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1 |
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The second material exam. Question number one. a |
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2 |
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corporation randomly selected or selects 150 |
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3 |
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salespeople and finds that 66% who have never |
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4 |
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taken self-improvement course would like such a |
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5 |
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00:00:29,930 |
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course. So in this case, currently, they select |
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6 |
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150 salespeople and find that 66% would |
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7 |
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00:00:52,170 |
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like or who have never taken this course. The firm |
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8 |
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did a similar study 10 years ago in which 60% of a |
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9 |
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random sample of 160 salespeople wanted a self |
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10 |
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00:01:10,290 |
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-improvement course. |
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11 |
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They select a random sample of 160 and tell that |
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12 |
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60% would like to take this course. So we have |
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13 |
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here two information about previous study and |
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14 |
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currently. So currently we have this information. |
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15 |
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The sample size was 150, with a proportion 66% for |
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16 |
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the people who would like to attend or take this |
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17 |
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course. Mid-Paiwan and Pai Tu represent the true |
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18 |
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proportion, it means the population proportion, of |
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19 |
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workers who would like to attend a self |
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20 |
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-improvement course in the recent study and the |
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21 |
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past studies in Taiwan. So recent, Paiwan. |
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22 |
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00:02:07,740 |
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And Pi 2 is the previous study. This weather, this |
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23 |
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proportion has changed from the previous study by |
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24 |
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using two approaches. Critical value approach and |
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25 |
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B value approach. So here we are talking about Pi |
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26 |
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1 equals Pi 2. Since the problem says that The |
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27 |
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proportion has changed. You don't know the exact |
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28 |
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direction, either greater than or smaller than. So |
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29 |
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this one should be Y1 does not equal Y2. So step |
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30 |
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one, you have to state the appropriate null and |
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31 |
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alternative hypothesis. |
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32 |
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Second step, compute the value of the test |
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33 |
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statistic. In this case, your Z statistic should |
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34 |
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be P1 minus P2 minus Pi 1 minus Pi 2, under the |
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35 |
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square root of P dash 1 minus P dash times 1 over |
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36 |
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N1 plus 1 over N1. Now, P1 and P2 are given under |
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37 |
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the null hypothesis Pi 1 minus Pi 2 is 0. So here |
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38 |
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we have to compute P dash, which is the overall |
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39 |
