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A leading magazine (like Barron's) reported at one time that theaverage number of weeks an individual is unemployed is 37 weeks.Assume that for the population ...

Question

A leading magazine (like Barron's) reported at one time that theaverage number of weeks an individual is unemployed is 37 weeks.Assume that for the population of all unemployed individuals thelength of unemployment is normally distributed where the populationmean length of unemployment is 37 weeks and that the populationstandard deviation is 3 weeks. Suppose you would like to select arandom sample of 33 unemployed individuals for a follow-upstudy.Find the probability that a single randomly

A leading magazine (like Barron's) reported at one time that the average number of weeks an individual is unemployed is 37 weeks. Assume that for the population of all unemployed individuals the length of unemployment is normally distributed where the population mean length of unemployment is 37 weeks and that the population standard deviation is 3 weeks. Suppose you would like to select a random sample of 33 unemployed individuals for a follow-up study. Find the probability that a single randomly selected value is less than 36. P(X < 36) = Find the probability that a sample of size n = 33 is randomly selected with a mean less than 36. P(M < 36) = Enter your answers as numbers accurate to 4 decimal places.



Answers

Barron's reported that the average number of weeks an individual is unemployed is 17.5 weeks (Barron's, February $18,2008 ) .$ Assume that for the population of all unemployed individuals the population mean length of unemployment is 17.5 weeks and that the population standard deviation is 4 weeks. Suppose you would like to select a random sample of 50 unemployed individuals for a follow-up study.
a. Show the sampling distribution of $\overline{x},$ the sample mean average for a sample of 50 unemployed individuals.
b. What is the probability that a simple random sample of 50 unemployed individuals will provide a sample mean within 1 week of the population mean?
c. What is the probability that a simple random sample of 50 unemployed individuals will provide a sample mean within 1$/ 2$ week of the population mean?

All right, so we're given data about the number of months. Certain sample is unemployed and we're also getting the number per month. So, for example, 1029 people reported they were unemployed for one month, 1686 at two months, et cetera. So first off, we're gonna make probability distribution, and I've totalled off the number of people unemployed here. That's 26,975. So for one month unemployed, we're gonna divide 1029 by 26,975. And when you do that, you'll get 0.0 381 for two months, we're going to do the same thing. 1686 divided by 26,975. That's going to be 0.6 to 5 and doing the same thing. For brevity sake, I'm not gonna explain. Every time we do this, I'm just gonna start writing them down for three. It's gonna be 0.841 for four. Zero points. 992 for 50.1293 for 60.17 to 5 for seven UM one point site. 0.1537 for 60 R for eight's my fault to certain 600.133490 point 08 62 and for 10 0.415 for part B. Need to test if what we got in part is a valid probability distribution. And to do that, we checked two things. Are all of these between zero and one? Yes, and we also need to make sure these all add up to one, and I will leave you two that I will leave that to you to verify. But this is a valid probability distribution part C. We need to find two probabilities, the 1st 1 of the probability that somebody's unemployed for two months or less. So that's gonna be the probability being unemployed for one month, plus the probability of being unemployed for two months. All right, the probability of being unemployed for 11 month excusing is 0.381 I believe. Yes, and for two months, it 0.6 to 5. You have these together you get 0.1006 now to find the probability that somebody's unemployed more than two months. While you can see that the opposite of this would be an unemployed for two months or less. So we're gonna take one and subtract probability that somebody is unemployed for two months or less. That's gonna be one minus 0.1006 which is 0.8994 That's a nine. So those are your two answers for part C party. Similar methodology. We're gonna find the probability that someone is unemployed for six months, plus the probability they're unemployed for seven months. Plus the probability that unemployed for unemployed for eight months, plus the probability that somebody's unemployed for nine months, lost the probability they're unemployed for 10 months. All right, let's look through the table to see what this So for six months, we get on retrospect. I should align this better. 0.7 are sorry. Deserve 117 to 5 plus for seven months. Zero point 1537 plus or eight months, 0.133 plus for nine months. 0.862 plus for 10 months. 0.415 If you add all these together, you'll get 0.4144 match. Her answer for party

