PL9fwy3NUQKwaIc7KWc8buW344941GStQL/I7SEpdLlzFg.srt

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Eventually I will give some practice problems for

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chapter eight. Generally speaking, there are three

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types of questions. The first type, multiple

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choice, so MCQ questions.

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The other type of problems will be true or false.

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Part B, Part C, three response problems.

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So three types of questions. Multiple choice, we

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have four answers. You have to select the correct one.

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True or false problems. And the last part, free

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response problems. Here we'll talk about one of

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these. I will cover multiple choice questions as

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well as true and false. Let's start with number

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one for multiple choice. The width of a confidence

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interval estimate for a proportion will be Here we

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are talking about the width of a confidence

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interval.

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Estimates for a proportion will be narrower for 99

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% confidence than for a 9%. For 95 confidence? No,

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because as we know that as the confidence level

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increases, the width becomes wider. So A is

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incorrect. Is this true? B. Wider for sample size

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of 100 than for a sample size of 50? False,

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because as sample size increases, the sampling

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error goes down. That means the width of the

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interval becomes smaller and smaller. Yes, for N.

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Part C. Narrower for 90% confidence, then for 95%

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confidence. That's correct. So C is the correct

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answer. Part D. Narrower when the sampling

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proportion is 50%. is incorrect because if we have

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smaller than 50%, we'll get smaller confidence,

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smaller weight of the confidence. So C is the

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correct answer. Any question? So C is the correct

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answer because as C level increases, the

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confidence interval becomes wider.

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Let's move to the second one.

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A 99% confidence interval estimate can be

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interpreted to mean that. Let's look at the

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interpretation of the 99% confidence interval.

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Part A.

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If all possible samples are taken and confidence

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interval estimates are developed, 99% of them

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would include the true population mean somewhere

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within their interval.

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Here we are talking about the population mean. It

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says that 99% of them of these intervals would

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include the true population mean somewhere within

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their interval. It's correct. Why false? Why is it

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false? This is the correct answer, because it's

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mentioned that 99% of these confidence intervals

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will contain the true population mean somewhere

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within their interval. So A is correct. Let's look

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at B. B says we have 99% confidence that we have

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selected a sample whose interval does include the

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population mean. Also, this one is correct. Again,

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it's mentioned that 99% confidence that we have

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selected a sample whose interval does include. So

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it's correct. So C is both of the above and D none

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of the above. So C is the correct answer. So

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sometimes maybe there is only one answer. Maybe in

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other problems, it might be two answers are

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correct. So for this one, B and C. I'm sorry, A

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and B are correct, so C is the correct answer.

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Number three, which of the following is not true

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about the student's T distribution? Here, we are

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talking about the not true statement about the

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student T distribution, A. It has more data in the

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tails and less in the center than does the normal

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distribution. That's correct because we mentioned

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last time that the T distribution, the tail is fatter

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than the Z normal. So that means it has more data

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in the tails and less data in the center. So

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that's correct.

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It is used to construct confidence intervals for

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the population mean when the population standard

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deviation is known. No, we use z instead of t, so

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this one is incorrect about t. It is bell-shaped

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and symmetrical, so that's true, so we are looking

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for the incorrect statement. D, as the number of

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degrees of freedom increases,

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the T distribution approaches the normal. That's

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true. So which one? B. So B is incorrect. So

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number four. Extra.

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Can you explain the average total compensation of

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CEOs in the service industry? Data were randomly

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collected from 18 CEOs and 19 employees. 97%

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confidence interval was calculated to be $281,

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$260,

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$5836,

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and $180. Which of the following interpretations

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is correct? Part number A. It says 97% of the

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sample data compensation value between these two

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values, correct or incorrect statement. Because it

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says 97% of the sample data. For the confidence

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interval, we are looking for the average, not

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for the population, not for the sample. So A is

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incorrect. Because A, it says here 97% of the

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sampling total. Sample total, we are looking for

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the average of the population. So A is incorrect

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statement. B, we are 97% confident that the mean

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of the sample. So it's false. Because the

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confidence about the entire population is about

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the population mean. So B is incorrect. C. In the

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population of the service industry, here we have

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97% of them will have a total compensation. Also,

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this one is incorrect because it mentions in the

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population. Here we are talking about total, but

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we are looking for the average. Now, part D. We

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are 97% confident that the average total

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compensation is between these two values. So this

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is the correct statement. So D is the

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correct statement. So for the confidence interval,

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we are looking for the population, number one. Number

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two, the average of that population. So D is the

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correct answer. Let's go back to part A. In part

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A, it says sample total. So this is incorrect.

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Next one. The mean of the sample. We are looking

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for the mean of the population. So B is incorrect.

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Part C. It mentions here the population, but total. So

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this one is incorrect. Finally here, we are 97%

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confident that the average total. So this one is

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true. So here we have the population and the

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average of that population. So it makes sense that

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this is the correct answer.

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Number five.

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Number five, a confidence interval. Confidence

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interval was used to estimate the proportion of

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statistics students that are females. A random

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sample of 72 statistics students generated the

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following 90% confidence interval, 0.438

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and 0.640.

