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+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 35 : Sampling And Inference"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 35.1, page no. 864"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Suppose the coin is unbiased\n",
+      "Then probability of getting the head in a toss = 1/2 \n",
+      "Then, expected no. of successes = a = 1/2∗400\n",
+      "Observed no. of successes = 216\n",
+      "The excess of observed value over expected value =  216\n",
+      "S. D. of simple sampling = (n∗p∗q)ˆ0.5 = c \n",
+      "Hence, z = (b−a)/c =  21.6\n",
+      "As z<1.96, the hypothesis is accepted at 5% level of significance\n"
+     ]
+    }
+   ],
+   "source": [
+    "print \"Suppose the coin is unbiased\"\n",
+    "print \"Then probability of getting the head in a toss = 1/2 \"\n",
+    "print \"Then, expected no. of successes = a = 1/2∗400\"\n",
+    "a = 1/2*400\n",
+    "print \"Observed no. of successes = 216\"\n",
+    "b = 216\n",
+    "print \"The excess of observed value over expected value = \",b-a\n",
+    "print \"S. D. of simple sampling = (n∗p∗q)ˆ0.5 = c \"\n",
+    "c = (400*0.5*0.5)**0.5\n",
+    "print \"Hence, z = (b−a)/c = \",(b-a)/c\n",
+    "print \"As z<1.96, the hypothesis is accepted at 5% level of significance\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 35.2, page no. 865"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Suppose the die is unbiased \n",
+      "Then probability of getting 5 or 6 with one die=1/3 \n",
+      "Then, expected no. of successes = a = 1/3∗9000 \n",
+      "Observed no. of successes = 3240\n",
+      "The excess of observed value over expected value =  240.0\n",
+      "S. D. of simple sampling = (n∗p∗q)ˆ0.5=c\n",
+      "Hence, z = (b−a)/c =  5.366563146\n",
+      "As z>2.58, the hypothesis has to be rejected at 1% level of significance\n"
+     ]
+    }
+   ],
+   "source": [
+    "print \"Suppose the die is unbiased \"\n",
+    "print \"Then probability of getting 5 or 6 with one die=1/3 \"\n",
+    "print \"Then, expected no. of successes = a = 1/3∗9000 \"\n",
+    "a = 1./3*9000\n",
+    "print \"Observed no. of successes = 3240\"\n",
+    "b = 3240\n",
+    "print \"The excess of observed value over expected value = \",b-a\n",
+    "print \"S. D. of simple sampling = (n∗p∗q)ˆ0.5=c\"\n",
+    "c = (9000*(1./3)*(2./3))**0.5\n",
+    "print \"Hence, z = (b−a)/c = \",(b-a)/c\n",
+    "print \"As z>2.58, the hypothesis has to be rejected at 1% level of significance\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 35.3, page no. 865"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "q = 1−p \n",
+      "Standard error of the population of families having a monthly income of rs. 250 or less=(p∗q/n)ˆ0.5 =  0.0148442858929\n",
+      "Hence taking 103/420 to be the estimate of families having a monthly income of rs. 250 or less, the limits are 20% and 29% approximately\n"
+     ]
+    }
+   ],
+   "source": [
+    "p = 206./840\n",
+    "print \"q = 1−p \"\n",
+    "q = 1-p\n",
+    "n = 840\n",
+    "print \"Standard error of the population of families having a monthly income of rs. 250 or less=(p∗q/n)ˆ0.5 = \",(p*q/n)**0.5\n",
+    "print \"Hence taking 103/420 to be the estimate of families having a monthly income of rs. 250 or less, the limits are 20% and 29% approximately\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 35.4, page no. 866"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "p=(n1∗p1+n2∗p2)/(n1+n2)\n",
+      "0.1904\n",
+      "q = 1−p \n",
+      "0.8096\n",
+      "e=(p∗q∗(1/n1+1/n2))ˆ0.5\n",
+      "0.0163590274093\n",
+      "z =  0.916924926202\n",
+      "As z<1, the difference between the proportions is not significant. \n"
+     ]
+    }
+   ],
+   "source": [
+    "n1 = 900\n",
+    "n2 = 1600\n",
+    "p1 = 20./100\n",
+    "p2 = 18.5/100\n",
+    "print \"p=(n1∗p1+n2∗p2)/(n1+n2)\"\n",
+    "p = (n1*p1+n2*p2)/(n1+n2)\n",
+    "print p\n",
+    "print \"q = 1−p \"\n",
+    "q = 1-p\n",
+    "print q\n",
+    "print \"e=(p∗q∗(1/n1+1/n2))ˆ0.5\"\n",
+    "e = (p*q*((1./n1)+(1./n2)))**0.5\n",
+    "print e\n",
+    "z = (p1-p2)/e\n",
+    "print \"z = \",z\n",
+    "print \"As z<1, the difference between the proportions is not significant. \""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 35.5, page no. 867"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "q1 = 1−p1 \n",
