5

The National Vital Statistics Reports for November 2011 states that the U,S, Select one answer. preterm birth rate for 2010 Was about 129. Preterm birth means prema...

Question

The National Vital Statistics Reports for November 2011 states that the U,S, Select one answer. preterm birth rate for 2010 Was about 129. Preterm birth means premature H0 points birth: Is the percentage of preterm births lower this vear? Suppose that this year we select random sample of 50 births. The conditions are not met for Use of a normal model since the expected number of preterm births is only 6 (12% of 50). So we ran simulation with p 0.12.Suppose that the sample of 50 babies has 4 that

The National Vital Statistics Reports for November 2011 states that the U,S, Select one answer. preterm birth rate for 2010 Was about 129. Preterm birth means premature H0 points birth: Is the percentage of preterm births lower this vear? Suppose that this year we select random sample of 50 births. The conditions are not met for Use of a normal model since the expected number of preterm births is only 6 (12% of 50). So we ran simulation with p 0.12. Suppose that the sample of 50 babies has 4 that are preterm This is 8% of the sample. 512 ouf 0/ 2,020 rardom eamous Iravo ann om erpooron 0l 0 Q8 & INDl , Tinb 25, 64, 000 005 410 0.15 020 035 0.30 Sampb proportons 450 0.,08 Do the data suggest that the percentage of preterm births in the USis lower than 12% this year? 4 0 Yes, because only 8% are preterm this year which is lower than 12%. B. 0 Yes; because sample proportions of 0.08 or less are unusual given the population proportion is 0.12, 0 No, because sample proportions of 0.08 or less are not Unusual if the proportion of preterm births in the population is 0.12, D. 0 No, because a 25.6% of the babies are preterm this year which is much higher than 12%,



Answers

In 2000, the Journal of the American Medical Association (JAMA) published a study that examined pregnancies that resulted in the birth of twins. Births were classified as preterm with intervention (induced labor or cesarean), preterm without procedures, or term/post-term. Researchers also classified the pregnancies by the level of prenatal medical care the mother received (inadequate, adequate, or intensive). The data, from the years 1995–1997, are summarized in the table on the next page. Figures are in thousands of births. (JAMA 284 [2000]:335–341)

(Table Can't Copy)

a) What percent of these mothers received inadequate medical care during their pregnancies?
b) What percent of all twin births were preterm?
c) Among the mothers who received inadequate medical care, what percent of the twin births were preterm?
d) Create an appropriate graph comparing the outcomes of these pregnancies by the level of medical care the mother received.
e) Write a few sentences describing the association between these two variables.

In this question, we're testing whether the births are uniformly distributed across seasons. So I have 120 observations of students births according to the season that they're born. That's 100 20 and for party were asked, What is the expected number of births in each season, assuming that there is a uniform distribution? So since there are four seasons, we expect 1/4 of the births to be in each season. So 1/4 of 120 for each season, so 1/4 of 100 20 is 30. So we're expecting 30 berths in each season for Part B were asked to compute the chi squared statistic. So you recall that this is the sum of so this will be the some of of the observed values minus C expected values squared, divided by the expected value. So we could make another column for this table and calculate this parameter for each season. So for the 1st 1 would do that just as an example. So observed is 25. That's the 25 birth in the winter, minus the expected, which is 30 squared over 30 and that comes out to 25/30 so that would be the 1st 1 And then you might do three more so that you have one for each season, and then you could some them. So I used a calculator to calculate the chi squared statistic for me gave me a value of 1.93 Now moving on to see, whereas how many degrees of freedom does this chi squared statistic have that is equal to end minus one degrees of freedom and minus one, and we have four categories, so it's four minus one gives us three degrees of freedom for D, Using a significance level of 0.5 we're to look up a critical value. So we have way have Alfa 0.5 and we also have our degrees of freedom. So what? These two things We could go to the table and check. So Alfa is the area in the upper tail, so 0.5 and we have three degrees of freedom. So are critical. Value is 7.815 and finally for E using the critical value were supposed to conclude about the no hypothesis. But the question didn't explicitly ask for the no hypothesis, but no hypothesis is that birther evenly distributed across season. And so we're comparing two values are chi squared statistic and the Critical Valley. And because they're chi squared, statistic is less than the critical value. We we failed to reject the null hypothesis, and we can say that we do not have sufficient evidence to believe that births are not distributed evenly across season.