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Based on the interval above the population

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proportion of females equals to 0.6. So here we

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have a confidence interval for the female proportion

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ranges between 0.438 up to 0.642. Based on this

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interval. Is the population proportion of females

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equal to 60%?

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So here we have from this point all the way up to

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0.6. Is the population proportion of females equal

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to 0.6? No. The answer is no, but now what?

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Number A. No, and we are 90% sure of it. No, the

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proportion is 54.17. See, maybe 60% is a

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believable value of the population proportion based on

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the information about it. He said yes, and we are 90%

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sure of it. So which one is correct? Farah. Which

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one is correct?

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B says the proportion is 54. 54 if we take the

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average of these two values, the answer is 54. But

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the true proportion is not the average of the two

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endpoints.

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So B is incorrect.

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If you look at A, the answer is no. And we

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mentioned before that this interval may or may not

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contain the true proportion, so A is incorrect.

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Now C, maybe. So C is the correct statement, maybe

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60% is a believable value of the population

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proportion based on the information about it. So C is

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the correct answer. A6, number six.

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Number six.

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So up to this point, we have the same information

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for the previous problem. Using the information

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about what total size sample would be necessary if

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we wanted to estimate the true proportion within

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plus or minus 0.108 using 95%

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confidence. Now here we are looking for the sample

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size that is required to estimate the true

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proportion to be within plus or minus 8% using

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95% confidence.

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The formula first, n equals z squared times p times 1 minus p,

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divided by e squared.

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Now, p is not given. So in this case, either we

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use a prior sample in order to estimate the sample

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proportion or use p to be 0.5. So in this case

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we have to use p to be 1 half. If you remember last

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time I gave you this equation. Z alpha over 2

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divided by 2 squared. So we have this equation.

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Because p is not given, just use p to be 1 half.

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Or you may use this equation. shortcut formula. In

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this case, here we are talking about 95%. So

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what's the value of Z? 196. 2 times E.

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E is 8%. So 196 divided by 2 times E, the quantity

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squared. Now the answer to this problem is 150. So

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approximately 150.

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So 150 is the correct answer. So again, here we

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used p to be 1 half because P is not given. And

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simple calculation results in 150 for the sample

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size. So P is the correct answer, 7.

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Number seven.

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Number seven. When determining the sample size

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necessary for estimating the true population

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mean, which factor is not considered when sampling

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with replacement? Now here, if you remember the

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formula for the sample size.

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Now, which factor is not considered when sampling

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without replacement? Now, the population

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size, the population size is not in this equation,

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correct answer is A, nine.

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A major department store chain is interested in

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estimating the average amount each credit and

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customers spent on their first visit to the

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chain's new store in the mall. 15 credit cards

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accounts were randomly sampled and analyzed with

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the following results. So here we have this

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information about the 15 data points. We have x

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bar.

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of $50.5.

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And S squared, the sample variance is 400.

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Construct a 95% confidence interval for the average

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amount its credit card customer spent on their

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first visit to the chain. It's a new store. It's

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in the mall, assuming the amount spent follows a

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normal distribution.

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In this case, we should use T instead of Z. So the

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formula should be X bar plus or minus T, alpha

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over 2S over root N.

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So X bar is 50.5. T, we should use the T table. In

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this case, here we are talking about 95%.

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So that means alpha is 5%, alpha over 2, 0.025.

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So now we are looking for 2.025, and degrees

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of freedom. N is 15. It says that 15 credit cards.

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So 15 credit cards accounts for random samples. So

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N equals 15. So since N is 15, degrees of freedom

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is 14. Now we may use the normal, the T table in

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order to find the value of T in 

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the upper tier actually. So what's the value if

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you have the table? So look at degrees of freedom

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14 under the probability of 0.025.

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So again, we are looking for degrees of freedom

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equal 14 under 2.5%.

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0.5 plus or minus 2

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.1448. S squared is given 400. Take the square root of

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this quantity 20 over root n over root 15. And the

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answer, just simple calculation will give

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this result, so D is the correct answer. So the

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answer should be 50.5 plus or minus 11.08. So D is

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the correct answer. So this one is straightforward

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calculation, gives part D to be the correct

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answer. Any question?

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11, 10?

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Private colleges and universities rely on money

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contributed by individuals and corporations for

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their operating expenses. Much of this money is

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put into a fund called an endowment, and the

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college spends only the interest earned by the

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fund. Now, here we have a recent It says that a

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recent survey of eight private colleges in the

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United States revealed the following endowment in

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millions of dollars. So we have this data. So it's

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raw data. Summary statistics yield the export to be

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180.

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So export.

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Now if this information is not given, you have to

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compute the average and standard deviation by the

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equations we know. But here, the mean and standard

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deviation are given. So just use this information

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anyway. Calculate a 95% confidence interval for the

283
00:24:27,480 --> 00:24:30,140
mean endowment of all private colleges in the

284
00:24:30,140 --> 00:24:34,520
United States, assuming a normal distribution for

285
00:24:34,520 --> 00:24:39,300
the endowment. Here we have 95%.

