PL9fwy3NUQKwaIc7KWc8buW344941GStQL/FS8UHlZfJpc.srt

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Last time, we talked about the types of samples

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and introduced 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, we also 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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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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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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case, if you, so that's your frame. Now, if you

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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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reasons, in this case, you will have coverage as

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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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Next is called non-response error. Suppose I 

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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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error refers to non-responsive, so you have to

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follow up, follow up. It means maybe sometimes you

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need to clarify the question you have in your 

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questionnaire so that the respondent understand

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what do you mean exactly by that question

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otherwise if you don't follow up it means it may

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be there is an error, and that error is called non

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-response. The other type of error is called

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measurement error, which is one of the most 

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important errors, and we have to avoid.

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It's called measurement error. Good questions

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elicit good responses. It means, suppose, for

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example, my question is, I feel this candidate is 

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good for us. What do you think? It's my question.

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I feel this candidate, candidate A, whatever he

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is, is good for us. What do you think? For sure

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there's abundant answer will be yes. I agree with 

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you. So that means you design the question in the 

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way that you will know they respond directly, that 

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he will answer yes or no depends on your design of 

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the question. So it means leading question. So 

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measurement error. So but if we have good

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questions, just ask any question for the 

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respondent, and let him or let his answer based on

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what exactly he thinks about it. So don't force 

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the respondent to answer the question in the

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direction you want to be. Otherwise you will get 

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something called Measurement Error. Do you think?

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Give me an example of Measurement Error. 

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Give me an example of Measurement Error. Just ask

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a question in a way that the respondent will 

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answer, I mean, his answer will be the same as you 

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think about it.

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Maybe I like coffee, do you like coffee or tea? So 

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maybe he will go with your answer. In this case

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it's measurement. Another example.

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Exactly.

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So it means that if you design a question in the 

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way that you will get the same answer you think

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about it, it means that you will have something 

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called measurement error. The last type is 

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sampling error. Sampling error always happens,

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always exists. For example, suppose you are around 

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50 students in this class. Suppose I select

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randomly 20 of you, and I am interested suppose in

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your age. Maybe for this sample.

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I will get an average of your age of 19 years

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someone

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select another sample from the same population

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with the same size maybe

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the average of your age is not equal to 19 years 

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maybe 19 years, 3 months

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Someone else maybe also select the same number of 

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students, but the average of the class might be 20

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years. So, the first one, second tier, each of them 

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has different sample statistics. I mean different

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sample means. This difference or this error

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actually is called sampling error and always

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happens. So, now we have five types of errors. One

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is called coverage error. In this case, you have

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a problem with the frame. The other type is called

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non-response error. It means you have a problem with

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following up. Measurement error. It means you have

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bad questionnaire design. The last type is called

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sampling error, and this one always happens and

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actually we would like to have this error. I mean

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this sampling error as small as possible. So, these are

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the steps you have to follow up when you design

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the questionnaire. 

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So again, for these types of errors, the first one 

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coverage error, or selection bias. This type of 

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error exists if some groups are excluded from the 

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frame, and have no chance of being selected. That's

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the first type of error, coverage error. So, it

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means there is a problem on the population frame.

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Non-response error bias. It means people who don't 

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respond may be different from those who do 

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respond. For example, suppose I have a sample of 

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tennis students. 

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And I got responses from number two, number five, 

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and number 10. So, I have these points of view for 

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these three students. Now, the other seven students 

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might be they have different opinions. So, the only 

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thing you have, the opinions of just the three, 

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and maybe the rest have different opinions, it 

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means in this case you will have something called

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non-responsiveness. Or the same as we said before,

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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 a 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. Finally, 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 a 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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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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of this. So x bar minus the mean of not x, x bar,

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then divided by sigma x bar. So this is my new,

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the score.

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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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we select a sample from is normal, I mean if the

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population is normally distributed, then you

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select a random sample of that population, it

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makes sense that the sample is also normal, so any

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statistic is computed from that sample is also

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normally distributed, so it makes sense. If the

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population is normal, then the shape is also

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normal. But if the population is unknown, you

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don't have any information about the underlying

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population, then you cannot say it's normal unless

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you have certain condition that we'll talk about

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maybe after 30 minutes. So, exactly, if the

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population is normal, then the shape is also

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normal, but otherwise, we have to think about it.