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B dash equals x1 plus x2 divided by n1 plus n2. |
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40 |
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Now these x's, I mean the number of successes are |
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41 |
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not given directly in this problem, but we can |
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42 |
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figure out the values of x1 and x2 by using this |
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43 |
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information, which is n1 equals 150 and b1 equals |
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44 |
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66%. Because we know that b1 equals x1 over n1. |
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45 |
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So, by using this equation, X1 equals N1 times V1. |
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46 |
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N1 150 times 66 percent, that will give 150 times |
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47 |
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66, so that's 99. So 150 times, it's 99. |
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48 |
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Similarly, X2 equals N2 times V2. N2 is given by |
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49 |
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160, so 160 times 60 percent, |
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50 |
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96. So the number of successes are 96 for the |
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51 |
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second, for the previous. Nine nine. |
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52 |
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So B dash equals x199 |
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53 |
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plus 96 divided by n1 plus n2, 350. And that will |
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54 |
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give the overall proportions divided by 310, 0 |
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55 |
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.629. |
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56 |
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So, this is the value of the overall proportion. |
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57 |
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Now, B dash equals 1.629. So, 1 times 1 minus B |
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58 |
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dash is 1 minus this value times 1 over N1, 1 over |
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59 |
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150 plus 1 over 160. Simple calculation will give |
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60 |
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The value of z, which is in this case 1.093. |
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61 |
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So just plug this information into this equation, |
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62 |
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you will get z value, which is 1.093. He asked to |
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63 |
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do this problem by using two approaches, critical |
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64 |
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value and b value. Let's start with the first one, |
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65 |
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b value approach. |
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66 |
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Now your B value or critical value, start with |
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67 |
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critical value. |
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68 |
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00:06:40,850 --> 00:06:46,490 |
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Now since we are taking about a two-sided test, so |
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69 |
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there are two critical values which are plus or |
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70 |
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minus Z alpha over. Alpha is given by five |
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71 |
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00:06:54,670 --> 00:06:56,990 |
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percent, so in this case |
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72 |
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00:06:59,630 --> 00:07:03,370 |
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is equal to plus or minus 1.96. |
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73 |
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00:07:05,930 --> 00:07:10,010 |
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Now, does this value, I mean does the value of |
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74 |
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this statistic which is 1.093 fall in the critical |
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75 |
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00:07:14,910 --> 00:07:22,730 |
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region? Now, my critical regions are above 196 or |
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76 |
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00:07:22,730 --> 00:07:28,130 |
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below negative 1.96. Now this value actually falls |
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77 |
