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Last time, we talked about the types of samples |
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and introduces two |
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types of samples. One is called non-probability |
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samples, and the other one is probability samples. |
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And also, we have discussed two types of non |
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-probability, which are judgment and convenience. |
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For the product samples also we produced four |
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types, random sample, systematic, stratified and |
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clustered sampling. That was last Sunday. Let's |
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see the comparison between these sampling data. A |
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simple, random sample, systematic random sample, |
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first, for these two techniques. First of all, |
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they are simple to use because we just use the |
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random tables, random number tables, or by using |
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any statistical software. But the disadvantage of |
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this technique |
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So it might be this sample is not representative |
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of the entire population. So this is the mainly |
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disadvantage of this sampling technique. So it can |
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be used unless the population is not symmetric or |
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the population is not heterogeneous. I mean if the |
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population has the same characteristics, then we |
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can use simple or systematic sample. But if there |
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are big differences or big disturbances between |
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the items of the population, I mean between or |
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among the individuals. In this case, stratified |
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sampling is better than using a simple random |
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sample. Stratified samples ensure representation |
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of individuals across the entire population. If |
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you remember last time we said A IUG population |
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can be splitted according to gender, either males |
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or females, or can be splitted according to |
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students' levels. First level, second level and |
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fourth level, and so on. The last type of sampling |
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was clusters. Cluster sampling is more cost |
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effective. Because in this case, you have to split |
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the population into many clusters, then you can |
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choose a random of these clusters. Also, it's less |
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efficient unless you use a large sample. For this |
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reason, it's more cost effective than using the |
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other sampling techniques. So, these techniques |
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are used based on the study you have. Sometimes |
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simple random sampling is fine and you can go |
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ahead and use it. Most of the time, stratified |
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random sampling is much better. So, it depends on |
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the population you have underlying your study. |
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That was what we talked about last Sunday. |
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Now, suppose we design a questionnaire or survey. |
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You have to know, number one, what's the purpose |
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50 |
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of the survey. In this case, you can determine the |
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frame of the population. Next, |
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survey |
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Is the survey based on a probability sample? If |
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the answer is yes, then go ahead and use one of |
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the non-probability sampling techniques either |
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similar than some certified cluster or systematic. |
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57 |
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Next, we have to distinguish between four types of |
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errors, at least now. One is called coverage |
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error. You have to ask yourself, is the frame |
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appropriate? I mean, frame appropriate means that |
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you have all the individual list, then you can |
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choose one of these. For example, suppose we |
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divide Gaza Strip into four governorates. North |
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Gaza, Gaza Middle Area, Khanon and Rafah. So we |
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have five. sections of five governments. In this |
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66 |
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case if you, so that's your frame. Now if you |
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67 |
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exclude one, for example, and that one is |
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important for you, but you exclude it for some |
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69 |
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reasons, in this case you will have coverage as |
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70 |
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well, because you excluded. one group out of five |
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and that group may be important for your study. |
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72 |
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Next is called non-response error. Suppose I |
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73 |
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attributed my questionnaire for 100 students and I |
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gave each one 30 minutes to answer the |
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questionnaire or to fill up the questionnaire, but |
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I didn't follow up. the response in this case it |
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might be you will get something error and that |
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78 |
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error refers to non-responsive so you have to |
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79 |
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follow up follow up it means maybe sometimes you |
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80 |
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need to clarify the question you have in your |
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81 |
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questionnaire so that the respondent understand |
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82 |
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what do you mean exactly by that question |
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83 |
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otherwise if you don't follow up it means it may |
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84 |
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be there is an error, and that error is called non |
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85 |
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-response. The other type of error is called |
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86 |
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measurement error, which is one of the most |
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87 |
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important errors, and we have to avoid. |
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88 |
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It's called measurement error. Good questions |
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89 |
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elicit good responses. It means suppose, for |
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90 |
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example, my question is, I feel this candidate is |
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91 |
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good for us. What do you think? It's my question. |
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92 |