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This is the first question. Now, there are two

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unknowns in this equation. We have to know the

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mean, Or x bar, so the mean of x bar is not given,

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the mean means the center. So the second question,

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what's the center of the distribution? In this

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case, the mean of x bar. So we are looking at

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what's the mean of x bar. The third question is

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sigma x bar is also unknown, spread. Now shape,

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center, spread, these are characteristics, these

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characteristics in this case sampling

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distribution, exactly which is called sampling

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distribution. So by sampling distribution we mean

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that, by sampling distribution, we mean that you

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have to know the center of distribution, I mean

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the mean of the statistic you are interested in.

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Second, the spread or the variability of the

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sample statistic also you are interested in. In

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addition to that, you have to know the shape of

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the statistic. So three things we have to know,

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center, spread and shape. So that's what we'll

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talk about now. So now sampling distribution is a

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distribution of all of the possible values of a

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sample statistic. This sample statistic could be

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sample mean, could be sample variance, could be

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sample proportion, because any population has

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mainly three characteristics, mean, standard

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deviation, and proportion.

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So again, a sampling distribution is a

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distribution of all of the possible values of a

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sample statistic or a given size sample selected

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from a population. For example, suppose you sample

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50 students from your college regarding their mean

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GPA. GPA means Graduate Point Average. Now, if you

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obtain many different samples of size 50, you will

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compute a different mean for each sample. As I

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mentioned here, I select a sample the same sizes,

282
00:21:43,540 --> 00:21:47,580
but we obtain different sample statistics, I mean

283
00:21:47,580 --> 00:21:54,260
different sample means. We are interested in the

284
00:21:54,260 --> 00:21:59,760
distribution of all potential mean GBA We might

285
00:21:59,760 --> 00:22:04,040
calculate for any given sample of 50 students. So

286
00:22:04,040 --> 00:22:09,440
let's focus into these values. So we have again a

287
00:22:09,440 --> 00:22:14,580
random sample of 50 sample means. So we have 1, 2,

288
00:22:14,700 --> 00:22:18,480
3, 4, 5, maybe 50, 6, whatever we have. So select

289
00:22:18,480 --> 00:22:22,660
a random sample of size 20. Maybe we repeat this

290
00:22:22,660 --> 00:22:28,590
sample 10 times. So we end with 10. different

291
00:22:28,590 --> 00:22:31,650
values of the simple means. Now we have new ten

292
00:22:31,650 --> 00:22:38,130
means. Now the question is, what's the center of

293
00:22:38,130 --> 00:22:42,590
these values, I mean for the means? What's the

294
00:22:42,590 --> 00:22:46,250
spread of the means? And what's the shape of the

295
00:22:46,250 --> 00:22:50,310
means? So these are the mainly three questions.

296
00:22:53,510 --> 00:22:56,810
For example, let's get just simple example and

297
00:22:56,810 --> 00:23:04,370
that we have only population of size 4. In the

298
00:23:04,370 --> 00:23:09,950
real life, the population size is much bigger than

299
00:23:09,950 --> 00:23:14,970
4, but just for illustration.

300
00:23:17,290 --> 00:23:20,190
Because size 4, I mean if the population is 4,

301
00:23:21,490 --> 00:23:24,950
it's a small population. So we can take all the

302
00:23:24,950 --> 00:23:27,610
values and find the mean and standard deviation.

303
00:23:28,290 --> 00:23:31,430
But in reality, we have more than that. So this

304
00:23:31,430 --> 00:23:37,390
one just for as example. So let's suppose that we

305
00:23:37,390 --> 00:23:42,790
have a population of size 4. So n equals 4.

306
00:23:46,530 --> 00:23:54,030
And we are interested in the ages. And suppose the

307
00:23:54,030 --> 00:23:58,930
values of X, X again represents H,

308
00:24:00,690 --> 00:24:01,810
and the values we have.

309
00:24:06,090 --> 00:24:08,930
So these are the four values we have.