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00:07:29,300 --> 00:07:32,420 |
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In the non-rejection region, so we don't reject |
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78 |
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00:07:32,420 |
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the null hypothesis. So my decision, don't reject |
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79 |
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the null hypothesis. That means there is not |
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80 |
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sufficient evidence to support the alternative |
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81 |
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00:07:43,420 --> 00:07:46,960 |
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which states that the proportion has changed from |
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82 |
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the previous study. So we don't reject the null |
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83 |
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hypothesis. It means there is not sufficient |
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84 |
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evidence to support the alternative hypothesis. |
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85 |
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00:07:58,270 |
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That means you cannot say that the proportion has |
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86 |
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00:08:02,010 |
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changed from the previous study. That by using |
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87 |
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00:08:05,530 |
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critical value approach. Now what's about p-value? |
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88 |
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00:08:11,830 --> 00:08:16,170 |
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In order to determine the p-value, |
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89 |
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00:08:19,460 --> 00:08:23,320 |
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We have to find the probability that the Z |
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90 |
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statistic fall in the rejection regions. So that |
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91 |
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means Z greater than my values 1093 or |
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92 |
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Z smaller than negative 1.093. |
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93 |
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1093 is the same as the left of negative, so they |
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94 |
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00:08:49,730 --> 00:08:52,810 |
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are the same because of symmetry. So just take 1 |
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95 |
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and multiply by 2. |
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96 |
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00:08:58,430 --> 00:09:03,070 |
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Now simple calculation will give the value of 0 |
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97 |
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00:09:03,070 --> 00:09:09,950 |
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.276 in chapter 6. So go back to chapter 6 to |
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98 |
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figure out how can we calculate the probability of |
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99 |
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00:09:13,290 --> 00:09:19,830 |
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Z greater than 1.0938. Now my B value is 0.276, |
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100 |
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00:09:20,030 --> 00:09:25,190 |
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always we reject the null hypothesis if my B value |
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101 |
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00:09:25,190 --> 00:09:29,050 |
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is smaller than alpha. Now this value is much much |
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102 |
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00:09:29,050 --> 00:09:31,210 |
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bigger than alpha, so we don't reject the null |
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103 |
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00:09:31,210 |
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hypothesis. So since my B value is much greater |
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104 |
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00:09:36,710 |
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than alpha, that means we don't reject the null |
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105 |
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00:09:42,650 --> 00:09:46,810 |
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hypothesis, so we reach the same conclusion, that |
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106 |
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00:09:46,810 --> 00:09:49,270 |
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there is not sufficient evidence to support the |
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107 |
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alternative. Also, we can perform the test by |
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108 |