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I feel this candidate, candidate A, whatever he |
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93 |
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is, is good for us. What do you think? For sure |
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94 |
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there's abundant answer will be yes. I agree with |
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95 |
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you. So that means you design the question in the |
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96 |
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way that you will know they respond directly that |
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97 |
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he will answer yes or no depends on your design of |
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98 |
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the question. So it means leading question. So |
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99 |
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measurement error. So but if we have good |
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100 |
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questions, just ask any question for the |
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101 |
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respondent and let him or let his answer based on |
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102 |
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what exactly he thinks about it. So don't force |
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103 |
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the respondent to answer the question in the |
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104 |
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direction you want to be. Otherwise you will get |
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105 |
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something called Measurement Error. Do you think? |
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106 |
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Give me an example of Measurement Error. |
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107 |
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Give me an example of Measurement Error. Just ask |
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108 |
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a question in a way that the respondent will |
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109 |
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answer I mean his answer will be the same as you |
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110 |
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think about it. |
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111 |
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Maybe I like coffee, do you like coffee or tea? So |
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112 |
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maybe he will go with your answer, in this case |
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113 |
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it's measurement. Another example. |
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114 |
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Exactly. |
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115 |
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So it means that if you design a question in the |
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116 |
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way that you will get the same answer you think |
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117 |
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about it, It means that you will have something |
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118 |
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called measurement error. The last type is |
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119 |
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sampling error. Sampling error always happens, |
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120 |
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always exists. For example, suppose you are around |
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121 |
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00:09:29,990 --> 00:09:33,150 |
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50 students in this class. Suppose I select |
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122 |
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randomly 20 of you. And I am interested suppose in |
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123 |
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your age. Maybe for this sample. |
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124 |
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I will get an average of your age of 19 years |
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125 |
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someone |
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126 |
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select another sample from the same population |
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127 |
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with the same size maybe |
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128 |
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the average of your age is not equal to 19 years |
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129 |
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maybe 19 years 3 months |
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130 |
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Someone else maybe also select the same number of |
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131 |
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students, but the average of the class might be 20 |
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132 |
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years. So the first one, second tier, each of them |
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133 |
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has different sample statistics, I mean different |
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134 |
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sample means. This difference or this error |
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135 |
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actually is called sampling error and always |
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136 |
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happens. So now we have five types of errors. One |
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137 |
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is called coverage error. In this case, you have |
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138 |
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problem with the frame. The other type is called |
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139 |
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non-response error. It means you have problem with |
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140 |
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following up. Measurement error, it means you have |
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141 |
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Bad questionnaire design, the last type is called |
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142 |
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semi-error and this one always happens and |
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143 |
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actually we would like to have this error, I mean |
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144 |
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this semi-error as small as possible. So these are |
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145 |
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the steps you have to follow up when you design |
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146 |
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the questionnaire. |
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147 |
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So again, for these types of errors, the first one |
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148 |
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coverage error or selection bias. This type of |
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149 |
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error exists if some groups are excluded from the |
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150 |
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frame and have no chance of being selected. That's |
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151 |
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the first type of error, coverage error. So it |
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152 |
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means there is a problem. on the population frame. |
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153 |
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Non-response error bias, it means people who don't |
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154 |
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respond may be different from those who do |
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155 |
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respond. For example, suppose I have a sample of |
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156 |
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tennis students. |
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157 |
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And I got responses from number two, number five, |
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158 |