286
00:24:39,300 --> 00:24:42,600
Now

287
00:24:42,600 --> 00:24:48,480
what's the sample size? It says that eight. So N

288
00:24:48,480 --> 00:24:53,900
is eight. So again, simple calculation. So 

289
00:24:53,900 --> 00:24:59,680
explore, plus or minus T, S over root N. So use

290
00:24:59,680 --> 00:25:04,200
the same idea for the previous one. And the answer

291
00:25:04,200 --> 00:25:10,420
for number 10 is part D. So D is the correct

292
00:25:10,420 --> 00:25:17,380
answer. So again, for eleven, D is the correct

293
00:25:17,380 --> 00:25:22,680
answer. For ten, D is the correct answer. Next. So

294
00:25:22,680 --> 00:25:26,280
this one is similar to the one we just did.

295
00:25:30,660 --> 00:25:31,260
Eleven.

296
00:25:47,140 --> 00:25:51,140
Here it says that rather than examine the records

297
00:25:51,140 --> 00:25:56,220
of all students, the dean took a random sample of

298
00:25:56,220 --> 00:26:01,340
size 200. So we have a large university. Here we

299
00:26:01,340 --> 00:26:04,860
took a representative sample of size 200.

300
00:26:26,980 --> 00:26:31,900
How many students would need to be assembled? It

301
00:26:31,900 --> 00:26:34,540
says that if the dean wanted to estimate the

302
00:26:34,540 --> 00:26:38,040
proportion of all students, the saving financial

303
00:26:38,040 --> 00:26:46,100
aid to within 3% with 99% probability. How many

304
00:26:46,100 --> 00:26:51,620
students would need to be sampled? So we have the

305
00:26:51,620 --> 00:26:56,920
formula, if you remember, n equals z y 1 minus y

306
00:26:56,920 --> 00:27:00,860
divided by e. So we have z squared.

307
00:27:03,640 --> 00:27:09,200
Now, y is not given. If Pi is not given, we have

308
00:27:09,200 --> 00:27:14,180
to look at either B or 0.5. Now in this problem,

309
00:27:15,000 --> 00:27:18,900
it says that the Dean selected 200 students, and he

310
00:27:18,900 --> 00:27:23,800
finds that out of this number, 118 of them are

311
00:27:23,800 --> 00:27:26,480
receiving financial aid. So based on this

312
00:27:26,480 --> 00:27:30,480
information, we can compute B. So B is x over n.

313
00:27:30,700 --> 00:27:34,840
It's 118 divided by 200. And this one gives?

314
00:27:41,090 --> 00:27:46,310
So in this case, out of 200 students, 118 of them

315
00:27:46,310 --> 00:27:49,630
are receiving financial aid. That means the

316
00:27:49,630 --> 00:27:53,730
proportion, the sample proportion, is 118 divided

317
00:27:53,730 --> 00:27:57,690
by 200, which is 0.59. So we have to use this

318
00:27:57,690 --> 00:28:03,830
information instead of pi. So n equals,

319
00:28:08,050 --> 00:28:15,120
now it's about 99%. 2.85. Exactly, it's one of

320
00:28:15,120 --> 00:28:21,380
these. We have 2.57 and

321
00:28:21,380 --> 00:28:30,220
2.8. It says 99%. So

322
00:28:30,220 --> 00:28:32,720
here we have 99%. So what's left?

323
00:28:42,180 --> 00:28:47,320
0.5 percent, this area. 0.5 to the right and 0.5

324
00:28:47,320 --> 00:28:52,500
to the left, so 0.005. Now if you look at 2.5 under

325
00:28:52,500 --> 00:28:57,280
7, the answer is 0.0051, the other one 0.0049.

326
00:28:59,840 --> 00:29:04,600
So either this one or the other value, so 2.57 or

327
00:29:04,600 --> 00:29:07,600
2.58, it's better to take the average of these

328
00:29:07,600 --> 00:29:13,320
two. Because 0.005 lies exactly between these two

329
00:29:13,320 --> 00:29:20,780
values. So the score in this case, either 2.75 or

330
00:29:20,780 --> 00:29:30,880
2.78, or the average. And the exact one, 2.7, I'm

331
00:29:30,880 --> 00:29:34,680
sorry, 2.576. The exact answer.

332
00:29:38,000 --> 00:29:40,700
It's better to use the average if you don't

333
00:29:40,700 --> 00:29:46,100
remember the exact answer. So it's the exact one.

334
00:29:47,480 --> 00:29:53,440
But 2.575 is okay. Now just use this equation, 2

335
00:29:53,440 --> 00:30:02,020
.575 times square, times 0.59.

336
00:30:03,900 --> 00:30:09,440
1 minus 0.59 divided by the error. It's three

337
00:30:09,440 --> 00:30:17,800
percent. So it's 0.0312 squared. So the answer in

338
00:30:17,800 --> 00:30:23,420
this case is part 2 

339
00:30:23,420 --> 00:30:30,300
.57 times 0.59 times 0.41 divided by 0.03 squared. The

340
00:30:30,300 --> 00:30:31,140
answer is part.

341
00:30:41,650 --> 00:30:46,530
You will get the exact answer if you use 2.576.

342
00:30:48,190 --> 00:30:51,230
You will get the exact answer. But anyway, if you

343
00:30:51,230 --> 00:30:53,310
use one of these, you will get an approximate answer

344
00:30:53,310 --> 00:30:56,430
to be 1784.