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using confidence interval approach, because here |
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109 |
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00:09:59,810 --> 00:10:02,850 |
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we are talking about two-tailed test. Your |
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110 |
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00:10:02,850 --> 00:10:06,670 |
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confidence interval is given by |
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111 |
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00:10:10,620 --> 00:10:17,280 |
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B1 minus B2 plus |
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112 |
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00:10:17,280 --> 00:10:23,720 |
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or minus Z alpha over 2 times B |
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113 |
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00:10:23,720 --> 00:10:30,120 |
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dash 1 minus B dash multiplied by 1 over N1 plus 1 |
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114 |
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00:10:30,120 --> 00:10:37,520 |
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over N2. By the way, this one |
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115 |
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00:10:37,520 --> 00:10:43,320 |
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called the margin of error. So z times square root |
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116 |
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00:10:43,320 --> 00:10:45,940 |
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of this sequence is called the margin of error, |
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117 |
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00:10:46,940 --> 00:10:52,280 |
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and the square root itself is called the standard |
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118 |
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00:10:52,280 --> 00:10:59,560 |
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error of the point estimate of pi 1 minus pi 2, |
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119 |
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00:10:59,720 --> 00:11:04,430 |
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which is P1 minus P2. So square root of b dash 1 |
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120 |
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00:11:04,430 --> 00:11:07,650 |
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minus b dash multiplied by 1 over n1 plus 1 over |
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121 |
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00:11:07,650 --> 00:11:12,270 |
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n2 is called the standard error of the estimate of |
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122 |
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00:11:12,270 --> 00:11:15,910 |
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pi 1 minus pi 2. So this is standard estimate of |
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123 |
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00:11:15,910 --> 00:11:21,750 |
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b1 minus b2. Simply, you will get the confidence |
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124 |
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00:11:21,750 --> 00:11:26,470 |
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interval to be between pi 1 minus the difference |
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125 |
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00:11:26,470 --> 00:11:32,620 |
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between the two proportions, 4 between negative. 0 |
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126 |
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00:11:32,620 --> 00:11:37,160 |
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.5 and |
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127 |
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00:11:37,160 --> 00:11:38,940 |
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0.7. |
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128 |
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00:11:44,060 --> 00:11:48,400 |
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Now this interval actually contains |
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129 |
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00:11:50,230 --> 00:11:54,250 |
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The value of 0, that means we don't reject the |
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130 |
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00:11:54,250 |
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null hypothesis. So since this interval starts |
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131 |
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00:11:57,570 |
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from negative, lower bound is negative 0.5, upper |
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132 |
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00:12:01,870 |
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bound is 0.17, that means 0 inside this interval, |
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133 |
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00:12:06,750 |
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I mean the confidence captures the value of 0, |
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134 |
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00:12:09,610 |
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that means we don't reject the null hypothesis. So |
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135 |
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00:12:13,810 --> 00:12:17,110 |