310
00:24:12,050 --> 00:24:16,910
Now simple calculation will

311
00:24:16,910 --> 00:24:19,910
give you the mean, the population mean.

312
00:24:25,930 --> 00:24:30,410
Just add these values and divide by the operation

313
00:24:30,410 --> 00:24:35,410
size, we'll get 21 years. And sigma, as we

314
00:24:35,410 --> 00:24:39,450
mentioned in chapter three, square root of this

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

317
00:24:49,450 --> 00:24:54,470
results. Now if you look at distribution of these

318
00:24:54,470 --> 00:24:58,430
values, Because as I mentioned, we are looking for

319
00:24:58,430 --> 00:25:03,810
center, spread, and shape. The center is 21 of the

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

322
00:25:14,770 --> 00:25:15,250
once.

323
00:25:17,930 --> 00:25:22,350
I mean, we have only one 18, so 18 divided one

324
00:25:22,350 --> 00:25:29,830
time over 425. 20% represent also 25%, the same as

325
00:25:29,830 --> 00:25:33,030
for 22 or 24. In this case, we have something

326
00:25:33,030 --> 00:25:37,530
called uniform distribution. In this case, the

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:

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
an 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 a bell

459
00:37:24,660 --> 00:37:31,290
shape. If we select our sample from a normal

460
00:37:31,290 --> 00:37:37,850
population with a mean equal to the population mean

461
00:37:37,850 --> 00:37:40,530
and standard deviation of the standard error equals

462
00:37:40,530 --> 00:37:48,170
sigma over the square root of n. So now, let's talk

463
00:37:48,170 --> 00:37:53,730
about the 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
the mean of X bar equals mu and sigma of X bar equals

472
00:38:27,070 --> 00:38:35,470
sigma over the square root of n. So again, if we sample from a 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 a 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 the square root of 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 the z-score, generally speaking, is 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 a standard normal

487
00:39:41,620 --> 00:39:48,020
distribution with a mean of zero and a variance of one. In

488
00:39:48,020 --> 00:39:52,860
this case, we are looking for the sampling

489
00:39:52,860 --> 00:39:59,350
-distribution of X bar. So Z equals 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 statistics. In the

492
00:40:10,770 --> 00:40:15,770
first one, we have x, for example, which 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 scores.

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 the square root of n, we will end with

500
00:40:51,240 --> 00:40:52,600
this equation 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 is the sample mean, mu is the

505
00:41:13,310 --> 00:41:15,990
population mean, sigma is the 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 over the square root of n.

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 expectation 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, an unbiased

521
00:42:40,620 --> 00:42:45,490
estimator X bar is called an 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 a good estimator of Mu. Because on average,

526
00:43:05,430 --> 00:43:08,230
The 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 a good 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
a normal distribution here with the population mean Mu

530
00:43:27,510 --> 00:43:30,550
and a 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, it 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 a sample size of 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 to 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 a 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 a smaller

550
00:44:59,860 --> 00:45:06,240
sample size. The red one for a 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 a mean of 368 and sigma of 15 and suppose

557
00:45:49,500 --> 00:45:52,900
we select a random sample of a size of 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 a 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 we can compute these two values.

571
00:47:03,820 --> 00:47:11,440
The problem says that the average is 368

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 x bar minus mu divided by

578
00:47:45,850 --> 00:47:50,850
sigma over the square root of n. Now cross multiplication, you

579
00:47:50,850 --> 00:47:55,970
will get x bar minus mu equals z sigma over the square root

580
00:47:55,970 --> 00:48:01,500
of n. That means x bar equals mu plus z sigma over

581
00:48:01,500 --> 00:48:04,440
the square root of 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 the square root of 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 an 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 on 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 the square root of 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 the other hand equals 316 plus 1.96

599
00:49:49,720 --> 00:49:56,100
sigma over the square root of n. 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 is 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 a standard deviation 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, it's normal. And we know that The mean of

611
00:50:55,240 --> 00:51:00,680
this population is 368 with a 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 a 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 mean 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 you 190.

633
00:53:22,740 --> 00:53:29,860
values between 362



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.