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5.6.8. Suppose a random sample of size n is drawn from the pdf...

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5.6.8. Suppose a random sample of size n is drawn from the pdf

5.6.8. Suppose a random sample of size n is drawn from the pdf



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A random sample is drawn from a normally distributed population of known standard deviation 5. Construct a $99.8 \%$ confidence interval for the population mean based on the information given (not all of the information given need be used). a. $n=16, \bar{x}-\infty, s=5.6$ b. $n=9, \bar{x}-\infty, s=5.6$

And the government has given an sample sizes. 2 5 sample means the samples and divisions of this Population means this and they were not able to say that population is to print from 600. I don't find european. Wanted to check that. So we go for the test here. Right, samples and division is given to go for the test. Semi activity pedometer. Is everything still you want to be expand negative view or what passed over and all that. To put the values here and call for The statistic that is coming out to be 91 point when I Like at one one. Is that these territories? All right now to the technology to integrate the P value for this. So we'll have You're listening individual .401 Total value. Their point of order one. Even there was no point before you know what And the required evenly now 50.10 point 24012 P. Yes, better than alcohol. Right? So therefore we failed to reject. That's not. We fail to reject national right? The things uh says that as the population means different from the center. No relation means is not different from 600 at all. 500.00.1 level of significant should be is not different from 600 at all. 500.1 level of significance. Or because there there is no exception evidence To say that population mean is different from the Sender and thereby put the .1 level. Thank you

So another application of the central limit theorem. Uh So if we take, well If we take a sample of size six from a population, what what has to be um what's the requirement of the sampling distribution to say that this is approximately normal, that the sampling distribution is approximately normal? Well, it has to be normal itself. The population needs to be normal. Uh huh. So the population, the original population distribution. Popular pop distribution. Yeah. Must be normal. That's what has to happen. Uh If we took more than six, let's say. And the This is for the case where up to 30, if you're less than 30, uh if N is less than 30, the population distribution has to be normal for the sampling distribution to be normal. But that's what the central s central limit theorem comes in. If this is the separate from the question, but kind of good to just piece together if um If N is greater than 30, Greater than or equal to 30, then the population means, or the population distribution can be pretty much anything we want uh than the sampling distribution uh will be normal. Yeah. Yeah, jimmy will be I will say approximately normal, approximately normal. And that's pretty cool. Regardless of the population distribution, the sampling distribution will always be approximately normal as end is bigger than 30. That's pretty cool. But again and again, the question here, The population distribution must be normal for the sampling distribution of size 6 to be normal. So there you go

All right, We've got a question here where we need to find a mean of random variable when the FX is r squared X e to the negative Rx and you wanna go ahead and reference back Exercise 63 which I've already solved for And you'll see what we do is we just substitute that ffx into our integral So we get to you value or I mean value, I should say, is equal to the integral From zero to infinity x r squared, multiplied by X e to the negative are ex dialects are here When we go ahead and expand out the integral and solve for it Once again, we're gonna set the limit equal to, um, Infinity. And then when we go ahead and expand out that integral we would get a to minus two e capital are, um little are to the negative capital. Our little are minus a two capital are little are e to the negative. Are capital are are minus capital r squared a lower case R squared and e to the negative. Our little are all over our okay, and that's just like I said, it's gonna happen when you go ahead and calculate out, um, that integral there. Okay? And once again, we're setting our limit as our approaches to infinity and then are integral. We could substitute this here as they are. Okay, When we get to this final value here we plug in infinity. You can see that we plug in infinity. There. That would cancel out. This will cancel. This will cancel out. And so you'll be left with a two over. Little are all right, and that will be the final answer there. If there is any confusion, make sure you go back to exercise 63 you'll see that video is a little bit more thoroughly explained on. We're just following on from that video. Well, I hope that clarifies the question. Thank you so much for watching

Along with vocabulary that we need to get used to when we work with the sampling distributions. To prepare for the inferential statistics, we also need to get used to some new notation. Now recall that the mean of the sampling distribution of X bar is new. So writing that in symbols, new sub expert is telling you it's the mean of the sampling distribution of X bar is equal to new. Is telling you the center of the sampling distribution for the sample mean is exactly the same as the center for the population that it was pulled from. Now. Also, the standard deviation of the sampling distribution of X. Bar is the standard deviation of the original population that it was pulled from divided by the square root of the sample size that you took in symbols. The standard deviation of the distribution of the sample mean is equal to the standard deviation of the individual numbers from the population. It was pulled from, divided by the square root of the sample size that you took.


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