345
00:30:58,590 --> 00:31:04,430
Any question? So in this case, we used the sample

346
00:31:04,430 --> 00:31:11,240
proportion instead of 0.5, because the dean

347
00:31:11,240 --> 00:31:14,120
selected a random sample of size 200, and he finds

348
00:31:14,120 --> 00:31:19,200
that 118 of them are receiving financial aid. That

349
00:31:19,200 --> 00:31:24,980
means the sample proportion is 118 divided by 200,

350
00:31:25,360 --> 00:31:30,420
which gives 0.59. So we have to use 59% as the

351
00:31:30,420 --> 00:31:38,360
sample proportion. Is it clear? Next, number

352
00:31:38,360 --> 00:31:38,760
three.

353
00:31:41,700 --> 00:31:45,860
An economist is interested in studying the incomes

354
00:31:45,860 --> 00:31:51,620
of consumers in a particular region. The 

355
00:31:51,620 --> 00:31:56,400
population standard deviation is known to be 1

356
00:31:56,400 --> 00:32:00,560
,000. A random sample of 50 individuals resulted

357
00:32:00,560 --> 00:32:06,460
in an average income of $15,000. What is the

358
00:32:06,460 --> 00:32:11,520
width of the 90% confidence interval? So here in

359
00:32:11,520 --> 00:32:16,560
this example, the population standard deviation

360
00:32:16,560 --> 00:32:21,480
sigma is known. So sigma is $1000.

361
00:32:24,600 --> 00:32:32,280
A random sample of size 50 is selected. This sample

362
00:32:32,280 --> 00:32:41,430
gives an average of $15,000 ask

363
00:32:41,430 --> 00:32:48,150
about what is the width of the 90% confidence

364
00:32:48,150 --> 00:32:55,630
interval. So again, here 

365
00:32:55,630 --> 00:32:58,710
we are asking about the width of the confidence

366
00:32:58,710 --> 00:33:02,570
interval. If we have a random sample of size 50,

367
00:33:03,320 --> 00:33:07,560
And that sample gives an average of $15,000. And

368
00:33:07,560 --> 00:33:10,940
we know that the population standard deviation is

369
00:33:10,940 --> 00:33:17,580
1,000. Now, what's the width of the 90% confidence

370
00:33:17,580 --> 00:33:21,800
interval? Any idea of this?

371
00:33:33,760 --> 00:33:40,020
So idea number one is fine. You may calculate the

372
00:33:40,020 --> 00:33:43,400
lower limit and upper limit. And the difference

373
00:33:43,400 --> 00:33:46,640
between these two gives the width. So idea number

374
00:33:46,640 --> 00:33:51,360
one, the width equals the distance between upper

375
00:33:51,360 --> 00:33:59,070
limit or limit minus lower limit. Now this

376
00:33:59,070 --> 00:34:03,270
distance gives a width, that's correct. Let's see.

377
00:34:04,710 --> 00:34:07,910
So in other words, you have to find the confidence

378
00:34:07,910 --> 00:34:12,070
interval by using this equation x bar plus or

379
00:34:12,070 --> 00:34:17,070
minus z sigma over root n, x bar is given.

380
00:34:20,190 --> 00:34:28,690
Now for 90% we know that z equals 1.645, sigma is

381
00:34:28,690 --> 00:34:32,670
1000 divided

382
00:34:32,670 --> 00:34:40,850
by root 50 plus or minus. By calculator, 1000

383
00:34:40,850 --> 00:34:45,010
times this number divided by root 50 will give

384
00:34:45,010 --> 00:34:49,190
around 

385
00:34:49,190 --> 00:34:50,730
232.6.

386
00:34:58,290 --> 00:35:05,790
So the upper limit is this value and lower limit

387
00:35:05,790 --> 00:35:09,650
is 14767.1.

388
00:35:11,350 --> 00:35:14,250
So now the upper limit and lower limit are

389
00:35:14,250 --> 00:35:18,590
computed. Now the difference between these two

390
00:35:18,590 --> 00:35:24,010
values will give the width. If you subtract these

391
00:35:24,010 --> 00:35:26,030
two values, what equals 15,000?

392
00:35:30,670 --> 00:35:37,190
And the answer is 465.13, around.

393
00:35:40,050 --> 00:35:45,550
Maybe I took two minutes to figure the answer, the

394
00:35:45,550 --> 00:35:49,350
right answer. But there is another one, another

395
00:35:49,350 --> 00:35:52,790
idea, maybe shorter. It'll take shorter time.

396
00:35:56,890 --> 00:36:00,230
It's correct, but straightforward. Just compute

397
00:36:00,230 --> 00:36:05,790
the lower and upper limits. And the width is the

398
00:36:05,790 --> 00:36:07,190
difference between these two values.

399
00:36:11,370 --> 00:36:16,050
If you look carefully at this equation, difference

400
00:36:16,050 --> 00:36:21,560
between these two values gives the width. Now

401
00:36:21,560 --> 00:36:25,880
let's imagine that the lower limit equals x bar

402
00:36:25,880 --> 00:36:28,920
minus

403
00:36:28,920 --> 00:36:36,340
the error term. And the upper limit is also x bar plus

404
00:36:36,340 --> 00:36:37,960
the error term.