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by using three different approaches, we end with |
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136 |
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00:12:17,110 --> 00:12:20,930 |
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the same decision and conclusion. That is, we |
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137 |
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00:12:20,930 --> 00:12:25,370 |
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don't reject null hypotheses. That's all for |
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138 |
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00:12:25,370 --> 00:12:26,110 |
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number one. |
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139 |
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00:12:31,450 --> 00:12:32,910 |
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Question number two. |
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140 |
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00:12:36,170 --> 00:12:40,450 |
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The excellent drug company claims its aspirin |
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141 |
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00:12:40,450 --> 00:12:43,610 |
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tablets will relieve headaches faster than any |
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142 |
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00:12:43,610 --> 00:12:47,470 |
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other aspirin on the market. So they believe that |
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143 |
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00:12:48,440 --> 00:12:52,220 |
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Their drug is better than the other drug in the |
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144 |
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00:12:52,220 --> 00:12:57,180 |
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market. To determine whether Excellence claim is |
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145 |
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00:12:57,180 --> 00:13:04,260 |
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valid, random samples of size 15 are chosen from |
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146 |
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00:13:04,260 --> 00:13:07,080 |
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aspirins made by Excellence and the sample drug |
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147 |
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00:13:07,080 --> 00:13:12,300 |
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combined. So sample sizes of 15 are chosen from |
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148 |
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00:13:12,300 --> 00:13:16,260 |
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each. So that means N1 equals 15 and N2 also |
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149 |
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00:13:16,260 --> 00:13:21,160 |
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equals 15. And aspirin is given to each of the 30 |
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150 |
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00:13:21,160 --> 00:13:23,520 |
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randomly selected persons suffering from |
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151 |
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00:13:23,520 --> 00:13:27,220 |
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headaches. So the total sample size is 30, because |
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152 |
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00:13:27,220 --> 00:13:30,780 |
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15 from the first company, and the second for the |
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153 |
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00:13:30,780 --> 00:13:36,860 |
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simple company. So they are 30 selected persons |
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154 |
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00:13:36,860 --> 00:13:40,280 |
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who are suffering from headaches. So we have |
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155 |
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00:13:40,280 --> 00:13:43,380 |
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information about number of minutes required for |
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156 |
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00:13:43,380 --> 00:13:47,720 |
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each to recover from the headache. is recorded, |
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157 |
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00:13:48,200 --> 00:13:51,500 |
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the sample results are. So here we have two |
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158 |
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00:13:51,500 --> 00:13:56,260 |
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groups, two populations. Company is called |
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159 |
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00:13:56,260 --> 00:13:58,420 |
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excellent company and other one simple company. |
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160 |
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00:13:59,120 --> 00:14:04,320 |
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The information we have, the sample means are 8.4 |
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161 |
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00:14:04,320 --> 00:14:08,260 |
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for the excellent and 8.9 for the simple company. |
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162 |