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And number 10. So I have these point of views for |
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159 |
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these three students. Now the other seven students |
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160 |
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might be they have different opinions. So the only |
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161 |
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thing you have, the opinions of just the three, |
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162 |
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and maybe the rest have different opinions, it |
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163 |
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means in this case you will have something called |
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164 |
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non-responsiveness. Or the same as we said before, |
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165 |
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if your question is designed in a correct way. The |
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other type, Sample Error, variations from sample |
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to sample will always exist. As I mentioned, here |
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we select six samples, each one has different |
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sample mean. The other type, Measurement Error, |
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due to weakness in question design. So that's the |
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type of survey errors. So one more time, average |
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error, it means you exclude a group or groups from |
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the frame. So in this case, suppose I excluded |
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these from my frame. So I just select the sample |
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from all of these except this portion, or these |
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two groups. Non-response error means you don't |
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have follow-up on non-responses. Sampling error, |
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random sample gives different sample statistics. |
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So it means random differences from sample to |
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sample. Final measurement error, bad or leading |
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questions. This is one of the most important ones |
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that you have to avoid. So that's the first part |
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of this chapter, assembling techniques. Do you |
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have any questions? Next, we'll talk about |
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assembling distributions. So far, up to this |
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point, I mean at the end of chapter 6, we |
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discussed the probability For example, of |
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computing X greater than, for example, 7. For |
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example, suppose X represents your score in |
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business statistics course. |
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And suppose we know that X is normally distributed |
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with mean of 80, standard deviation of 10. My |
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question was, in chapter 6, what's the probability |
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that the student scores more than 70? Suppose we |
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select randomly one student, and the question is, |
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what's the probability that his score, so just for |
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one individual, for one student, his score is |
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above 70? In that case, if you remember, we |
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transform from normal distribution to standard |
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normal distribution by using this equation, which |
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is x minus the mean divided by sigma. The mean, |
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this one, it means the mean of x. And sigma is |
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also x. Now suppose instead of saying what's the |
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probability that a selected student scores more |
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than 70 or above 70, suppose we select a random |
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sample of 20 or whatever it is, 20 students from |
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this class, and I'm interested in the probability |
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that the average score of these 20 students is |
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above 70. Now look at the difference between two |
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portions. First one, we select a student randomly, |
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211 |
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and we are asking about what's the probability |
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that this selected student can score above 70. The |
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other one, we |
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select a random sample of size 20. And we are |
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interested on the probability that the average of |
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these scores is above 70. |
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And again, suppose X is normally distributed with |
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218 |
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mean 80, sigma is 10. It makes sense that we have |
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219 |
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to transform again from normal distribution, |
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220 |
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standardized normal distribution, by using exactly |
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221 |
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the same technique.The same z-score but with |
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222 |
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different statistics. Here we use x, just x, the |
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score of the student. Now we have to use something |
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other called x bar. I'm interested in the average |
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225 |
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of this. So x bar minus the mean of not x, x bar, |
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226 |
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then divided by sigma x bar. So this is my new, |
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227 |
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the score. |
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228 |
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Here, there are three questions. Number one, |
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what's the shape of the distribution of X bar? So, |
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we are asking about the shape of the distribution. |
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It might be normal. If the entire population that |
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232 |
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we select a sample from is normal, I mean if the |
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233 |
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population is normally distributed, then you |
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234 |
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select a random sample of that population, it |
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235 |
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makes sense that the sample is also normal, so any |
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236 |
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statistic is computed from that sample is also |
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237 |
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|
normally distributed, so it makes sense. If the |
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238 |
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population is normal, then the shape is also |
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239 |
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normal. But if the population is unknown, you |
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240 |
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|
don't have any information about the underlying |
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241 |
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population, then you cannot say it's normal unless |
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242 |
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you have certain condition that we'll talk about |
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243 |
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|
maybe after 30 minutes. So, exactly, if the |
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244 |
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|