405
00:36:41,460 --> 00:36:46,580
Now if we add this, or if we subtract 2 from 1,

406
00:36:47,900 --> 00:36:52,560
you will get the upper limit minus the lower limit equals

407
00:36:52,560 --> 00:36:55,740
x

408
00:36:55,740 --> 00:37:07,280
bar cancels with 2x bar. If you subtract, w minus 

409
00:37:07,280 --> 00:37:10,960
equals 2e.

410
00:37:12,520 --> 00:37:18,060
Upper limit minus lower limit is the width. So w,

411
00:37:18,760 --> 00:37:

445
00:40:25,610 --> 00:40:28,390
have to construct the confidence interval, then

446
00:40:28,390 --> 00:40:32,910
subtract the upper limit from the lower limit, you

447
00:40:32,910 --> 00:40:38,030
will get the width of the interval. The other way,

448
00:40:38,610 --> 00:40:42,150
just compute the error and multiply the answer by

449
00:40:42,150 --> 00:40:48,210
2, you will get the same result. Number 13.

450
00:40:56,020 --> 00:41:00,980
13th says that the head librarian at the Library

451
00:41:00,980 --> 00:41:04,780
of Congress has asked her assistant for an

452
00:41:04,780 --> 00:41:07,980
interval estimate of a mean number of books

453
00:41:07,980 --> 00:41:12,720
checked out each day. The assistant provides the

454
00:41:12,720 --> 00:41:23,000
following interval estimate. From 740 to 920 books

455
00:41:23,000 --> 00:41:28,360
per day. If the head librarian knows that the

456
00:41:28,360 --> 00:41:33,880
population standard deviation is 150 books shipped

457
00:41:33,880 --> 00:41:37,420
outwardly, approximately how large a sample did

458
00:41:37,420 --> 00:41:40,200
her assistant use to determine the interval

459
00:41:40,200 --> 00:41:46,540
estimate? So the information we have is the

460
00:41:46,540 --> 00:41:50,860
following. We have information about the

461
00:41:50,860 --> 00:41:51,700
confidence interval.

462
00:42:01,440 --> 00:42:02,800
920 books.

463
00:42:05,940 --> 00:42:08,700
And sigma is known to be 150.

464
00:42:12,980 --> 00:42:17,980
That's all we have. He asked about how large a

465
00:42:17,980 --> 00:42:20,880
sample did her assistant use to determine the

466
00:42:20,880 --> 00:42:21,820
interval estimate.

467
00:42:26,740 --> 00:42:31,850
Look at the answers. A is 2. B is 3, C is 12, it

468
00:42:31,850 --> 00:42:33,950
cannot be determined from the information given.

469
00:42:37,190 --> 00:42:43,190
Now, in order to find the number, the sample, we

470
00:42:43,190 --> 00:42:48,350
need Sigma or E squared. Confidence is not given.

471
00:42:50,550 --> 00:43:00,140
Sigma is okay. We can find the error. The error is

472
00:43:00,140 --> 00:43:07,940
just W divided by 2. So the error is fine. I mean,

473
00:43:08,100 --> 00:43:12,200
E is fine. E is B minus A or upper limit minus

474
00:43:12,200 --> 00:43:17,100
lower limit divided by 2. So width divided by 2.

475
00:43:17,240 --> 00:43:20,740
So this is fine. But you don't have information

476
00:43:20,740 --> 00:43:21,780
about Z.

477
00:43:25,020 --> 00:43:29,550
We are looking for N. So Z is not I mean, cannot

478
00:43:29,550 --> 00:43:32,810
be computed because the confidence level is not

479
00:43:32,810 --> 00:43:39,830
given. So the information is determined

480
00:43:39,830 --> 00:43:46,170
from the information given. Make sense? So we

481
00:43:46,170 --> 00:43:50,790
cannot compute this score. Z is fine. Z is 150.

482
00:43:51,330 --> 00:43:54,310
The margin of error, we can compute the margin by

483
00:43:54,310 --> 00:43:59,090
using this interval, the width. Divide by two

484
00:43:59,090 --> 00:44:05,790
gives the same result. Now for number 14, we have

485
00:44:05,790 --> 00:44:11,330
the same information. But here,

486
00:44:14,450 --> 00:44:22,030
she asked her assistant to use 25 days. So now, n

487
00:44:22,030 --> 00:44:24,990
is 25. We have the same information about the

488
00:44:24,990 --> 00:44:25,310
interval.

489
00:44:32,020 --> 00:44:33,300
And sigma is 150.

490
00:44:36,300 --> 00:44:40,800
So she asked her assistant to use 25 days of data

491
00:44:40,800 --> 00:44:43,860
to construct the interval estimate. So n is 25.

492
00:44:44,980 --> 00:44:48,300
What confidence level can she attach to the

493
00:44:48,300 --> 00:44:53,500
interval estimate? Now in this case, we are asking

494
00:44:53,500 --> 00:44:56,240
about confidence, not z.