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00:14:09,040 --> 00:14:13,280 |
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With the standard deviations for the sample are 2 |
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163 |
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00:14:13,280 --> 00:14:18,340 |
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.05 and 2.14 respectively for excellent and simple |
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164 |
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00:14:18,340 --> 00:14:21,480 |
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and as we mentioned the sample sizes are the same |
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165 |
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00:14:21,480 --> 00:14:26,380 |
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are equal 15 and 15. Now we are going to test at |
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166 |
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00:14:26,380 --> 00:14:32,540 |
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five percent level of significance test whether to |
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167 |
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00:14:32,540 --> 00:14:35,560 |
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determine whether excellence aspirin cure |
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168 |
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00:14:35,560 --> 00:14:39,140 |
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headaches significantly faster than simple |
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169 |
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00:14:39,140 --> 00:14:46,420 |
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aspirin. Now faster it means Better. Better it |
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170 |
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00:14:46,420 --> 00:14:49,480 |
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means the time required to relieve headache is |
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171 |
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00:14:49,480 --> 00:14:53,920 |
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smaller there. So you have to be careful in this |
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172 |
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00:14:53,920 --> 00:15:00,800 |
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case. If we assume that Mu1 is the mean time |
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|
173 |
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00:15:00,800 --> 00:15:05,120 |
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required for excellent aspirin. So Mu1 for |
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174 |
|
00:15:05,120 --> 00:15:05,500 |
|
excellent. |
|
|
|
175 |
|
00:15:17,260 --> 00:15:21,540 |
|
So Me1, mean time required for excellence aspirin, |
|
|
|
176 |
|
00:15:22,780 --> 00:15:28,860 |
|
and Me2, mean time required for simple aspirin. So |
|
|
|
177 |
|
00:15:28,860 --> 00:15:32,760 |
|
each one, Me1, is smaller than Me3. |
|
|
|
178 |
|
00:15:41,140 --> 00:15:45,960 |
|
Since Me1 represents the time required to relieve |
|
|
|
179 |
|
00:15:45,960 --> 00:15:51,500 |
|
headache by using excellent aspirin and this one |
|
|
|
180 |
|
00:15:51,500 --> 00:15:55,460 |
|
is faster faster it means it takes less time in |
|
|
|
181 |
|
00:15:55,460 --> 00:15:59,620 |
|
order to recover from headache so mu1 should be |
|
|
|
182 |
|
00:15:59,620 --> 00:16:06,400 |
|
smaller than mu2 we are going to use T T is x1 bar |
|
|
|
183 |
|
00:16:06,400 --> 00:16:11,380 |
|
minus x2 bar minus the difference between the two |
|
|
|
184 |
|
00:16:11,380 --> 00:16:14,720 |
|
population proportions divided by |
|
|
|
185 |
|
00:16:17,550 --> 00:16:22,070 |
|
S squared B times 1 over N1 plus 1 over N2. |
|
|
|
186 |
|
00:16:25,130 --> 00:16:30,470 |
|
S squared B N1 |
|
|
|
187 |
|
00:16:30,470 --> 00:16:35,330 |
|
minus 1 S1 squared plus N2 minus 1 S2 squared |
|
|
|
188 |
|
00:16:35,330 --> 00:16:41,990 |
|
divided by N1 plus N2 minus 1. Now, a simple |
|
|
|
189 |
|
00:16:41,990 --> 00:16:44,030 |
|
calculation will give the following results. |
|
|
|
190 |
|
00:16:59,660 --> 00:17:03,080 |
|
So again, we have this data. Just plug this |
|
|
|
191 |
|
00:17:03,080 --> 00:17:06,620 |
|
information here to get the value of S square B. |
|
|
|
192 |
|
00:17:07,740 --> 00:17:13,120 |
|
And finally, you will end with this result. |
|
|
|
193 |
|
00:17:18,220 --> 00:17:24,920 |
|
S squared B equals 2 |
|
|
|
194 |
|
00:17:24,920 --> 00:17:27,240 |
|
.095 squared. |
|
|
|
195 |
|
00:17:30,140 --> 00:17:35,920 |
|
Your T statistic equals negative |
|
|
|
196 |
|
00:17:42,790 --> 00:17:48,370 |
|
So that's your T-statistic value. So just plug the |
|
|
|
197 |
|
00:17:48,370 |
|
values in 1 and 2, this 1 squared and this 2 |
|
|
|
198 |
|
00:17:51,210 |
|
squared into this equation, you will get this |
|
|
|
199 |
|
00:17:53,350 |
|
value. So 2.059 squared, that is 4.239. |
|
|
|
200 |
|
00:18:07,670 |
|
Here you can use either the critical value |
|
|
|
201 |
|
00:18:10,690 |
|
approach, Or B value. Let's do a critical value. |
|
|
|
202 |
|
00:18:21,920 --> 00:18:27,460 |
|
Since the alternative is the lower tail, one-sided |
|
|
|
203 |
|
00:18:27,460 --> 00:18:31,820 |
|
lower tail, so your B value, your critical value |
|
|
|
204 |
|
00:18:31,820 --> 00:18:37,630 |
|
is negative, T alpha, and there is a freedom. So |
|
|
|
205 |
|
00:18:37,630 --> 00:18:47,630 |
|
this is equal to negative T, 5% with 28 degrees of |
|
|
|
206 |
|
00:18:47,630 --> 00:18:55,270 |
|
freedom. By using the table you have 28, |
|
|
|
207 |
|
00:18:56,030 --> 00:19:00,070 |
|
28 |
|
|
|
208 |
|
00:19:00,070 --> 00:19:12,790 |
|
under 5%, so 28 under 5%, so |
|
|
|
209 |
|
00:19:12,790 --> 00:19:20,870 |
|
1.701, negative 1.701. |
|
|
|
210 |
|
00:19:23,750 --> 00:19:28,290 |
|
Now, we reject the null hypothesis if |
|
|
|
211 |
|
00:19:33,770 --> 00:19:42,890 |
|
region. Now again, since it's lower TL, so your |
|
|
|
212 |
|
00:19:42,890 |
|
rejection region is below negative 1.701. |
|
|
|
213 |
|
00:19:51,230 |
|