population is normal, then the shape is also |
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245 |
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normal, but otherwise, we have to think about it. |
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246 |
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This is the first question. Now, there are two |
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247 |
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unknowns in this equation. We have to know the |
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248 |
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|
mean, Or x bar, so the mean of x bar is not given, |
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249 |
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|
the mean means the center. So the second question, |
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250 |
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|
what's the center of the distribution? In this |
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251 |
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|
case, the mean of x bar. So we are looking at |
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252 |
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|
what's the mean of x bar. The third question is |
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253 |
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|
sigma x bar is also unknown, spread. Now shape, |
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254 |
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|
center, spread, these are characteristics, these |
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255 |
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|
characteristics in this case sampling |
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256 |
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|
distribution, exactly which is called sampling |
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257 |
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distribution. So by sampling distribution we mean |
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258 |
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|
that, by sampling distribution, we mean that you |
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259 |
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|
have to know the center of distribution, I mean |
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260 |
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|
the mean of the statistic you are interested in. |
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261 |
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|
Second, the spread or the variability of the |
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262 |
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sample statistic also you are interested in. In |
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263 |
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|
addition to that, you have to know the shape of |
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264 |
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|
the statistic. So three things we have to know, |
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265 |
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|
center, spread and shape. So that's what we'll |
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266 |
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|
talk about now. So now sampling distribution is a |
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267 |
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|
distribution of all of the possible values of a |
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268 |
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|
sample statistic. This sample statistic could be |
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269 |
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|
sample mean, could be sample variance, could be |
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270 |
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|
sample proportion, because any population has |
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271 |
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|
mainly three characteristics, mean, standard |
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272 |
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deviation, and proportion. |
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273 |
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|
So again, a sampling distribution is a |
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274 |
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|
distribution of all of the possible values of a |
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275 |
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|
sample statistic or a given size sample selected |
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276 |
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|
from a population. For example, suppose you sample |
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277 |
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|
50 students from your college regarding their mean |
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278 |
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|
GPA. GPA means Graduate Point Average. Now, if you |
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279 |
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|
obtain many different samples of size 50, you will |
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280 |
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|
compute a different mean for each sample. As I |
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281 |
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|
mentioned here, I select a sample the same sizes, |
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282 |
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|
but we obtain different sample statistics, I mean |
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283 |
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|
different sample means. We are interested in the |
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284 |
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|
distribution of all potential mean GBA We might |
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285 |
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|
calculate for any given sample of 50 students. So |
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286 |
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|
let's focus into these values. So we have again a |
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287 |
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|
random sample of 50 sample means. So we have 1, 2, |
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288 |
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|
3, 4, 5, maybe 50, 6, whatever we have. So select |
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289 |
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|
a random sample of size 20. Maybe we repeat this |
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290 |
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|
sample 10 times. So we end with 10. different |
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291 |
|
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|
values of the simple means. Now we have new ten |
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292 |
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|
means. Now the question is, what's the center of |
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293 |
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|
these values, I mean for the means? What's the |
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294 |
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|
spread of the means? And what's the shape of the |
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295 |
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|
means? So these are the mainly three questions. |
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296 |
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|
For example, let's get just simple example and |
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297 |
|
00:22:56,810 --> 00:23:04,370 |
|
that we have only population of size 4. In the |
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298 |
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00:23:04,370 --> 00:23:09,950 |
|
real life, the population size is much bigger than |
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299 |
|
00:23:09,950 --> 00:23:14,970 |
|
4, but just for illustration. |
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|
300 |
|
00:23:17,290 --> 00:23:20,190 |
|
Because size 4, I mean if the population is 4, |
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301 |
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00:23:21,490 --> 00:23:24,950 |
|
it's a small population. So we can take all the |
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|
302 |
|
00:23:24,950 --> 00:23:27,610 |
|
values and find the mean and standard deviation. |
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|
303 |
|
00:23:28,290 --> 00:23:31,430 |
|
But in reality, we have more than that. So this |
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|
304 |
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00:23:31,430 --> 00:23:37,390 |
|
one just for as example. So let's suppose that we |
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305 |
|
00:23:37,390 --> 00:23:42,790 |
|
have a population of size 4. So n equals 4. |
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|
306 |
|
00:23:46,530 --> 00:23:54,030 |
|
And we are interested in the ages. And suppose the |
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307 |
|
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|
values of X, X again represents H, |
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|
308 |
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00:24:00,690 --> 00:24:01,810 |
|
and the values we have. |
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|
309 |
|
00:24:06,090 --> 00:24:08,930 |