+      "q2=1−p2\n",
+      "e=((p1∗q1/n1)+(p2*q2/n2))ˆ0.5\n",
+      "Hence, it is likely that real difference will be hidden.\n"
+     ]
+    }
+   ],
+   "source": [
+    "p1 = 0.3\n",
+    "p2 = 0.25\n",
+    "print \"q1 = 1−p1 \"\n",
+    "q1 = 1-p1\n",
+    "print \"q2=1−p2\"\n",
+    "q2 = 1-p2\n",
+    "n1 = 1200\n",
+    "n2 = 900\n",
+    "print \"e=((p1∗q1/n1)+(p2*q2/n2))ˆ0.5\"\n",
+    "e = ((p1*q1/n1)+(p2*q2/n2))**0.5\n",
+    "z = (p1-p2)/e\n",
+    "print \"Hence, it is likely that real difference will be hidden.\" "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 35.6, page no. 868"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 11,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "m and n represents mean and number of objects in sample respectively \n",
+      "z=(m−M)/(d/(nˆ0.5)\n",
+      "z =  2.7950310559\n",
+      "As z>1.96, it cannot be regarded as a random sample\n"
+     ]
+    }
+   ],
+   "source": [
+    "print \"m and n represents mean and number of objects in sample respectively \"\n",
+    "m = 3.4\n",
+    "n = 900.\n",
+    "M = 3.25\n",
+    "d = 1.61\n",
+    "print \"z=(m−M)/(d/(nˆ0.5)\"\n",
+    "z = (m-M)/(d/(n**0.5))\n",
+    "print \"z = \",z\n",
+    "print \"As z>1.96, it cannot be regarded as a random sample\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 35.9, page no. 871"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 12,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "m1 and n1 represents mean and no. of objects in sample 1\n",
+      "m2 and n2 represents mean and no. of objects in sample 2\n",
+      "On the hypothesis that the samples are drawn from the same population of d = 2.5, we get\n",
+      "z =  -5.16397779494\n",
+      "Since |z|>1.96, thus samples cannot be regarded as drawn from the same population\n"
+     ]
+    }
+   ],
+   "source": [
+    "print \"m1 and n1 represents mean and no. of objects in sample 1\"\n",
+    "print \"m2 and n2 represents mean and no. of objects in sample 2\"\n",
+    "m1 = 67.5\n",
+    "m2 = 68.\n",
+    "n1 = 1000.\n",
+    "n2 = 2000.\n",
+    "d = 2.5\n",
+    "print \"On the hypothesis that the samples are drawn from the same population of d = 2.5, we get\"\n",
+    "z = (m1-m2)/(d*((1/n1)+(1/n2))**0.5)\n",
+    "print \"z = \",z\n",
+    "print \"Since |z|>1.96, thus samples cannot be regarded as drawn from the same population\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 35.10, page no. 872"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 13,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "m1, d1 and n1 denotes mean, deviation and no.of objects in first sample \n",
+      "m2, d2 and n2 denotes mean, deviation and no.of objects in second sample\n",
+      "S. E. of the difference of the mean heights is  0.0703246933872\n",
+      "-0.7\n",
+      "|m1−m2|>10e, this is highly significant. hence, the data indicates that the sailors are on the average taller than the soldiers.\n"
+     ]
+    }
+   ],
+   "source": [
+    "print \"m1, d1 and n1 denotes mean, deviation and no.of objects in first sample \"\n",
+    "m1 = 67.85\n",
+    "d1 = 2.56\n",
+    "n1 = 6400.\n",
+    "print \"m2, d2 and n2 denotes mean, deviation and no.of objects in second sample\"\n",
+    "m2 = 68.55\n",
+    "d2 = 2.52\n",
+    "n2 = 1600.\n",
+    "print \"S. E. of the difference of the mean heights is \",\n",
+    "e = ((d**2/n1)+(d2**2/n2))**0.5\n",
+    "print e\n",
+    "print m1-m2\n",
+    "print \"|m1−m2|>10e, this is highly significant. hence, the data indicates that the sailors are on the average taller than the soldiers.\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 35.12, page no. 874"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 18,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "First of row denotes the different values of sample \n",
+      "The second row denotes the corresponding deviation\n",
+      "The third row denotes the corresponding square of deviation\n",
+      "The sum of second row elements =  10.0\n",
+      "The sum of third row elements =  66.0\n",
+      "let m be the mean \n",
+      "let d be the standard deviation\n",
+      "1.84522581263\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy\n",
+    "\n",
+    "A = numpy.zeros((3,9))\n",
+    "n = 9\n",