Now, does this value fall in the rejection region? |
|
|
|
214 |
|
00:19:56,510 |
|
It falls in the non-rejection region. So the |
|
|
|
215 |
|
00:20:02,350 |
|
answer is Don't reject the null hypothesis. That |
|
|
|
216 |
|
00:20:08,040 --> 00:20:11,380 |
|
means we don't have sufficient evidence to support |
|
|
|
217 |
|
00:20:11,380 |
|
the excellent drug company claim which states that |
|
|
|
218 |
|
00:20:16,300 |
|
their aspirin tablets relieve headaches faster |
|
|
|
219 |
|
00:20:21,380 |
|
than the simple one. So that's by using a critical |
|
|
|
220 |
|
00:20:28,540 --> 00:20:33,230 |
|
value approach because this value falls in the non |
|
|
|
221 |
|
00:20:33,230 --> 00:20:36,450 |
|
-rejection region, so we don't reject the null |
|
|
|
222 |
|
00:20:36,450 |
|
hypothesis. |
|
|
|
223 |
|
00:20:44,130 |
|
Or you maybe use the B-value approach. |
|
|
|
224 |
|
00:20:53,070 |
|
Now, since the alternative is µ1 smaller than µ2, |
|
|
|
225 |
|
00:20:57,640 |
|
So B value is probability of T smaller than |
|
|
|
226 |
|
00:21:03,260 |
|
negative 0 |
|
|
|
227 |
|
00:21:08,820 |
|
.653. |
|
|
|
228 |
|
00:21:14,300 |
|
So we are looking for this probability B of Z |
|
|
|
229 |
|
00:21:18,420 |
|
smaller than negative 0.653. |
|
|
|
230 |
|
00:21:23,210 |
|
The table you have gives the area in the upper |
|
|
|
231 |
|
00:21:27,050 |
|
tail. So this is the same as beauty greater than. |
|
|
|
232 |
|
00:21:37,790 |
|
Because the area to the right of 0.653 is the same |
|
|
|
233 |
|
00:21:44,350 |
|
as the area to the left of negative 0.75. Because |
|
|
|
234 |
|
00:21:48,070 |
|
of symmetry. Just look at the tea table. Now, |
|
|
|
235 |
|
00:21:53,070 |
|
smaller than negative, means this area is actually |
|
|
|
236 |
|
00:22:00,810 |
|
the same as the area to the right of the same |
|
|
|
237 |
|
00:22:02,690 |
|
value, but on the other side. So these two areas |
|
|
|
238 |
|
00:22:07,330 |
|
are the same. So it's the same as D of T greater |
|
|
|
239 |
|
00:22:11,890 --> 00:22:17,710 |
|
than 0.653. If you look at the table for 28 |
|
|
|
240 |
|
00:22:17,710 --> 00:22:19,150 |
|
degrees of freedom, |
|
|
|
241 |
|
00:22:22,300 --> 00:22:23,520 |
|
That's your 28. |
|
|
|
242 |
|
00:22:27,580 |
|
I am looking for the value of 0.653. The first |
|
|
|
243 |
|
00:22:32,720 |
|
value here is 0.683. The other one is 0.8. It |
|
|
|
244 |
|
00:22:38,420 |
|
means my value is below this one. If you go back |
|
|
|
245 |
|
00:22:43,600 |
|
here, |
|
|
|
246 |
|
00:22:46,700 |
|
so it should be to the left of this value. Now |
|
|
|
247 |
|
00:22:52,610 |
|
here 25, then 20, 20, 15 and so on. So it should |
|
|
|
248 |
|
00:22:57,170 |
|
be greater than 25. So your B value actually is |
|
|
|
249 |
|
00:23:01,930 |
|
greater than 25%. As we mentioned before, T table |
|
|
|
250 |
|
00:23:08,570 |
|
does not give the exact B value. So approximately |
|
|
|
251 |
|
00:23:12,010 |
|
my B value is greater than 25%. This value |
|
|
|
252 |
|
00:23:17,290 |
|
actually is much bigger than 5%. So again, we |
|
|
|
253 |
|
00:23:22,400 |
|
reject, we don't reject the null hypothesis. So |
|
|
|
254 |
|
00:23:27,480 --> 00:23:30,600 |
|
again, to compute the B value, it's probability of |
|
|
|
255 |
|
00:23:30,600 |
|
T smaller than the value of the statistic, which |
|
|
|
256 |
|
00:23:37,320 |
|
is negative 0.653. The table you have gives the |
|
|
|
257 |
|
00:23:42,040 |
|
area to the right. |
|
|
|
258 |
|
00:23:46,980 |
|
So this probability is the same as B of T greater |
|
|
|
259 |
|
00:23:50,700 |
|
than 0.653. So by using this table, you will get |
|
|
|
260 |
|
00:23:55,920 |
|
approximate value of B, which is greater than 25%. |
|
|
|
261 |
|
00:24:00,100 |
|
Always, as we mentioned, we reject the null |
|
|
|
262 |
|
00:24:02,960 |
|
hypothesis if my B value is smaller than alpha. In |
|
|
|
263 |
|
00:24:06,660 |
|
this case, this value is greater than alpha, so we |
|
|
|
264 |
|
00:24:08,920 |
|
don't reject the null. So we reach the same |
|
|
|
265 |
|
00:24:11,480 --> 00:24:15,640 |
|
decision as by using the critical value approach. |
|
|
|
266 |
|
00:24:17,040 --> 00:24:23,360 |
|
Any question? So that's for number two. Question |
|
|
|
267 |
|
00:24:23,360 |
|
number three. |
|
|
|
268 |
|
00:24:32,120 |
|
To test the effectiveness of a business school |
|
|
|
269 |
|
00:24:35,820 |
|
preparation course, eight students took a general |
|
|
|
270 |
|
00:24:41,640 |
|
business test before and after the course. Let X1 |
|
|
|
271 |
|
00:24:47,210 |
|
denote before, |
|
|
|
272 |
|
00:24:53,010 |
|
and X2 after. |
|
|
|
273 |
|
00:24:59,630 |
|
And the difference is X2 minus X1. |
|
|
|
274 |
|
00:25:14,780 |
|
The mean of the difference equals 50. And the |
|
|
|
275 |
|
00:25:19,540 |
|
standard deviation of the difference is 65.03. So |
|
|
|
276 |
|
00:25:25,540 |
|
sample statistics are sample mean for the |
|
|
|
277 |
|
00:25:28,900 |
|
difference and sample standard deviation of the |
|
|
|
278 |
|
00:25:32,040 |
|
difference. So these two values are given. Test to |
|
|
|
279 |
|
00:25:36,860 |
|
determine the effectiveness of a business school |
|
|
|
280 |
|
00:25:40,200 |
|
preparation course. So what's your goal? An |
|
|
|
281 |
|
00:25:45,960 --> 00:25:48,120 |
|
alternative, null equals zero. An alternative |
|
|
|
282 |
|
00:25:48,120 --> 00:25:52,340 |
|
should |
|
|
|
283 |
|
00:25:52,340 --> 00:25:58,360 |
|
be greater than zero. Because D is X2 minus X1. So |
|
|
|
284 |
|
00:25:58,360 --> 00:26:02,840 |
|
effective, it means after is better than before. |
|
|
|
285 |
|
00:26:03,680 --> 00:26:08,420 |
|
So my score after taking the course is better than |
|
|
|
286 |
|
00:26:08,420 --> 00:26:12,080 |
|
before taking the course. So X in UD is positive. |
|
|
|
287 |
|
00:26:19,090 --> 00:26:27,510 |
|
T is D bar minus 0 divided by SD over square root |
|
|
|
288 |
|
00:26:27,510 --> 00:26:41,090 |
|
of A. D bar is 50 divided by 65 divided |
|
|
|
289 |
|
00:26:41,090 --> 00:26:54,490 |
|
by Square root of 8. So 50 divided by square |
|
|
|
290 |
|
00:26:54,490 --> 00:26:57,910 |
|
root of 8, 2.17. |
|
|
|
291 |
|
00:27:04,070 --> 00:27:09,570 |
|
Now Yumi used the critical value approach. So my |
|
|
|
292 |
|
00:27:09,570 --> 00:27:10,930 |
|
critical value is T alpha. |
|
|
|
293 |
|
00:27:13,680 --> 00:27:20,140 |
|
And degrees of freedom is 7. It's upper 10. So |
|
|
|
294 |
|