495
00:45:00,930 --> 00:45:03,530
You have to distinguish between confidence level

496
00:45:03,530 --> 00:45:08,130
and z. We use z, I'm sorry, we use z level to

497
00:45:08,130 --> 00:45:13,350
compute the z score. Now, which one is correct? 99

498
00:45:13,350 --> 00:45:21,670
.7, 99, 98, 95.4. Let's see. Now, what's the

499
00:45:21,670 --> 00:45:25,070
average? I'm sorry, the formula is x bar plus or

500
00:45:25,070 --> 00:45:29,270
minus z sigma over root n. What's the average? In

501
00:45:29,270 --> 00:45:34,710
this case, this is the formula we have. We are

502
00:45:34,710 --> 00:45:38,770
looking about this one. Now, also there are two

503
00:45:38,770 --> 00:45:43,250
ways to solve this problem. Either focus on the

504
00:45:43,250 --> 00:45:47,950
aorta, or just find a continuous interval by

505
00:45:47,950 --> 00:45:55,830
itself. So let's focus on this one. Z sigma over

506
00:45:55,830 --> 00:45:56,230
root of.

507
00:45:59,620 --> 00:46:05,380
And we have x bar. What's the value of x bar? x

508
00:46:05,380 --> 00:46:15,240
bar 740 plus x

509
00:46:15,240 --> 00:46:16,400
bar 830.

510
00:46:25,380 --> 00:46:31,740
1660 divided by 2, 830. Now, z equals, I don't

511
00:46:31,740 --> 00:46:40,660
know, sigma, sigma is 150, n is 5. So here we have

512
00:46:40,660 --> 00:46:41,600
30 sigma.

513
00:46:44,980 --> 00:46:51,560
Now, what's the value of sigma? 36, so we have x

514
00:46:51,560 --> 00:46:54,900
bar, now the value of x bar.

515
00:47:02,330 --> 00:47:10,530
So we have x bar 830 plus or minus 30 there.

516
00:47:15,290 --> 00:47:18,030
Now, if you look carefully at this equation,

517
00:47:19,550 --> 00:47:24,570
what's the value of z in order to have this

518
00:47:24,570 --> 00:47:29,630
confidence interval, which is 740 and 920?

519
00:47:36,170 --> 00:47:40,730
So, Z should be...

520
00:47:40,730 --> 00:47:46,290
What's the value of Z? Now, 830 minus 3Z equals

521
00:47:46,290 --> 00:47:46,870
this value.

522
00:47:49,830 --> 00:47:53,390
So, Z equals...

523
00:47:53,390 --> 00:47:56,450
3.

524
00:47:56,830 --> 00:48:03,540
So, Z is 3. That's why. Now, Z is 3. What do you

525
00:48:03,540 --> 00:48:05,180
think the corresponding C level?

526
00:48:11,460 --> 00:48:16,560
99.7% If

527
00:48:16,560 --> 00:48:27,080
you remember for the 68 empirical rule 68, 95, 99

528
00:48:27,080 --> 00:48:33,760
.7% In chapter 6 we said that 99.7% of the data

529
00:48:33,760 --> 00:48:37,220
falls within three standard deviations of the

530
00:48:37,220 --> 00:48:41,980
mean. So if these three, I am sure that we are

531
00:48:41,980 --> 00:48:50,340
using 99.7% for the confidence level. So for this

532
00:48:50,340 --> 00:48:53,280
particular example here, we have new information

533
00:48:53,280 --> 00:48:57,280
about the sample size. So N is 25.

534
00:49:01,630 --> 00:49:06,190
So just simple calculation x bar as I mentioned

535
00:49:06,190 --> 00:49:11,510
here. The average is lower limit plus upper limit

536
00:49:11,510 --> 00:49:18,270
divided by 2. So x bar equals 830. So now your

537
00:49:18,270 --> 00:49:25,130
confidence interval is x bar plus or minus z sigma

538
00:49:25,130 --> 00:49:31,070
over root n. z sigma over root n, z is unknown,

539
00:49:32,190 --> 00:49:37,030
sigma is 150, n is 25, which is 5, square root of

540
00:49:37,030 --> 00:49:48,390
it, so we'll have 3z. So now x bar 830 minus 3z,

541
00:49:49,610 --> 00:49:53,870
this is the lower limit, upper limit 830 plus 3z.

542
00:49:55,480 --> 00:49:59,000
Now, the confidence interval is given by 740 and

543
00:49:59,000 --> 00:50:09,020
920. Just use the lower limit. 830 minus 3z equals

544
00:50:09,020 --> 00:50:10,820
740.

545
00:50:12,300 --> 00:50:18,280
Simple calculation here. 830 minus 740 is 90,

546
00:50:18,660 --> 00:50:22,340
equals 3z. That means z equals 3.

547
00:50:26,070 --> 00:50:29,750
Now the z value is 3, it means the confidence is

548
00:50:29,750 --> 00:50:33,530
99.7, so the correct answer is A.

