**"
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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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"### Z检验实例2:"
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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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"某机床厂加工一种零件,根据经验知道,该厂加工零件的椭圆度近似服从正态分布,其总体均值为p=0.081mm,总体标准差为=0.025。今换一种新机床进行加抽取n=200个零件进行检验,得到的椭圆度为0.076mm。试问新机床加工零件的椭圆度的均值与以前有无显著差异?(a=0.05)"
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]
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},
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{
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"attachments": {},
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"H0: μ = 0.081;H1:μ ≠ 0.081; α = 0.05; n = 200\n",
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"<br>**检验统计量**\n",
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"$$\n",
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"z = \\frac{\\overline{x}-μ_0}{σ/\\sqrt{n}} \n",
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"= \n",
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"\\frac{0.076-0.081}{0.025/\\sqrt{200}} = -2.83\n",
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"$$\n",
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"<br>\n",
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"决策:\n",
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"<br>-2.83在-1.96左侧,也就是p值的面积小于α/2,在α = 0.05的水平上拒绝H0\n",
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"<br>结论:\n",
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"<br>有证据表明新机床加工的零件的椭圆度与以前有显著差异\n",
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"<img src=\"assets/20201115151902.png\" width=\"50%\">"
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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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"### Z检验实例3:\n",
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"根据过去大量资料,某厂生产的灯泡的使用寿命服从正态分布N~(1020,100^2)。现从最近生产的一批产品中随机抽取16只,测得样本平均寿命为1080小时。试在005的显著性水平下判断这批产品的使用寿命是否有显著提高?(a=0.05)"
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},
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{
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"attachments": {},
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"H0: μ ≤ 1020;H1:μ > 1020; α = 0.05; n = 16\n",
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"<br>**检验统计量(单侧)**\n",
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"$$\n",
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"z = \\frac{\\overline{x}-μ_0}{σ/\\sqrt{n}} \n",
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"= \n",
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"\\frac{1080-1020}{100/\\sqrt{14}} = 2.4\n",
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"$$\n",
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"1-0.05=0.95,其临界值没有,相近的是0.9505和0.9495,那么把它们两相加除以2作为0.95的临界值,(1.6+1.6)/2+(0.04+0.05)/2=1.645\n",
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"<br>\n",
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"决策:\n",
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"<br>2.4在1.645右侧,也就是p值的面积小于α,在α = 0.05的水平上拒绝H0\n",
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"<br>结论:\n",
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"<br>有证据表明新生产的灯泡的使用寿命有显著提高\n",
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"<img src=\"assets/20201115152913.png\" width=\"50%\">"
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]
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},
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{
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"cell_type": "code",
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"cell_type": "code",
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"execution_count": null,
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"execution_count": null,
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