Question 27. We're told that the weekly average unemployment benefit in the US is $238 and we're also told that a researcher in Virginia state suspects that the state average there is a less than $238. So for part A were asked to develop a hypothesis test for this situation. So we can describe the Virginia state researchers view as the alternative hypothesis that Mu is less than 238 and therefore the no hypothesis is mu is greater than or equal to 238. So that's your hypothesis test for Part B. We're told that we have a sample from Virginia State of size 100 and the average of the sample is $231 and the sample standard deviation is $80 and we're asked with the T values, so we're not given the population standard deviation. So we can say that sample averages are distributed according to the T distribution, so tea is there test statistic, and that is equal to the sample average minus the population average over the sample standard deviation divided by the square root of sample size and that comes out to minus 0.88 and then were asked to calculate the P value. So just to show you what's happening here, so this is the T distribution. It's symmetrical and this is a lower tail test because the alternative hypothesis suggests that the mean is lower than the no hypothesis. So basically we have a test statistic that comes out to minus 0.88 and we're basically asking the question. Does it fall in some critical region in the lower tail such that it's low enough or extreme enough to reject the no hypothesis? So the critical region would it be defined by Alfa, which we haven't been given yet? And so the P value is the area to the left of our test statistics of the red area, so it amounts to comparing the P value to the Alfa value to see which is smaller. The P value of smaller than our test statistic is more extreme and lower in, which would lead us to rejecting the no hypothesis. But of course, the tea table gives upper tail values. So we're looking at areas in the upper tail but our test statistic if we just take the opposite of it. So plus 0.88 we would find their the area that we're looking for. So this is also equal to the P here yet, so these two tails are identical, so we can go to the table and use plus 0.88 to find RPI value. But one thing we need before we go to the table is the degrees of freedom, and that's simply and minus one, and that comes out to 99. So then, when we go to the tea table, we want to roll. That is 99 degrees of freedom, and our test statistic of 0.88 falls between these two values, which means that our P values between 0.1 and 0.2 so that's the T value. So Part C says at an Alfa equal to 0.5 what do you conclude? And we can simply say that the P value is greater than Alfa for this situation, and therefore we failed to reject the no hypothesis then for party, whereas to repeat the hypothesis test using the critical value method so we have to find her critical value. So Alfa is 0.5 and the critical value then has gotten from the table. So again it's 99 degrees of freedom. Upper tail area of 0.5 corresponds to Alfa and therefore critical values 1.660 But remember that this is a lower tail test, so it's actually minus 60.66 three. Let me just check that daily again minus 1.660 So on the left tail, critical value is equal to minus 1.66 and then to use the critical value method, we simply compare our test statistic to the critical value. And our test statistic is equal to from part B minus 0.88 and that is bigger than minus 1.66 which is the critical value. And therefore, in this case, we also feel to reject the null hypothesis

In this problem, the number of unemployed will cost millions is given. The first question asked is finding being number unemployed a millions to find that first with fine diese some Isha Thanks that this be some off all the number off unemployed broke us the number of unemployed workers they rented noted by eggs. So Summation X is equal to 82.65 The former lover me experienced summation X divided by end. So this is it will do 82.65 Do you read it? But the information given is off 10 years. So what n is then? So these equal to 8.265 and it this a millions 8.265 millions now for that is asked which years had employment unemployment No zest to be mean from dinky one day that we can see that the years doesn't two and 2004 had unemployment no zest to demean the years 2002 Add to 1000 full Now in the next caution. It is for us to find this time that aviation for drink even Decca 25 day. But we've been fined summation X quipped that this some off all the X squared values summation it's good is equal to 7. 32 buying buttons. You go 97 The form level Standard division is another move. Submission X squared minus and multiplying by X Birth square divided by end minus one. So mission excluded 72 point Once you don't 97 minus our anything will do then. So big day. My people like by eggs that we about to be 8.265 We have to take its square that this 8.265 square be ready by end minus one over any Stine so n minus one is none. Therefore, our standard deviation is equal to 2.334 Now, in the next caution This us How many off these years is unemployment? The din. But stand the aviation off the me So on. So that first we will find exact minus s and expect. Plus s. It's very minus s will be eight point no. 65 minus our X is 2.334 So expert minus s is it will do 5.931 Thank you. Expect Plessis is it point 465 Yes, 2.334 So exactness s is equal to 10 point. Find 99 as we can usually see from the give ended up that it off the 10 years the number of unemployed Brokaw's is between 51 91 and 10.599 So in a year's the number off I have probable cause is within one standard deviation off. You mean now? In the next caution, it is asked in harmony off these years is unemployment bidding three standard deviations off the mean one so that we find X behind us three years and expect plus to be Yes, it's about minus two years is it will do but 20.263 and expect last two years is it will do 15.267 because in the given data, that all thank yours for in this range that this between 1.263 and 15.267 So in 10 years, the number of unemployed workers is bidding three standard deviations off the mean

Mhm. All right. So we got a large sample uh sample mean is going to be equal to the population means. So we can think that this is normally distributed uh Fada X or sorry, sigma X is going to equal sigma and divided by the square root of the sample size, which is gonna be 92 over route 50. This is approximately equal to 13 point oh one For Z score we have Z is equal to X minus mu X, divided by sigma X, Which gets us plus or -20, divided by 13.01. And this is plus or minus 1.54 And we'll find out what this corresponding p value is. So we start off at 0.938 to minus point oh +618 And that gets us 0.8764 or 87.64%.


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