00:27:20,140 |
|
it's plus. So it's T alpha 0, 5. And DF is 7, |
|
|
|
295 |
|
00:27:27,320 |
|
because N equals 8. Now by using the table, at 7 |
|
|
|
296 |
|
00:27:33,820 |
|
degrees of freedom, |
|
|
|
297 |
|
00:27:38,220 |
|
so at 7, |
|
|
|
298 |
|
00:27:53,560 |
|
So my T value is greater than the |
|
|
|
299 |
|
00:28:03,380 |
|
critical region, so we reject the null hypothesis. |
|
|
|
300 |
|
00:28:10,740 |
|
The rejection region starts from 1.9895 and this |
|
|
|
301 |
|
00:28:17,700 |
|
value actually greater than 1.8. So since it falls |
|
|
|
302 |
|
00:28:24,800 |
|
in the rejection region, then we reject the null |
|
|
|
303 |
|
00:28:30,320 |
|
hypothesis. It means that taking the course, |
|
|
|
304 |
|
00:28:36,370 |
|
improves your score. So we have sufficient |
|
|
|
305 |
|
00:28:39,690 |
|
evidence to support the alternative hypothesis. |
|
|
|
306 |
|
00:28:44,330 |
|
That's for number three. The other part, the other |
|
|
|
307 |
|
00:28:50,650 --> 00:28:51,130 |
|
part. |
|
|
|
308 |
|
00:28:54,290 --> 00:28:58,550 |
|
A statistician selected a sample of 16 receivable |
|
|
|
309 |
|
00:28:58,550 --> 00:29:03,530 |
|
accounts. He reported that the sample information |
|
|
|
310 |
|
00:29:04,690 --> 00:29:07,790 |
|
indicated the mean of the population ranges from |
|
|
|
311 |
|
00:29:07,790 --> 00:29:12,730 |
|
these two values. So we have lower and upper |
|
|
|
312 |
|
00:29:12,730 --> 00:29:21,910 |
|
limits, which are given by 4739. |
|
|
|
313 |
|
00:29:36,500 --> 00:29:42,400 |
|
So the mean of the population ranges between these |
|
|
|
314 |
|
00:29:42,400 --> 00:29:47,880 |
|
two values. And in addition to that, we have |
|
|
|
315 |
|
00:29:47,880 --> 00:29:55,920 |
|
information about the sample standard deviation is |
|
|
|
316 |
|
00:29:55,920 --> 00:29:56,340 |
|
400. |
|
|
|
317 |
|
00:29:59,500 --> 00:30:03,260 |
|
The statistician neglected to report what |
|
|
|
318 |
|
00:30:03,260 --> 00:30:07,440 |
|
confidence level he had used. So we don't know C |
|
|
|
319 |
|
00:30:07,440 |
|
level. So C level is unknown, which actually is 1 |
|
|
|
320 |
|
00:30:14,180 |
|
minus alpha. |
|
|
|
321 |
|
00:30:20,980 |
|
Based on the above information, what's the |
|
|
|
322 |
|
00:30:25,360 --> 00:30:28,380 |
|
confidence level? So we are looking for C level. |
|
|
|
323 |
|
00:30:29,380 --> 00:30:34,160 |
|
Now just keep in mind the confidence interval is |
|
|
|
324 |
|
00:30:34,160 --> 00:30:38,200 |
|
given and we are looking for C level. |
|
|
|
325 |
|
00:30:42,920 --> 00:30:46,600 |
|
So this area actually is alpha over 2 and other |
|
|
|
326 |
|
00:30:46,600 --> 00:30:49,940 |
|
one is alpha over 2, so the area between is 1 |
|
|
|
327 |
|
00:30:49,940 --> 00:30:50,440 |
|
minus alpha. |
|
|
|
328 |
|
00:30:53,340 --> 00:30:58,620 |
|
Now since the sample size equal |
|
|
|
329 |
|
00:31:01,950 --> 00:31:10,010 |
|
16, N equals 16, so N equals 16, so your |
|
|
|
330 |
|
00:31:10,010 --> 00:31:12,490 |
|
confidence interval should be X bar plus or minus |
|
|
|
331 |
|
00:31:12,490 --> 00:31:14,610 |
|
T, S over root N. |
|
|
|
332 |
|
00:31:19,350 --> 00:31:26,390 |
|
Now, C level can be determined by T, and we know |
|
|
|
333 |
|
00:31:26,390 --> 00:31:28,130 |
|
that this quantity, |
|
|
|
334 |
|
00:31:30,730 --> 00:31:36,970 |
|
represents the margin of error. So, E equals TS |
|
|
|
335 |
|
00:31:36,970 --> 00:31:42,950 |
|
over root N. Now, since the confidence interval is |
|
|
|
336 |
|
00:31:42,950 --> 00:31:50,270 |
|
given, we know from previous chapters that the |
|
|
|
337 |
|
00:31:50,270 --> 00:31:53,970 |
|
margin equals the difference between upper and |
|
|
|
338 |
|
00:31:53,970 --> 00:31:59,560 |
|
lower divided by two. So, half distance of lower |
|
|
|
339 |
|
00:31:59,560 --> 00:32:06,320 |
|
and upper gives the margin. So that will give 260 |
|
|
|
340 |
|
00:32:06,320 --> 00:32:17,620 |
|
.2. So that's E. So now E is known to be 260.2 |
|
|
|
341 |
|
00:32:17,620 |
|
equals to S is given by 400 and N is 16. |
|
|
|
342 |
|
00:32:26,800 |
|
Now, simple calculation will give the value of T, |
|
|
|
343 |
|
00:32:30,060 |
|
which is the critical value. |
|
|
|
344 |
|
00:32:35,280 |
|
So, my T equals 2.60. |
|
|
|
345 |
|
00:32:41,960 |
|
Actually, this is T alpha over 2. Now, the value |
|
|
|
346 |
|
00:32:47,220 |
|
of the critical value is known to be 2.602. What's |
|
|
|
347 |
|
00:32:52,400 --> 00:32:56,520 |
|
the corresponding alpha over 2? Now look at the |
|
|
|
348 |
|
00:32:56,520 --> 00:32:59,660 |
|
table, at 15 degrees of freedom, |
|
|
|
349 |
|
00:33:02,720 --> 00:33:10,680 |
|
look at 15, at this value 2.602, at this value. |
|
|
|
350 |
|
00:33:12,640 --> 00:33:19,880 |
|
So, 15 degrees of freedom, 2.602, so the |
|
|
|
351 |
|
00:33:19,880 --> 00:33:21,940 |
|
corresponding alpha over 2, not alpha. |
|
|
|
352 |
|
00:33:24,610 --> 00:33:31,830 |
|
it's 1% so my alpha over 2 is |
|
|
|
353 |
|
00:33:31,830 |
|
1% so alpha is 2% so the confidence level is 1 |
|
|
|
354 |
|
00:33:43,110 |
|
minus alpha so 1 minus alpha is 90% so c level is |
|
|
|
355 |
|
00:33:50,510 |
|
98% so that's level or the confidence level. So |
|
|
|
356 |
|
00:33:59,410 --> 00:34:03,990 |
|
again, maybe this is a tricky question. |
|
|
|
357 |
|
00:34:07,330 --> 00:34:10,530 |
|
But at least you know that if the confidence |
|
|
|
358 |
|
00:34:10,530 --> 00:34:15,270 |
|
interval is given, you can determine the margin of |
|
|
|
359 |
|
00:34:15,270 --> 00:34:18,930 |
|
error by the difference between lower and upper |
|
|
|
360 |
|
00:34:18,930 --> 00:34:23,310 |
|
divided by two. Then we know this term represents |
|
|
|
361 |
|
00:34:23,310 --> 00:34:27,150 |
|
this margin. So by using this equation, we can |
|
|
|
362 |
|
00:34:27,150 --> 00:34:29,770 |
|
compute the value of T, I mean the critical value. |
|
|
|
363 |
|
00:34:30,670 --> 00:34:35,290 |
|
So since the critical value is given or is |
|
|
|
364 |
|
00:34:35,290 --> 00:34:38,590 |
|
computed, we can determine the corresponding alpha |
|
|
|
365 |
|
00:34:38,590 --> 00:34:45,390 |
|
over 2. So alpha over 2 is 1%. So your alpha is |
|
|
|
366 |
|
00:34:45,390 --> 00:34:51,710 |
|
2%. So my C level is 98%. That's |
|
|
|
367 |
|
00:34:51,710 |
|
all. Any questions? We're done, Muhammad. |
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|
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|