|
So these are the four values we have. |
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|
310 |
|
00:24:12,050 --> 00:24:16,910 |
|
Now simple calculation will |
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|
311 |
|
00:24:16,910 --> 00:24:19,910 |
|
give you the mean, the population mean. |
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|
312 |
|
00:24:25,930 --> 00:24:30,410 |
|
Just add these values and divide by the operation |
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|
313 |
|
00:24:30,410 --> 00:24:35,410 |
|
size, we'll get 21 years. And sigma, as we |
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|
314 |
|
00:24:35,410 --> 00:24:39,450 |
|
mentioned in chapter three, square root of this |
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|
|
315 |
|
00:24:39,450 --> 00:24:44,550 |
|
quantity will give 2.236 |
|
|
|
316 |
|
00:24:44,550 --> 00:24:49,450 |
|
years. So simple calculation will give these |
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|
317 |
|
00:24:49,450 --> 00:24:54,470 |
|
results. Now if you look at distribution of these |
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|
318 |
|
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|
values, Because as I mentioned, we are looking for |
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|
319 |
|
00:24:58,430 --> 00:25:03,810 |
|
center, spread, and shape. The center is 21 of the |
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|
320 |
|
00:25:03,810 --> 00:25:09,430 |
|
exact population. The variation is around 2.2. |
|
|
|
321 |
|
00:25:10,050 --> 00:25:14,770 |
|
Now, the shape of distribution. Now, 18 represents |
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|
322 |
|
00:25:14,770 --> 00:25:15,250 |
|
once. |
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|
323 |
|
00:25:17,930 --> 00:25:22,350 |
|
I mean, we have only one 18, so 18 divided one |
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|
324 |
|
00:25:22,350 --> 00:25:29,830 |
|
time over 425. 20% represent also 25%, the same as |
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|
325 |
|
00:25:29,830 --> 00:25:33,030 |
|
for 22 or 24. In this case, we have something |
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|
326 |
|
00:25:33,030 --> 00:25:37,530 |
|
called uniform distribution. In this case, the |
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|
|
327 |
|
00:25:37,530 --> 00:25:43,330 |
|
proportions are the same. So, the mean, not |
|
|
|
328 |
|
00:25:43,330 --> 00:25:48,030 |
|
normal, it's uniform distribution. The mean is 21, |
|
|
|
329 |
|
00:25:48,690 --> 00:25:52,490 |
|
standard deviation is 2.236, and the distribution |
|
|
|
330 |
|
00:25:52,490 --> 00:25:58,840 |
|
is uniform. Okay, so that's center, spread and |
|
|
|
331 |
|
00:25:58,840 --> 00:26:02,920 |
|
shape of the true population we have. Now suppose |
|
|
|
332 |
|
00:26:02,920 --> 00:26:03,520 |
|
for example, |
|
|
|
333 |
|
00:26:06,600 --> 00:26:12,100 |
|
we select a random sample of size 2 from this |
|
|
|
334 |
|
00:26:12,100 --> 00:26:12,620 |
|
population. |
|
|
|
335 |
|
00:26:15,740 --> 00:26:21,500 |
|
So we select a sample of size 2. We have 18, 20, |
|
|
|
336 |
|
00:26:21,600 --> 00:26:25,860 |
|
22, 24 years. We have four students, for example. |
|
|
|
337 |
|
00:26:27,760 --> 00:26:31,140 |
|
And we select a sample of size two. So the first |
|
|
|
338 |
|
00:26:31,140 --> 00:26:40,820 |
|
one could be 18 and 18, 18 and 20, 18 and 22. So |
|
|
|
339 |
|
00:26:40,820 --> 00:26:47,400 |
|
we have 16 different samples. So number of samples |
|
|
|
340 |
|
00:26:47,400 --> 00:26:54,500 |
|
in this case is 16. Imagine that we have five. I |
|
|
|
341 |
|
00:26:54,500 --> 00:27:00,220 |
|
mean the population size is 5 and so on. So the |
|
|
|
342 |
|
00:27:00,220 --> 00:27:06,000 |
|
rule is number |
|
|
|
343 |
|
00:27:06,000 --> 00:27:13,020 |
|
of samples in this case and the volume is million. |
|
|
|
344 |
|
00:27:14,700 --> 00:27:19,140 |
|
Because we have four, four squared is sixteen, |
|
|
|
345 |
|
00:27:19,440 --> 00:27:26,940 |
|
that's all. 5 squared, 25, and so on. Now, we have |
|
|
|
346 |
|
00:27:26,940 --> 00:27:31,740 |
|
16 different samples. For sure, we will have |
|
|
|
347 |
|
00:27:31,740 --> 00:27:37,940 |
|
different sample means. Now, for the first sample, |
|
|
|
348 |
|
00:27:39,560 --> 00:27:47,200 |
|
18, 18, the average is also 18. The next one, 18, |
|
|
|
349 |
|
00:27:47,280 --> 00:27:50,040 |
|
20, the average is 19. |
|
|
|
350 |
|
00:27:54,790 --> 00:27:59,770 |
|
20, 18, 24, the average is 21, and so on. So now |
|
|
|
351 |
|
00:27:59,770 --> 00:28:05,450 |
|
we have 16 sample means. Now this is my new |
|
|
|
352 |
|
00:28:05,450 --> 00:28:10,510 |
|
values. It's my sample. This sample has different |
|
|
|
353 |
|
00:28:10,510 --> 00:28:16,050 |
|
sample means. Now let's take these values and |
|
|
|
354 |
|
00:28:16,050 --> 00:28:23,270 |
|
compute average, sigma, and the shape of the |
|
|
|
355 |
|
00:28:23,270 --> 00:28:29,200 |
|
distribution. So again, we have a population of |
|
|
|
356 |
|
00:28:29,200 --> 00:28:35,240 |
|
size 4, we select a random cell bone. of size 2 |
|
|
|
357 |
|
00:28:35,240 --> 00:28:39,060 |
|
from that population, we end with 16 random |
|
|
|
358 |
|
00:28:39,060 --> 00:28:43,620 |
|
samples, and they have different sample means. |
|
|
|
359 |
|
00:28:43,860 --> 00:28:46,700 |
|
Might be two of them are the same. I mean, we have |
|
|
|
360 |
|
00:28:46,700 --> 00:28:52,220 |
|
18 just repeated once, but 19 repeated twice, 23 |
|
|
|
361 |
|
00:28:52,220 --> 00:28:59,220 |
|
times, 24 times, and so on. 22 three times, 23 |
|
|
|
362 |
|
00:28:59,220 --> 00:29:04,270 |
|
twice, 24 once. So it depends on The sample means |
|
|
|
363 |
|
00:29:04,270 --> 00:29:07,210 |
|
you have. So we have actually different samples. |
|
|
|
364 |
|
00:29:14,790 --> 00:29:18,970 |
|
For example, let's look at 24 and 22. What's the |
|
|
|
365 |
|
00:29:18,970 --> 00:29:22,790 |
|
average of these two values? N divided by 2 will |
|
|
|
366 |
|
00:29:22,790 --> 00:29:24,290 |
|
give 22. |
|
|
|
367 |
|
00:29:33,390 --> 00:29:35,730 |
|
So again, we have 16 sample means. |
|
|
|
368 |
|
00:29:38,610 --> 00:29:41,550 |
|
Now look first at the shape of the distribution. |
|
|
|
369 |
|
00:29:43,110 --> 00:29:47,490 |
|
18, as I mentioned, repeated once. So 1 over 16. |
|
|
|
370 |
|
00:29:48,430 --> 00:29:57,950 |
|
19 twice. 23 times. 1 four times. 22 three times. |
|
|
|
371 |
|
00:29:58,940 --> 00:30:03,340 |
|
then twice then once now the distribution was |
|
|
|
372 |
|
00:30:03,340 --> 00:30:07,960 |
|
uniform remember now it becomes normal |
|
|
|
373 |
|
00:30:07,960 --> 00:30:10,780 |
|
distribution so the first one x1 is normal |
|
|
|
374 |
|
00:30:10,780 --> 00:30:16,340 |
|
distribution so it has normal distribution so |
|
|
|
375 |
|
00:30:16,340 --> 00:30:20,040 |
|
again the shape of x1 looks like normal |
|
|
|
376 |
|
00:30:20,040 --> 00:30:26,800 |
|
distribution we need to compute the center of X |
|
|
|
377 |
|
00:30:26,800 --> 00:30:32,800 |
|
bar, the mean of X bar. We have to add the values |
|
|
|
378 |
|
00:30:32,800 --> 00:30:36,380 |
|
of X bar, the sample mean, then divide by the |
|
|
|
379 |
|
00:30:36,380 --> 00:30:42,800 |
|
total number of size, which is 16. So in this |
|
|
|
380 |
|
00:30:42,800 --> 00:30:51,720 |
|
case, we got 21, which is similar to the one for |
|
|
|
381 |
|
00:30:51,720 --> 00:30:55,950 |
|
the entire population. So this is the first |
|
|
|
382 |
|
00:30:55,950 --> 00:30:59,930 |
|
unknown parameter. The mu of x bar is the same as |
|
|
|
383 |
|
00:30:59,930 --> 00:31:05,490 |
|
the population mean mu. The second one, the split |
|
|
|
384 |
|
00:31:05,490 --> 00:31:13,450 |
|
sigma of x bar by using the same equation |
|
|
|
385 |
|
00:31:13,450 --> 00:31:17,170 |
|
we have, sum of x bar in this case minus the mean |
|
|
|
386 |
|
00:31:17,170 --> 00:31:21,430 |
|
of x bar squared, then divide this quantity by the |
|
|
|
387 |
|
00:31:21,430 --> 00:31:26,270 |
|
capital I which is 16 in this case. So we will end |
|
|
|
388 |
|
00:31:26,270 --> 00:31:28,510 |
|
with 1.58. |
|
|
|
389 |
|
00:31:31,270 --> 00:31:36,170 |
|
Now let's compare population standard deviation |
|
|
|
390 |
|
00:31:36,170 --> 00:31:42,210 |
|
and the sample standard deviation. First of all, |
|
|
|
391 |
|
00:31:42,250 --> 00:31:45,050 |
|
you see that these two values are not the same. |
|
|
|
392 |
|
00:31:47,530 --> 00:31:50,370 |
|
The population standard deviation was 2.2, around |
|
|
|
393 |
|
00:31:50,370 --> 00:31:57,310 |
|
2.2. But for the sample, for the sample mean, it's |
|
|
|
394 |
|
00:31:57,310 --> 00:32:02,690 |
|
1.58, so that means sigma of X bar is smaller than |
|
|
|
395 |
|
00:32:02,690 --> 00:32:03,710 |
|
sigma of X. |
|
|
|
396 |
|
00:32:07,270 --> 00:32:12,010 |
|
It means exactly, the variation of X bar is always |
|
|
|
397 |
|
00:32:12,010 --> 00:32:15,770 |
|
smaller than the variation of X, always. |
|
|
|
398 |
|
00:32:20,420 --> 00:32:26,480 |
|
So here is the comparison. The distribution was |
|
|
|
399 |
|
00:32:26,480 --> 00:32:32,000 |
|
uniform. It's no longer uniform. It looks like a |
|
|
|
400 |
|
00:32:32,000 --> 00:32:36,440 |
|
bell shape. The mean of X is 21, which is the same |
|
|
|
401 |
|
00:32:36,440 --> 00:32:40,440 |
|
as the mean of X bar. But the standard deviation |
|
|
|
402 |
|
00:32:40,440 --> 00:32:44,200 |
|
of the population is larger than the standard |
|
|
|
403 |
|
00:32:44,200 --> 00:32:48,060 |
|
deviation of the sample mean or the average. |
|
|
|
404 |
|
00:32:53,830 --> 00:32:58,090 |
|
Different samples of the same sample size from the |
|
|
|
405 |
|
00:32:58,090 --> 00:33:00,790 |
|
same population will yield different sample means. |
|
|
|
406 |
|
00:33:01,450 --> 00:33:06,050 |
|
We know that. If we have a population and from |
|
|
|
407 |
|
00:33:06,050 --> 00:33:08,570 |
|
that population, so we have this big population, |
|
|
|
408 |
|
00:33:10,250 --> 00:33:15,010 |
|
from this population suppose we selected 10 |
|
|
|
409 |
|
00:33:15,010 --> 00:33:19,850 |
|
samples, sample 1 with size 50. |
|
|
|
410 |
|
00:33:21,540 --> 00:33:26,400 |
|
Another sample, sample 2 with the same size. All |
|
|
|
411 |
|
00:33:26,400 --> 00:33:29,980 |
|
the way, suppose we select 10 samples, sample 10, |
|
|
|
412 |
|
00:33:30,860 --> 00:33:38,460 |
|