549
00:50:44,690 --> 00:50:49,390
The other way, you can use that one, by using the

550
00:50:53,010 --> 00:50:55,830
margin of error, which is the difference between

551
00:50:55,830 --> 00:50:58,270
these two divided by two, you will get the same

552
00:50:58,270 --> 00:51:02,630
result. So there are two methods, one of these

553
00:51:02,630 --> 00:51:05,830
straightforward one. The other one, as you

554
00:51:05,830 --> 00:51:13,550
mentioned, is the error term. It's B minus upper

555
00:51:13,550 --> 00:51:19,550
limit minus lower limit divided by two. Upper

556
00:51:19,550 --> 00:51:27,450
limit is 920. Minus 74 divided by 2. What's the

557
00:51:27,450 --> 00:51:28,370
value for this one?

558
00:51:34,570 --> 00:51:40,610
90. So the margin of error is 90. So E equals E.

559
00:51:41,070 --> 00:51:43,790
Sigma or N equals?

560
00:51:47,110 --> 00:51:50,810
All squared. So by using this equation you can get

561
00:51:50,810 --> 00:51:56,860
your result. So, N is 25, Z is unknown, Sigma is

562
00:51:56,860 --> 00:52:05,520
150, R is 90. This one squared. You will get the

563
00:52:05,520 --> 00:52:10,020
same Z-score. Make sense?

564
00:52:17,770 --> 00:52:21,810
Because if you take z to be three times one-fifth

565
00:52:21,810 --> 00:52:25,150
divided by nine squared, you will get the same

566
00:52:25,150 --> 00:52:30,790
result for z value. So both will give the same

567
00:52:30,790 --> 00:52:35,790
result. So that's for the multiple choice

568
00:52:35,790 --> 00:52:42,430
problems. Any question? Let's move to the section

569
00:52:42,430 --> 00:52:46,370
number two, true or false problems.

570
00:52:47,810 --> 00:52:48,790
Number one,

571
00:52:51,850 --> 00:52:57,950
a race car driver

572
00:52:57,950 --> 00:53:03,670
tested his car for time from 0 to 60 mileage per

573
00:53:03,670 --> 00:53:09,390
hour. And in 20 tests, obtained an average of 4.85

574
00:53:09,390 --> 00:53:16,660
seconds, with a standard deviation of 1.47 seconds. 95

575
00:53:16,660 --> 00:53:23,440
confidence interval for the 0 to 60 time is 4.62

576
00:53:23,440 --> 00:53:29,540
seconds up to 5.18. I think straightforward. Just

577
00:53:29,540 --> 00:53:33,440
simple calculation, it will give the right answer.

578
00:53:36,660 --> 00:53:40,640
x bar n,

579
00:53:41,360 --> 00:53:43,620
so we have to use this equation.

580
00:53:48,220 --> 00:53:54,020
You can do it. So it says the answer is false. You

581
00:53:54,020 --> 00:53:58,340
have to check this result. So it's 4.5 plus or

582
00:53:58,340 --> 00:54:03,460
minus T. We have to find T. S is given to be 147

583
00:54:03,460 --> 00:54:10,120
divided by root 20. Now, to find T, we have to use

584
00:54:10,120 --> 00:54:18,480
the t-distribution with 19 degrees of freedom. By this value here, you'll get the

585
00:54:18,480 --> 00:54:22,160
exact answer. Part number two.

586
00:54:24,980 --> 00:54:32,380
Given a sample mean of 2.1. So x bar is 2.1.

587
00:54:33,680 --> 00:54:34,680
Excuse me?

588
00:54:38,500 --> 00:54:45,920
Because n is small. Now, this sample, This sample

589
00:54:45,920 --> 00:54:52,220
gives an average of 4.85, and standard deviation

590
00:54:52,220 --> 00:55:02,420
based on this sample. So S, so X bar, 4.85, and S

591
00:55:02,420 --> 00:55:09,640
is equal to 1.47. So this is not sigma, because it

592
00:55:09,640 --> 00:55:15,210
says that 20 tests, so the sample size is 20. This

593
00:55:15,210 --> 00:55:19,390
sample gives an average of this amount and

594
00:55:19,390 --> 00:55:21,350
standard deviation of this amount.

595
00:55:29,710 --> 00:55:34,610
We are looking for the

596
00:55:34,610 --> 00:55:40,470
confidence interval, and we have two cases. First

597
00:55:40,470 --> 00:55:43,630
case, if sigma is known,

598
00:55:47,220 --> 00:55:48,240
Sigma is unknown.

599
00:55:51,520 --> 00:55:58,440
Now for this example, sigma is unknown. So since

600
00:55:58,440 --> 00:56:05,740
sigma is unknown, we have to use the t-distribution if

601
00:56:05,740 --> 00:56:09,940
the distribution is normal or if N is large

602
00:56:09,940 --> 00:56:14,380
enough. Now for this example, N is 20. So we have

603
00:56:14,380 --> 00:56:17,860
to assume that the population is approximately

604
00:56:17,860 --> 00:56:23,660
normal. So we have to use the t-distribution. So my confidence

605
00:56:23,660 --> 00:56:26,100
interval should be x bar plus or minus 3s over

606
00:56:26,100 --> 00:56:32,560
root n. Now, number two. Given a sample mean of 2

607
00:56:32,560 --> 00:56:36,180
.1 and a population standard deviation. I

608
00:56:36,180 --> 00:56:38,720
mentioned that population standard deviation is

609
00:56:38,720 --> 00:56:43,900
given. So sigma is 0.7. So sigma is known in this

610
00:56:43,900 --> 00:56:49,170
example. So in part two, sigma is given. Now, from

611
00:56:49,170 --> 00:56:50,890
a sample of 10 data points,

612
00:56:53,730 --> 00:56:56,190
we are looking for 90% confidence interval.