+    "print \"First of row denotes the different values of sample \"\n",
+    "A[0,:] = [45,47,50,52,48,47,49,53,51]\n",
+    "print \"The second row denotes the corresponding deviation\"\n",
+    "for i in range(0,9):\n",
+    "    A[1,i] = A[0,i]-48\n",
+    "print \"The third row denotes the corresponding square of deviation\"\n",
+    "for i in range(0,9):\n",
+    "    A[2,i] = A[1,i]**2\n",
+    "print \"The sum of second row elements = \",\n",
+    "a  =0\n",
+    "for i in range(0,9):\n",
+    "    a = a+A[1,i]\n",
+    "print a\n",
+    "print \"The sum of third row elements = \",\n",
+    "b = 0\n",
+    "for i in range(0,9):\n",
+    "    b = b+A[2,i]\n",
+    "print b\n",
+    "print \"let m be the mean \"\n",
+    "m = 48+a/n\n",
+    "print \"let d be the standard deviation\"\n",
+    "d = ((b/n)-(a/n)**2)**0.5\n",
+    "t = (m-47.5)*(n-1)**0.5/d\n",
+    "print t"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 35.13, page no. 876"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 19,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "d and n represents the deviation and no. objects in given sample\n",
+      "Taking the hypothesis that the product is not inferior i.e. there is no significant difference between m and M\n",
+      "3.15\n",
+      "Degrees of freedom= \n",
+      "9.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "print \"d and n represents the deviation and no. objects in given sample\"\n",
+    "n = 10.\n",
+    "d = 0.04\n",
+    "m = 0.742\n",
+    "M = 0.700\n",
+    "print \"Taking the hypothesis that the product is not inferior i.e. there is no significant difference between m and M\"\n",
+    "t = (m-M)*(n-1)**0.5/d\n",
+    "print t\n",
+    "print \"Degrees of freedom= \"\n",
+    "f = n-1\n",
+    "print f"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 34.15, page no. 878"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 21,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The first row denotes the boy no. \n",
+      "The second row denotes the marks in test I(x1)\n",
+      "The third row denotes the marks in test I(x2)\n",
+      "The fourth row denotes the difference of marks in two tests(d)\n",
+      "The fifth row denotes the (d−1)\n",
+      "The sixth row denotes the square of elements of fourth row \n",
+      "[[  1.   2.   3.   4.   5.   6.   7.   8.   9.  10.  11.]\n",
+      " [ 23.  20.  19.  21.  18.  20.  18.  17.  23.  16.  19.]\n",
+      " [ 24.  19.  22.  18.  20.  22.  20.  20.  23.  20.  17.]\n",
+      " [  1.  -1.   3.  -3.   2.   2.   2.   3.   0.   4.  -2.]\n",
+      " [  0.  -2.   2.  -4.   1.   1.   1.   2.  -1.   3.  -3.]\n",
+      " [  1.   1.   9.   9.   4.   4.   4.   9.   0.  16.   4.]]\n",
+      "The sum of elements of fourth row=\n",
+      "11.0\n",
+      "The sum of elements of sixth row= \n",
+      "61.0\n",
+      "Standard deviation\n",
+      "d =  2.46981780705\n",
+      "t =  1.48063606712\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy\n",
+    "\n",
+    "A = numpy.zeros((6,11))\n",
+    "n = 11\n",
+    "print \"The first row denotes the boy no. \"\n",
+    "A[0,:] = [1,2,3,4,5,6,7,8,9,10,11]\n",
+    "print \"The second row denotes the marks in test I(x1)\"\n",
+    "A[1,:] = [23,20,19,21,18,20,18,17,23,16,19]\n",
+    "print \"The third row denotes the marks in test I(x2)\"\n",
+    "A[2,:] = [24,19,22,18,20,22,20,20,23,20,17]\n",
+    "print \"The fourth row denotes the difference of marks in two tests(d)\"\n",
+    "for i in range (0,11):\n",
+    "    A[3,i] = A[2,i]-A[1,i]\n",
+    "print \"The fifth row denotes the (d−1)\"\n",
+    "for i in range (0,11):\n",
+    "    A[4,i] = A[3,i]-1\n",
+    "print \"The sixth row denotes the square of elements of fourth row \"\n",
+    "for i in range(0,11):\n",
+    "    A[5,i] = A[3,i]**2\n",
+    "print A\n",
+    "a = 0\n",
+    "print \"The sum of elements of fourth row=\"\n",
+    "for i in range(0,11):\n",
+    "    a = a+A[3,i]\n",
+    "print a\n",
+    "b = 0\n",
+    "print \"The sum of elements of sixth row= \"\n",
+    "for i in range(0,11):\n",
+    "    b = b + A[5,i]\n",
+    "print b\n",
+    "print \"Standard deviation\"\n",
+    "d = (b/(n-1))**0.5\n",
+    "t = (1-0)*(n)**0.5/2.24\n",
+    "print \"d = \",d\n",
+    "print \"t = \",t"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.10"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}