also with the same sample size. Each one will have |
|
|
|
413 |
|
00:33:38,460 --> 00:33:42,960 |
|
different average, different sample. Maybe the |
|
|
|
414 |
|
00:33:42,960 --> 00:33:48,520 |
|
first one has 70, 68, suppose the last one has 71. |
|
|
|
415 |
|
00:33:49,880 --> 00:33:56,040 |
|
So again, different samples of the same size, I |
|
|
|
416 |
|
00:33:56,040 --> 00:34:00,460 |
|
got size of 15, from the same population will |
|
|
|
417 |
|
00:34:00,460 --> 00:34:04,940 |
|
yield different sample means. This is one fact. |
|
|
|
418 |
|
00:34:05,900 --> 00:34:10,440 |
|
Now, a measure of the variability, which means |
|
|
|
419 |
|
00:34:10,440 --> 00:34:17,060 |
|
sigma. and I'm interested in x bar. So a measure |
|
|
|
420 |
|
00:34:17,060 --> 00:34:20,400 |
|
of variability in the mean from sample to sample |
|
|
|
421 |
|
00:34:20,400 --> 00:34:23,580 |
|
is given by something called, instead of saying |
|
|
|
422 |
|
00:34:23,580 --> 00:34:27,100 |
|
standard deviation of the sample mean, we have |
|
|
|
423 |
|
00:34:27,100 --> 00:34:31,780 |
|
this expression, standard error of the mean. So |
|
|
|
424 |
|
00:34:31,780 --> 00:34:37,920 |
|
this one is called standard error of the mean, or |
|
|
|
425 |
|
00:34:37,920 --> 00:34:41,880 |
|
sigma of x bar. So again, it's called standard |
|
|
|
426 |
|
00:34:43,590 --> 00:34:46,590 |
|
error of the mean or simply just say standard |
|
|
|
427 |
|
00:34:46,590 --> 00:34:50,850 |
|
error. So it's better to distinguish between |
|
|
|
428 |
|
00:34:50,850 --> 00:34:57,510 |
|
population standard deviation sigma and sigma of x |
|
|
|
429 |
|
00:34:57,510 --> 00:35:01,530 |
|
bar which is the standard error of the mean. So we |
|
|
|
430 |
|
00:35:01,530 --> 00:35:06,900 |
|
have sigma And sigma of x bar. This one is |
|
|
|
431 |
|
00:35:06,900 --> 00:35:12,380 |
|
standard error of x bar. And always sigma of x bar |
|
|
|
432 |
|
00:35:12,380 --> 00:35:17,300 |
|
is smaller than sigma, unless n equals one. Later |
|
|
|
433 |
|
00:35:17,300 --> 00:35:21,640 |
|
we'll see why if n is one, then the two quantities |
|
|
|
434 |
|
00:35:21,640 --> 00:35:29,600 |
|
are the same. Now, sigma of x bar |
|
|
|
435 |
|
00:35:32,340 --> 00:35:37,820 |
|
In this case, it's 158 equals sigma over root n. I |
|
|
|
436 |
|
00:35:37,820 --> 00:35:42,480 |
|
mean, for this specific example, if we divide |
|
|
|
437 |
|
00:35:42,480 --> 00:35:50,620 |
|
sigma, which is 2.236 divided by n, and n 2, you |
|
|
|
438 |
|
00:35:50,620 --> 00:35:53,440 |
|
will get 1.58. |
|
|
|
439 |
|
00:35:57,620 --> 00:36:03,640 |
|
So we got mu x bar equal mu. The spread is sigma |
|
|
|
440 |
|
00:36:03,640 --> 00:36:07,060 |
|
over root. Now, if you compare sigma x bar and |
|
|
|
441 |
|
00:36:07,060 --> 00:36:10,680 |
|
sigma, always sigma of x bar is smaller than |
|
|
|
442 |
|
00:36:10,680 --> 00:36:14,460 |
|
sigma, unless m equals 1. And in reality, we don't |
|
|
|
443 |
|
00:36:14,460 --> 00:36:17,860 |
|
have a sample of size 1. So always the sample size |
|
|
|
444 |
|
00:36:17,860 --> 00:36:23,500 |
|
is greater than 1. So always sigma of x bar is |
|
|
|
445 |
|
00:36:23,500 --> 00:36:28,660 |
|
smaller than sigma of the standard deviation of |
|
|
|
446 |
|
00:36:28,660 --> 00:36:33,180 |
|
normalization. Now if you look at the relationship |
|
|
|
447 |
|
00:36:33,180 --> 00:36:36,380 |
|
between the standard error of X bar and the sample |
|
|
|
448 |
|
00:36:36,380 --> 00:36:41,760 |
|
size, we'll see that as the sample size increases, |
|
|
|
449 |
|
00:36:42,500 --> 00:36:46,440 |
|
sigma of X bar decreases. So if we have large |
|
|
|
450 |
|
00:36:46,440 --> 00:36:51,200 |
|
sample size, I mean instead of selecting a random |
|
|
|
451 |
|
00:36:51,200 --> 00:36:53,520 |
|
sample of size 2, if you select a random sample of |
|
|
|
452 |
|
00:36:53,520 --> 00:36:56,900 |
|
size 3 for example, you will get sigma of X bar |
|
|
|
453 |
|
00:36:56,900 --> 00:37:03,140 |
|
less than 1.58. So note that standard error of the |
|
|
|
454 |
|
00:37:03,140 --> 00:37:09,260 |
|
mean decreases as the sample size goes up. So as n |
|
|
|
455 |
|
00:37:09,260 --> 00:37:13,000 |
|
increases, sigma of x bar goes down. So there is |
|
|
|
456 |
|
00:37:13,000 --> 00:37:17,440 |
|
inverse relationship between the standard error of |
|
|
|
457 |
|
00:37:17,440 --> 00:37:21,900 |
|
the mean and the sample size. So now we answered |
|
|
|
458 |
|
00:37:21,900 --> 00:37:24,660 |
|
the three questions. The shape looks like bell |
|
|
|
459 |
|
00:37:24,660 --> 00:37:31,290 |
|
shape. if we select our sample from normal |
|
|
|
460 |
|
00:37:31,290 --> 00:37:37,850 |
|
population with mean equals the population mean |
|
|
|
461 |
|
00:37:37,850 --> 00:37:40,530 |
|
and standard deviation of standard error equals |
|
|
|
462 |
|
00:37:40,530 --> 00:37:48,170 |
|
sigma over square root of that. So now, let's talk |
|
|
|
463 |
|
00:37:48,170 --> 00:37:53,730 |
|
about sampling distribution of the sample mean if |
|
|
|
464 |
|
00:37:53,730 --> 00:37:59,170 |
|
the population is normal. So now, my population is |
|
|
|
465 |
|
00:37:59,170 --> 00:38:03,830 |
|
normally distributed, and we are interested in the |
|
|
|
466 |
|
00:38:03,830 --> 00:38:06,430 |
|
sampling distribution of the sample mean of X bar. |
|
|
|
467 |
|
00:38:07,630 --> 00:38:11,330 |
|
If the population is normally distributed with |
|
|
|
468 |
|
00:38:11,330 --> 00:38:14,870 |
|
mean mu and standard deviation sigma, in this |
|
|
|
469 |
|
00:38:14,870 --> 00:38:18,590 |
|
case, the sampling distribution of X bar is also |
|
|
|
470 |
|
00:38:18,590 --> 00:38:22,870 |
|
normally distributed, so this is the shape. with |
|
|
|
471 |
|
00:38:22,870 --> 00:38:27,070 |
|
mean of X bar equals mu and sigma of X bar equals |
|
|
|
472 |
|
00:38:27,070 --> 00:38:35,470 |
|
sigma over 2. So again, if we sample from normal |
|
|
|
473 |
|
00:38:35,470 --> 00:38:39,650 |
|
population, I mean my sampling technique, I select |
|
|
|
474 |
|
00:38:39,650 --> 00:38:44,420 |
|
a random sample from a normal population. Then if |
|
|
|
475 |
|
00:38:44,420 --> 00:38:47,640 |
|
we are interested in the standard distribution of |
|
|
|
476 |
|
00:38:47,640 --> 00:38:51,960 |
|
X bar, then that distribution is normally |
|
|
|
477 |
|
00:38:51,960 --> 00:38:56,000 |
|
distributed with mean equal to mu and standard |
|
|
|
478 |
|
00:38:56,000 --> 00:39:02,540 |
|
deviation sigma over mu. So that's the shape. It's |
|
|
|
479 |
|
00:39:02,540 --> 00:39:05,670 |
|
normal. The mean is the same as the population |
|
|
|
480 |
|
00:39:05,670 --> 00:39:09,030 |
|
mean, and the standard deviation of x bar equals |
|
|
|
481 |
|
00:39:09,030 --> 00:39:16,130 |
|
sigma over root n. So now let's go back to the z |
|
|
|
482 |
|
00:39:16,130 --> 00:39:21,130 |
|
-score we discussed before. If you remember, I |
|
|
|
483 |
|
00:39:21,130 --> 00:39:25,150 |
|
mentioned before |
|
|
|
484 |
|
00:39:25,150 --> 00:39:32,720 |
|
that z-score, generally speaking, X minus the mean |
|
|
|
485 |
|
00:39:32,720 --> 00:39:34,740 |
|
of X divided by sigma X. |
|
|
|
486 |
|
00:39:37,640 --> 00:39:41,620 |
|
And we know that Z has standard normal |
|
|
|
487 |
|
00:39:41,620 --> 00:39:48,020 |
|
distribution with mean zero and variance one. In |
|
|
|
488 |
|
00:39:48,020 --> 00:39:52,860 |
|
this case, we are looking for the semi |
|
|
|
489 |
|
00:39:52,860 --> 00:39:59,350 |
|
-distribution of X bar. So Z equal X bar. minus |
|
|
|
490 |
|
00:39:59,350 --> 00:40:06,050 |
|
the mean of x bar divided by sigma of x bar. So |
|
|
|
491 |
|
00:40:06,050 --> 00:40:10,770 |
|
the same equation, but different statistic. In the |
|
|
|
492 |
|
00:40:10,770 --> 00:40:15,770 |
|
first one, we have x, for example, represents the |
|
|
|
493 |
|
00:40:15,770 --> 00:40:20,370 |
|
score. Here, my sample statistic is the sample |
|
|
|
494 |
|
00:40:20,370 --> 00:40:22,890 |
|
mean, which represents the average of the score. |
|
|
|
495 |
|
00:40:23,470 --> 00:40:29,460 |
|
So x bar, minus its mean, I mean the mean of x |
|
|
|
496 |
|
00:40:29,460 --> 00:40:37,280 |
|
bar, divided by its standard error. So x bar minus |
|
|
|
497 |
|
00:40:37,280 --> 00:40:41,000 |
|
the mean of x bar divided by sigma of x bar. By |
|
|
|
498 |
|
00:40:41,000 --> 00:40:48,020 |
|
using that mu of x bar equals mu, and sigma of x |
|
|
|
499 |
|
00:40:48,020 --> 00:40:51,240 |
|
bar equals sigma over root n, we will end with |
|
|
|
500 |
|
00:40:51,240 --> 00:40:52,600 |
|
this mu z square. |
|
|
|
501 |
|
00:40:56,310 --> 00:41:00,790 |
|
So this equation will be used instead of using the |
|
|
|
502 |
|
00:41:00,790 --> 00:41:04,650 |
|
previous one. So z square equals sigma, I'm sorry, |
|
|
|
503 |
|
00:41:04,770 --> 00:41:08,470 |
|
z equals x bar minus the mean divided by sigma |
|
|
|
504 |
|
00:41:08,470 --> 00:41:13,310 |
|
bar, where x bar the sample mean, mu the |
|
|
|
505 |
|
00:41:13,310 --> 00:41:15,990 |
|
population mean, sigma population standard |
|
|
|
506 |
|
00:41:15,990 --> 00:41:19,810 |
|
deviation, and n is the sample size. So that's the |
|
|
|
507 |
|
00:41:19,810 --> 00:41:22,490 |
|
difference between chapter six, |
|
|
|
508 |
|
00:41:25,110 --> 00:41:32,750 |
|
And that one we have only x minus y by sigma. Here |
|
|
|
509 |
|
00:41:32,750 --> 00:41:36,450 |
|
we are interested in x bar minus the mean of x bar |
|
|
|
510 |
|
00:41:36,450 --> 00:41:40,290 |
|
which is mu. And sigma of x bar equals sigma of r. |
|
|
|
511 |
|
00:41:47,970 --> 00:41:52,010 |
|
Now when we are saying that mu of x bar equals mu, |
|
|
|
512 |
|
00:41:54,530 --> 00:42:01,690 |
|
That means the expected value of |
|
|
|
513 |
|
00:42:01,690 --> 00:42:05,590 |
|
the sample mean equals the population mean. When |
|
|
|
514 |
|
00:42:05,590 --> 00:42:08,610 |
|
we are saying mean of X bar equals mu, it means |
|
|
|
515 |
|
00:42:08,610 --> 00:42:13,270 |
|
the expected value of X bar equals mu. In other |
|
|
|
516 |
|
00:42:13,270 --> 00:42:20,670 |
|
words, the expected of X bar equals mu. If this |
|
|
|
517 |
|
00:42:20,670 --> 00:42:27,900 |
|
happens, we say that X bar is an unbiased |
|
|
|
518 |
|
00:42:27,900 --> 00:42:31,420 |
|
estimator |
|
|
|
519 |
|
00:42:31,420 --> 00:42:35,580 |
|
of |
|
|
|
520 |
|
00:42:35,580 --> 00:42:40,620 |
|
mu. So this is a new definition, unbiased |
|
|
|
521 |
|
00:42:40,620 --> 00:42:45,490 |
|
estimator X bar is called unbiased estimator if |
|
|
|
522 |
|
00:42:45,490 --> 00:42:49,410 |
|