613
00:56:58,790 --> 00:57:07,230
90% confidence interval will have a width of 2.36.

614
00:57:16,460 --> 00:57:19,180
What is two times the sampling error?

615
00:57:22,500 --> 00:57:28,040
So the answer is given. So the error here, error A

616
00:57:28,040 --> 00:57:30,160
equals W.

617
00:57:32,060 --> 00:57:34,120
So the answer is 1.16.

618
00:57:40,520 --> 00:57:45,220
So he asked about given a sample, 90% confidence

619
00:57:45,220 --> 00:57:50,540
interval will have a width of 2.36. Let's see if

620
00:57:50,540 --> 00:57:54,780
the exact width is 2.36 or not. Now we have x bar

621
00:57:54,780 --> 00:58:03,240
plus or minus z, sigma 1.8. x bar is 2.1, plus or

622
00:58:03,240 --> 00:58:08,660
minus. Now what's the error? 1.18.

623
00:58:11,230 --> 00:58:16,370
This amount without calculation, or you just use

624
00:58:16,370 --> 00:58:19,590
this straightforward calculation here we are

625
00:58:19,590 --> 00:58:23,530
talking about z about 90 percent, so this amount, 1

626
00:58:23,530 --> 00:58:30,330
.645 times sigma divided by root n, for sure this

627
00:58:30,330 --> 00:58:35,430
quantity equals 1.18. But you don't need to do that

628
00:58:35,430 --> 00:58:40,570
because the width is given to be 2.36. So E is 1

629
00:58:40,570 --> 00:58:46,430
.18. So that saves time in order to compute the

630
00:58:46,430 --> 00:58:55,190
error term. So now 2.1 minus 1.8. 2.1 plus 1.8.

631
00:58:56,350 --> 00:58:59,070
That F, the width, is 2.3

667
01:02:25,600 --> 01:02:30,580
statistics can be used to estimate the true

668
01:02:30,580 --> 01:02:33,400
population parameter, either X bar as a point

669
01:02:33,400 --> 01:02:34,900
estimate or P.

670
01:02:41,380 --> 01:02:48,000
So that's true. Number eight. The T distribution

671
01:02:48,000 --> 01:02:51,100
is used to develop a confidence interval estimate

672
01:02:51,100 --> 01:02:54,240
of the population mean when the population

673
01:02:54,240 --> 01:02:57,200
standard deviation is unknown. That's correct

674
01:02:57,200 --> 01:03:01,240
because we are using T distribution if sigma is

675
01:03:01,240 --> 01:03:03,740
not given and here we have to assume the

676
01:03:03,740 --> 01:03:07,960
population is normal. 9.

677
01:03:11,540 --> 01:03:15,180
The standardized normal distribution is used to

678
01:03:15,180 --> 01:03:17,340
develop a confidence interval estimate of the

679
01:03:17,340 --> 01:03:20,700
population proportion when the sample size is

680
01:03:20,700 --> 01:03:22,820
large enough or sufficiently large.

681
01:03:28,640 --> 01:03:32,640
The width

682
01:03:32,640 --> 01:03:37,720
of a confidence interval equals twice the sampling

683
01:03:37,720 --> 01:03:42,570
error. The weight equals twice the sample, so

684
01:03:42,570 --> 01:03:46,370
that's true. A population parameter is used to

685
01:03:46,370 --> 01:03:50,650
estimate a confidence interval? No way. Because we

686
01:03:50,650 --> 01:03:53,570
use statistics to estimate the confidence

687
01:03:53,570 --> 01:03:58,130
interval. These are statistics. So we are using

688
01:03:58,130 --> 01:04:02,390
statistics to construct the confidence interval.

689
01:04:04,190 --> 01:04:10,080
Number 12. Holding the sample size fixed. In

690
01:04:10,080 --> 01:04:14,560
increasing level, the level of confidence in a

691
01:04:14,560 --> 01:04:17,520
confidence interval will necessarily lead to wider

692
01:04:17,520 --> 01:04:20,500
confidence interval. That's true. Because as C

693
01:04:20,500 --> 01:04:24,840
level increases, Z becomes large, so we have large

694
01:04:24,840 --> 01:04:29,670
width, so the confidence becomes wider. Last one,

695
01:04:30,550 --> 01:04:33,150
holding the weight of a confidence interval fixed

696
01:04:33,150 --> 01:04:36,190
and increasing the level of confidence can be

697
01:04:36,190 --> 01:04:40,090
achieved with lower sample size with large sample

698
01:04:40,090 --> 01:04:44,830
size. So it's false. So that's for section two.

699
01:04:46,230 --> 01:04:49,970
One section is left, free response problems or

700
01:04:49,970 --> 01:04:52,990
questions, you can do it at home. So next time,

701
01:04:53,070 --> 01:04:57,530
inshallah, we'll start chapter nine. That's all.