this condition is satisfied. I mean, if the mean |
|
|
|
523 |
|
00:42:49,410 --> 00:42:54,450 |
|
of X bar or if the expected value of X bar equals |
|
|
|
524 |
|
00:42:54,450 --> 00:42:57,790 |
|
the population mean, in this case, we say that X |
|
|
|
525 |
|
00:42:57,790 --> 00:43:02,450 |
|
bar is good estimator of Mu. Because on average, |
|
|
|
526 |
|
00:43:05,430 --> 00:43:08,230 |
|
Expected value of X bar equals the population |
|
|
|
527 |
|
00:43:08,230 --> 00:43:14,970 |
|
mean, so in this case X bar is L by estimator of |
|
|
|
528 |
|
00:43:14,970 --> 00:43:20,410 |
|
Mu. Now if you compare the two distributions, |
|
|
|
529 |
|
00:43:22,030 --> 00:43:27,510 |
|
normal distribution here with population mean Mu |
|
|
|
530 |
|
00:43:27,510 --> 00:43:30,550 |
|
and standard deviation for example sigma. |
|
|
|
531 |
|
00:43:33,190 --> 00:43:40,590 |
|
That's for the scores, the scores. Now instead of |
|
|
|
532 |
|
00:43:40,590 --> 00:43:43,690 |
|
the scores above, we have x bar, the sample mean. |
|
|
|
533 |
|
00:43:44,670 --> 00:43:48,590 |
|
Again, the mean of x bar is the same as the |
|
|
|
534 |
|
00:43:48,590 --> 00:43:52,990 |
|
population mean. Both means are the same, mu of x |
|
|
|
535 |
|
00:43:52,990 --> 00:43:57,130 |
|
bar equals mu. But if you look at the spread of |
|
|
|
536 |
|
00:43:57,130 --> 00:44:00,190 |
|
the second distribution, this is more than the |
|
|
|
537 |
|
00:44:00,190 --> 00:44:03,350 |
|
other one. So that's the comparison between the |
|
|
|
538 |
|
00:44:03,350 --> 00:44:05,530 |
|
two populations. |
|
|
|
539 |
|
00:44:07,050 --> 00:44:13,390 |
|
So again, to compare or to figure out the |
|
|
|
540 |
|
00:44:13,390 --> 00:44:17,910 |
|
relationship between sigma of x bar and the sample |
|
|
|
541 |
|
00:44:17,910 --> 00:44:22,110 |
|
size. Suppose we have this blue normal |
|
|
|
542 |
|
00:44:22,110 --> 00:44:28,590 |
|
distribution with sample size say 10 or 30, for |
|
|
|
543 |
|
00:44:28,590 --> 00:44:28,870 |
|
example. |
|
|
|
544 |
|
00:44:32,220 --> 00:44:37,880 |
|
As n gets bigger and bigger, sigma of x bar |
|
|
|
545 |
|
00:44:37,880 --> 00:44:41,800 |
|
becomes smaller and smaller. If you look at the |
|
|
|
546 |
|
00:44:41,800 --> 00:44:44,760 |
|
red one, maybe if the red one has n equal 100, |
|
|
|
547 |
|
00:44:45,700 --> 00:44:48,780 |
|
we'll get this spread. But for the other one, we |
|
|
|
548 |
|
00:44:48,780 --> 00:44:55,240 |
|
have larger spread. So as n increases, sigma of x |
|
|
|
549 |
|
00:44:55,240 --> 00:44:59,860 |
|
bar decreases. So this, the blue one for smaller |
|
|
|
550 |
|
00:44:59,860 --> 00:45:06,240 |
|
sample size. The red one for larger sample size. |
|
|
|
551 |
|
00:45:06,840 --> 00:45:11,120 |
|
So again, as n increases, sigma of x bar goes down |
|
|
|
552 |
|
00:45:11,120 --> 00:45:12,040 |
|
four degrees. |
|
|
|
553 |
|
00:45:21,720 --> 00:45:29,480 |
|
Next, let's use this fact to |
|
|
|
554 |
|
00:45:29,480 --> 00:45:37,440 |
|
figure out an interval for the sample mean with 90 |
|
|
|
555 |
|
00:45:37,440 --> 00:45:42,140 |
|
% confidence and suppose the population we have is |
|
|
|
556 |
|
00:45:42,140 --> 00:45:49,500 |
|
normal with mean 368 and sigma of 15 and suppose |
|
|
|
557 |
|
00:45:49,500 --> 00:45:52,900 |
|
we select a random sample of a 25 and the question |
|
|
|
558 |
|
00:45:52,900 --> 00:45:57,600 |
|
is find symmetrically distributed interval around |
|
|
|
559 |
|
00:45:57,600 --> 00:46:03,190 |
|
the mean That will include 95% of the sample means |
|
|
|
560 |
|
00:46:03,190 --> 00:46:08,610 |
|
when mu equals 368, sigma is 15, and your sample |
|
|
|
561 |
|
00:46:08,610 --> 00:46:13,830 |
|
size is 25. So in this case, we are looking for |
|
|
|
562 |
|
00:46:13,830 --> 00:46:17,150 |
|
the |
|
|
|
563 |
|
00:46:17,150 --> 00:46:19,110 |
|
estimation of the sample mean. |
|
|
|
564 |
|
00:46:23,130 --> 00:46:24,970 |
|
And we have this information, |
|
|
|
565 |
|
00:46:28,910 --> 00:46:31,750 |
|
Sigma is 15 and N is 25. |
|
|
|
566 |
|
00:46:35,650 --> 00:46:38,890 |
|
The problem mentioned there, we have symmetric |
|
|
|
567 |
|
00:46:38,890 --> 00:46:48,490 |
|
distribution and this area is 95% bisymmetric and |
|
|
|
568 |
|
00:46:48,490 --> 00:46:52,890 |
|
we have only 5% out. So that means half to the |
|
|
|
569 |
|
00:46:52,890 --> 00:46:56,490 |
|
right and half to the left. |
|
|
|
570 |
|
00:46:59,740 --> 00:47:02,640 |
|
And let's see how can we compute these two values. |
|
|
|
571 |
|
00:47:03,820 --> 00:47:11,440 |
|
The problem says that the average T68 |
|
|
|
572 |
|
00:47:11,440 --> 00:47:18,660 |
|
for this data and the standard deviation sigma of |
|
|
|
573 |
|
00:47:18,660 --> 00:47:28,510 |
|
15. He asked about what are the values of x bar. I |
|
|
|
574 |
|
00:47:28,510 --> 00:47:32,430 |
|
mean, we have to find the interval of x bar. Let's |
|
|
|
575 |
|
00:47:32,430 --> 00:47:36,130 |
|
see. If you remember last time, z score was x |
|
|
|
576 |
|
00:47:36,130 --> 00:47:41,130 |
|
minus mu divided by sigma. But now we have x bar. |
|
|
|
577 |
|
00:47:41,890 --> 00:47:45,850 |
|
So your z score should be minus mu divided by |
|
|
|
578 |
|
00:47:45,850 --> 00:47:50,850 |
|
sigma over root n. Now cross multiplication, you |
|
|
|
579 |
|
00:47:50,850 --> 00:47:55,970 |
|
will get x bar minus mu equals z sigma over root |
|
|
|
580 |
|
00:47:55,970 --> 00:48:01,500 |
|
n. That means x bar equals mu plus z sigma over |
|
|
|
581 |
|
00:48:01,500 --> 00:48:04,440 |
|
root n. Exactly the same equation we got in |
|
|
|
582 |
|
00:48:04,440 --> 00:48:09,840 |
|
chapter six, but there, in that one we have x |
|
|
|
583 |
|
00:48:09,840 --> 00:48:13,700 |
|
equals mu plus z sigma. Now we have x bar equals |
|
|
|
584 |
|
00:48:13,700 --> 00:48:18,200 |
|
mu plus z sigma over root n, because we have |
|
|
|
585 |
|
00:48:18,200 --> 00:48:23,000 |
|
different statistics. It's x bar instead of x. Now |
|
|
|
586 |
|
00:48:23,000 --> 00:48:28,510 |
|
we are looking for these two values. Now let's |
|
|
|
587 |
|
00:48:28,510 --> 00:48:29,410 |
|
compute z-score. |
|
|
|
588 |
|
00:48:32,450 --> 00:48:36,830 |
|
The z-score for this point, which has area of 2.5% |
|
|
|
589 |
|
00:48:36,830 --> 00:48:41,930 |
|
below it, is the same as the z-score, but in the |
|
|
|
590 |
|
00:48:41,930 --> 00:48:48,670 |
|
opposite direction. If you remember, we got this |
|
|
|
591 |
|
00:48:48,670 --> 00:48:49,630 |
|
value, 1.96. |
|
|
|
592 |
|
00:48:52,790 --> 00:48:58,080 |
|
So my z-score is negative 1.96 to the left. and 1 |
|
|
|
593 |
|
00:48:58,080 --> 00:49:08,480 |
|
.9621 so now my x bar in the lower limit in this |
|
|
|
594 |
|
00:49:08,480 --> 00:49:17,980 |
|
side in the left side equals mu which is 368 minus |
|
|
|
595 |
|
00:49:17,980 --> 00:49:29,720 |
|
1.96 times sigma which is 15 Divide by root, 25. |
|
|
|
596 |
|
00:49:30,340 --> 00:49:34,980 |
|
So that's the value of the sample mean in the |
|
|
|
597 |
|
00:49:34,980 --> 00:49:39,740 |
|
lower limit, or lower bound. On the other hand, |
|
|
|
598 |
|
00:49:42,320 --> 00:49:49,720 |
|
expand our limit to our hand equals 316 plus 1.96 |
|
|
|
599 |
|
00:49:49,720 --> 00:49:56,100 |
|
sigma over root. Simple calculation will give this |
|
|
|
600 |
|
00:49:56,100 --> 00:49:56,440 |
|
result. |
|
|
|
601 |
|
00:49:59,770 --> 00:50:06,870 |
|
The first X bar for the lower limit is 362.12, the |
|
|
|
602 |
|
00:50:06,870 --> 00:50:10,050 |
|
other 373.1. |
|
|
|
603 |
|
00:50:11,450 --> 00:50:17,170 |
|
So again for this data, for this example, the mean |
|
|
|
604 |
|
00:50:17,170 --> 00:50:23,030 |
|
was, the population mean was 368, the population |
|
|
|
605 |
|
00:50:23,030 --> 00:50:26,310 |
|
has non-degradation of 15, we select a random |
|
|
|
606 |
|
00:50:26,310 --> 00:50:31,070 |
|
sample of size 25, Then we end with this result |
|
|
|
607 |
|
00:50:31,070 --> 00:50:41,110 |
|
that 95% of all sample means of sample size 25 are |
|
|
|
608 |
|
00:50:41,110 --> 00:50:44,810 |
|
between these two values. It means that we have |
|
|
|
609 |
|
00:50:44,810 --> 00:50:49,530 |
|
this big population and this population is |
|
|
|
610 |
|
00:50:49,530 --> 00:50:55,240 |
|
symmetric, is normal. And we know that The mean of |
|
|
|
611 |
|
00:50:55,240 --> 00:51:00,680 |
|
this population is 368 with sigma of 15. |
|
|
|
612 |
|
00:51:02,280 --> 00:51:08,320 |
|
We select from this population many samples. Each |
|
|
|
613 |
|
00:51:08,320 --> 00:51:11,600 |
|
one has size of 25. |
|
|
|
614 |
|
00:51:15,880 --> 00:51:20,940 |
|
Suppose, for example, we select 100 samples, 100 |
|
|
|
615 |
|
00:51:20,940 --> 00:51:27,260 |
|
random samples. So we end with different sample |
|
|
|
616 |
|
00:51:27,260 --> 00:51:27,620 |
|
means. |
|
|
|
617 |
|
00:51:33,720 --> 00:51:39,820 |
|
So we have 100 new sample means. In this case, you |
|
|
|
618 |
|
00:51:39,820 --> 00:51:46,320 |
|
can say that 95 out of these, 95 out of 100, it |
|
|
|
619 |
|
00:51:46,320 --> 00:51:52,560 |
|
means 95, one of these sample means. have values |
|
|
|
620 |
|
00:51:52,560 --> 00:52:01,720 |
|
between 362.12 and 373.5. And what's remaining? |
|
|
|
621 |
|
00:52:03,000 --> 00:52:07,940 |
|
Just five of these sample means would be out of |
|
|
|
622 |
|
00:52:07,940 --> 00:52:13,220 |
|
this interval either below 362 or above the upper |
|
|
|
623 |
|
00:52:13,220 --> 00:52:17,720 |
|
limit. So you are 95% sure that |
|
|
|
624 |
|
00:52:21,230 --> 00:52:24,350 |
|
The sample means lies between these two points. |
|
|
|
625 |
|
00:52:25,410 --> 00:52:29,470 |
|
So, 5% of the sample means will be out. Make |
|
|
|
626 |
|
00:52:29,470 --> 00:52:37,510 |
|
sense? Imagine that I have selected 200 samples. |
|
|
|
627 |
|
00:52:40,270 --> 00:52:46,330 |
|
Now, how many X bar will be between these two |
|
|
|
628 |
|
00:52:46,330 --> 00:52:54,140 |
|
values? 95% of these 200. So how many 95%? How |
|
|
|
629 |
|
00:52:54,140 --> 00:52:56,060 |
|
many means in this case? |
|
|
|
630 |
|
00:52:58,900 --> 00:53:04,600 |
|
95% out of 200 is 190. |
|
|
|
631 |
|
00:53:05,480 --> 00:53:12,200 |
|
190. Just multiply. 95 multiplies by 200. It will |
|
|
|
632 |
|
00:53:12,200 --> 00:53:13,160 |
|
give 190. |
|
|
|
633 |
|
00:53:22,740 --> 00:53:29,860 |
|
values between 362 |
|
|
|
634 |
|
00:53:29,860 --> 00:53:38,680 |
|
.12 and 373.8. Take any value, will have any value |
|
|
|
635 |
|
00:53:38,680 --> 00:53:42,200 |
|
between these two values. |
|
|
|
636 |
|
00:53:48,000 --> 00:53:51,180 |
|
In the previous one, we assumed that the |
|
|
|
637 |
|
00:53:51,180 --> 00:53:55,480 |
|
population is normal distribution. If we go back a |
|
|
|
638 |
|
00:53:55,480 --> 00:54:01,060 |
|
little bit here, we assumed the population is |
|
|
|
639 |
|
00:54:01,060 --> 00:54:05,600 |
|
normal. If the population is normal, then the |
|
|
|
640 |
|
00:54:05,600 --> 00:54:07,840 |
|
standard distribution of X bar is also normal |
|
|
|
641 |
|
00:54:07,840 --> 00:54:10,840 |
|
distribution with minimum standard deviation of |
|
|
|
642 |
|
00:54:10,840 --> 00:54:16,300 |
|
sigma over R. Now, the second case is Suppose we |
|
|
|
643 |
|
00:54:16,300 --> 00:54:18,340 |
|
are looking for the sampling distribution of the |
|
|
|
644 |
|
00:54:18,340 --> 00:54:23,100 |
|
sample mean if the population is not normal. You |
|
|
|
645 |
|
00:54:23,100 --> 00:54:25,320 |
|
don't have any information about your population, |
|
|
|
646 |
|
00:54:26,160 --> 00:54:29,940 |
|
and you are looking for the sampling distribution |
|
|
|
647 |
|
00:54:29,940 --> 00:54:37,800 |
|
of X bar. In this case, we can apply a |
|
|
|
648 |
|
00:54:37,800 --> 00:54:42,040 |
|
new theorem called central limit theorem. This |
|
|
|
649 |
|
00:54:42,040 --> 00:54:47,120 |
|
theorem says that Even if the population is not |
|
|
|
650 |
|
00:54:47,120 --> 00:54:54,240 |
|
normally distributed. In this case, the sampling |
|
|
|
651 |
|
00:54:54,240 --> 00:54:59,400 |
|
distribution of the sample means will be not |
|
|
|
652 |
|
00:54:59,400 --> 00:55:03,480 |
|
exactly normal, but approximately normally as long |
|
|
|
653 |
|
00:55:03,480 --> 00:55:07,320 |
|
as the sample size is large enough. I mean if you |
|
|
|
654 |
|
00:55:07,320 --> 00:55:10,100 |
|
select a random sample from unknown population and |
|
|
|
655 |
|
00:55:10,100 --> 00:55:12,600 |
|
that population is not symmetric, is not normal. |
|
|
|
656 |
|
00:55:13,880 --> 00:55:17,560 |
|
In this case, you can say that the sampling |
|
|
|
657 |
|
00:55:17,560 --> 00:55:24,100 |
|
distribution of X bar is approximately normal as |
|
|
|
658 |
|
00:55:24,100 --> 00:55:27,940 |
|
long as the sample size is large enough, with the |
|
|
|
659 |
|
00:55:27,940 --> 00:55:31,280 |
|
same population, with the same mean, population |
|
|
|
660 |
|
00:55:31,280 --> 00:55:34,140 |
|
mean is mu, and same standard deviation sigma over |
|
|
|
661 |
|
00:55:34,140 --> 00:55:40,960 |
|
root N. So now, there is a condition here. If the |
|
|
|
662 |
|
00:55:40,960 --> 00:55:43,520 |
|
population is not normal, I mean if you select a |
|
|
|
663 |
|
00:55:43,520 --> 00:55:46,960 |
|
random sample from unknown population, the |
|
|
|
664 |
|
00:55:46,960 --> 00:55:51,480 |
|
condition is N is large. If N is large enough, |
|
|
|
665 |
|
00:55:52,160 --> 00:55:56,580 |
|
then you can apply this theorem, central limit |
|
|
|
666 |
|
00:55:56,580 --> 00:56:00,160 |
|
theorem, which says that as the sample size goes |
|
|
|
667 |
|
00:56:00,160 --> 00:56:03,640 |
|
larger and larger, or gets larger and larger, then |
|
|
|
668 |
|
00:56:03,640 --> 00:56:06,860 |
|
the standard distribution of X bar is |
|
|
|
669 |
|
00:56:06,860 --> 00:56:14,090 |
|
approximately normal in this Again, look at the |
|
|
|
670 |
|
00:56:14,090 --> 00:56:19,630 |
|
blue curve. Now, this one looks like skewed |
|
|
|
671 |
|
00:56:19,630 --> 00:56:20,850 |
|
distribution to the right. |
|
|
|
672 |
|
00:56:24,530 --> 00:56:28,730 |
|
Now, as the sample gets large enough, then it |
|
|
|
673 |
|
00:56:28,730 --> 00:56:33,470 |
|
becomes normal. So, the sample distribution |
|
|
|
674 |
|
00:56:33,470 --> 00:56:37,350 |
|
becomes almost normal regardless of the shape of |
|
|
|
675 |
|
00:56:37,350 --> 00:56:41,570 |
|
the population. I mean if you sample from unknown |
|
|
|
676 |
|
00:56:41,570 --> 00:56:46,590 |
|
population, and that one has either right skewed |
|
|
|
677 |
|
00:56:46,590 --> 00:56:52,130 |
|
or left skewed, if the sample size is large, then |
|
|
|
678 |
|
00:56:52,130 --> 00:56:55,810 |
|
the sampling distribution of X bar becomes almost |
|
|
|
679 |
|
00:56:55,810 --> 00:57:01,530 |
|
normal distribution regardless of the… so that's |
|
|
|
680 |
|
00:57:01,530 --> 00:57:06,830 |
|
the central limit theorem. So again, if the |
|
|
|
681 |
|
00:57:06,830 --> 00:57:10,980 |
|
population is not normal, The condition is only |
|
|
|
682 |
|
00:57:10,980 --> 00:57:15,360 |
|
you have to select a large sample. In this case, |
|
|
|
683 |
|
00:57:15,960 --> 00:57:19,340 |
|
the central tendency mu of X bar is same as mu. |
|
|
|
684 |
|
00:57:20,000 --> 00:57:24,640 |
|
The variation is also sigma over root N. |
|
|
|
685 |
|
00:57:28,740 --> 00:57:32,120 |
|
So again, standard distribution of X bar becomes |
|
|
|
686 |
|
00:57:32,120 --> 00:57:38,620 |
|
normal as N. The theorem again says If we select a |
|
|
|
687 |
|
00:57:38,620 --> 00:57:42,500 |
|
random sample from unknown population, then the |
|
|
|
688 |
|
00:57:42,500 --> 00:57:44,560 |
|
standard distribution of X part is approximately |
|
|
|
689 |
|
00:57:44,560 --> 00:57:53,580 |
|
normal as long as N gets large enough. Now the |
|
|
|
690 |
|
00:57:53,580 --> 00:57:57,100 |
|
question is how large is large enough? |
|
|
|
691 |
|
00:58:00,120 --> 00:58:06,530 |
|
There are two cases, or actually three cases. For |
|
|
|
692 |
|
00:58:06,530 --> 00:58:11,310 |
|
most distributions, if you don't know the exact |
|
|
|
693 |
|
00:58:11,310 --> 00:58:18,670 |
|
shape, n above 30 is enough to use or to apply |
|
|
|
694 |
|
00:58:18,670 --> 00:58:22,290 |
|
that theorem. So if n is greater than 30, it will |
|
|
|
695 |
|
00:58:22,290 --> 00:58:24,650 |
|
give a standard distribution that is nearly |
|
|
|
696 |
|
00:58:24,650 --> 00:58:29,070 |
|
normal. So if my n is large, it means above 30, or |
|
|
|
697 |
|
00:58:29,070 --> 00:58:33,450 |
|
30 and above this. For fairly symmetric |
|
|
|
698 |
|
00:58:33,450 --> 00:58:35,790 |
|
distribution, I mean for nearly symmetric |
|
|
|
699 |
|
00:58:35,790 --> 00:58:38,630 |
|
distribution, the distribution is not exactly |
|
|
|
700 |
|
00:58:38,630 --> 00:58:42,910 |
|
normal, but approximately normal. In this case, N |
|
|
|
701 |
|
00:58:42,910 --> 00:58:46,490 |
|
to be large enough if it is above 15. So, N |
|
|
|
702 |
|
00:58:46,490 --> 00:58:48,770 |
|
greater than 15 will usually have same |
|
|
|
703 |
|
00:58:48,770 --> 00:58:50,610 |
|
distribution as almost normal. |
|
|
|
704 |
|
00:58:55,480 --> 00:58:57,840 |
|
For normal population, as we mentioned, of |
|
|
|
705 |
|
00:58:57,840 --> 00:59:00,740 |
|
distributions, the semantic distribution of the |
|
|
|
706 |
|
00:59:00,740 --> 00:59:02,960 |
|
mean is always. |
|
|
|
707 |
|
00:59:06,680 --> 00:59:12,380 |
|
Okay, so again, there are three cases. For most |
|
|
|
708 |
|
00:59:12,380 --> 00:59:16,280 |
|
distributions, N to be large, above 30. In this |
|
|
|
709 |
|
00:59:16,280 --> 00:59:20,460 |
|
case, the distribution is nearly normal. For |
|
|
|
710 |
|
00:59:20,460 --> 00:59:24,300 |
|
fairly symmetric distributions, N above 15 gives |
|
|
|
711 |
|
00:59:24,660 --> 00:59:28,960 |
|
almost normal distribution. But if the population |
|
|
|
712 |
|
00:59:28,960 --> 00:59:32,400 |
|
by itself is normally distributed, always the |
|
|
|
713 |
|
00:59:32,400 --> 00:59:35,800 |
|
sample mean is normally distributed. So that's the |
|
|
|
714 |
|
00:59:35,800 --> 00:59:37,300 |
|
three cases. |
|
|
|
715 |
|
00:59:40,040 --> 00:59:47,480 |
|
Now for this example, suppose we have a |
|
|
|
716 |
|
00:59:47,480 --> 00:59:49,680 |
|
population. It means we don't know the |
|
|
|
717 |
|
00:59:49,680 --> 00:59:52,900 |
|
distribution of that population. And that |
|
|
|
718 |
|
00:59:52,900 --> 00:59:57,340 |
|
population has mean of 8. Standard deviation of 3. |
|
|
|
719 |
|
00:59:58,200 --> 01:00:01,200 |
|
And suppose a random sample of size 36 is |
|
|
|
720 |
|
01:00:01,200 --> 01:00:04,780 |
|
selected. In this case, the population is not |
|
|
|
721 |
|
01:00:04,780 --> 01:00:07,600 |
|
normal. It says A population, so you don't know |
|
|
|
722 |
|
01:00:07,600 --> 01:00:12,340 |
|
the exact distribution. But N is large. It's above |
|
|
|
723 |
|
01:00:12,340 --> 01:00:15,060 |
|
30, so you can apply the central limit theorem. |
|
|
|
724 |
|
01:00:15,920 --> 01:00:20,380 |
|
Now we ask about what's the probability that a |
|
|
|
725 |
|
01:00:20,380 --> 01:00:25,920 |
|
sample means. is between what's the probability |
|
|
|
726 |
|
01:00:25,920 --> 01:00:29,240 |
|
that the same element is between these two values. |
|
|
|
727 |
|
01:00:32,180 --> 01:00:36,220 |
|
Now, the difference between this lecture and the |
|
|
|
728 |
|
01:00:36,220 --> 01:00:39,800 |
|
previous ones was, here we are interested in the |
|
|
|
729 |
|
01:00:39,800 --> 01:00:44,440 |
|
exponent of X. Now, even if the population is not |
|
|
|
730 |
|
01:00:44,440 --> 01:00:47,080 |
|
normally distributed, the central limit theorem |
|
|
|
731 |
|
01:00:47,080 --> 01:00:51,290 |
|
can be abused because N is large enough. So now, |
|
|
|
732 |
|
01:00:51,530 --> 01:00:57,310 |
|
the mean of X bar equals mu, which is eight, and |
|
|
|
733 |
|
01:00:57,310 --> 01:01:02,170 |
|
sigma of X bar equals sigma over root N, which is |
|
|
|
734 |
|
01:01:02,170 --> 01:01:07,150 |
|
three over square root of 36, which is one-half. |
|
|
|
735 |
|
01:01:11,150 --> 01:01:17,210 |
|
So now, the probability of X bar greater than 7.8, |
|
|
|
736 |
|
01:01:17,410 --> 01:01:21,890 |
|
smaller than 8.2, Subtracting U, then divide by |
|
|
|
737 |
|
01:01:21,890 --> 01:01:26,210 |
|
sigma over root N from both sides, so 7.8 minus 8 |
|
|
|
738 |
|
01:01:26,210 --> 01:01:30,130 |
|
divided by sigma over root N. Here we have 8.2 |
|
|
|
739 |
|
01:01:30,130 --> 01:01:33,230 |
|
minus 8 divided by sigma over root N. I will end |
|
|
|
740 |
|
01:01:33,230 --> 01:01:38,150 |
|
with Z between minus 0.4 and 0.4. Now, up to this |
|
|
|
741 |
|
01:01:38,150 --> 01:01:43,170 |
|
step, it's in U, for chapter 7. Now, Z between |
|
|
|
742 |
|
01:01:43,170 --> 01:01:47,630 |
|
minus 0.4 up to 0.4, you have to go back. And use |
|
|
|
743 |
|
01:01:47,630 --> 01:01:51,030 |
|
the table in chapter 6, you will end with this |
|
|
|
744 |
|
01:01:51,030 --> 01:01:54,530 |
|
result. So the only difference here, you have to |
|
|
|
745 |
|
01:01:54,530 --> 01:01:55,790 |
|
use sigma over root N. |
|
|
|
|