Everyone. So today we've been given another scenario on asked a couple of questions about it. And so, just like with, I'm actually just going to write down all of the numbers that we've been given so I can help you identify which ones are important on which on. So we're told that the average birth weight I'm gonna put B two B for birth weight in the US is 3300 grams. And what were then told is that the average birth weight, um for babies off women living in poverty. Um, I'm just gonna write women impulsivity to be Why pay is 200 at Sorry, 2800 grams. And this is I'm literally reading through the paragraph. Justo, Sorry, Just let you know what I progress about. It's talking about the mean birth weight off babies and especially looking at the mean birth weight of babies who are born to women who live in poverty and so were given a bunch of different nation on the baby's birth weights. And so I'm initially we definitely have more numbers in this paragraph than we need. And so, in this first instance, I'm just trying to help you see which ones are important and which ones aren't. Which is why I'm writing all these numbers. And we're told that the standard deviation for this average here is 500 grams. Now, that doesn't necessarily mean that the every son of deviation for the average birth weight in the U. S. Is 500. It specifically says the standard deviation is for babies off, um, women in poverty. And then it's talking about how the hospital has looked at a new prenatal care program in the Hope Teoh improve the birth weight of babies born to women in poverty. And so they do this, uh, discourse. And then I take a sample and I take 25 randomly selected babies. So 25. And that's the number in our sample. And they take babies off women who live in poverty. Andi, they took part of program and the mean birth weight for those babies. So expo was 3000 and 75 grams. Now, the questions asking us, is there a significant improvement in the birth weight of babies born poor women? And we're told to use a significance level of no point naughty. Okay, so Obviously we have a one to many values here, and important thing to notice is that actually this value up here at the top the average birth weight for babies in the US isn't important to this question for us. We're only looking at babies who boards women in poverty. And so this is some context for the question. It's saying that women in poverty have a lower birth weight on average than the general in the US But in terms of this hypothesis test, we actually don't need this value. But all And so what we're gonna do is we're gonna use these values here. I'm gonna start doing hypothesis test. And to do that, we gotta start by using this handy. Formula two car codes that star so said style is equal to X for which we're told is 3000 and 75 minus mu, which were using 2800. And then we given 500 grams of standard deviation. Have a Rouen, which is route 25. So, people this NJ can't connect that you're gonna get out a value of 2.75 actually, what I should have done First of all, after we roll out, this number is we should have, um, were in high potties down. So what are our hope all season? This question Well, we're looking at one of this study has improved the birth way off babies born to women in poverty. And so no hypothesis is that it hasn't improved it and therefore that it would be in line with the average. So I know hypothesis. We told you to right here tells us that we reckon the population, I mean for the birth weight of babies born to women of poverty is 2800. And I offered about this. Is that actually this study? This course improved that the birth weight increased. And so now we have as that star value and we can use this to find a P value on RPI value for this which we confined using the tables or our calculators If you vote for this is no 0.0 Syria to nine eight on we given this significance level half, which is no point no to. And what we find is that this number is far smaller. They're not significant level. So what that means is because we have P is less than Alfa. We reject Haitian. We found significant evidence to suggest that this, uh, prenatal program improves on increases the that mean birth weight off babies born to women in poverty on because we I found it significant, um, improvement according to the note point not to level of significance, we can safely reject fictional. And so, in terms of answering the specific questions, I realized that this is, um, just a normal hypothesis test I've done here and were given in the textbook is in specific questions to answer. And so to start with what has to define the prouder, that's what I was talking about up here. When we're talking about which is important, what are we measuring? So what we're measuring just right away here is we're measuring the mean birth weight, and I'll write it properly rather than using my acronyms. Mean birth weight off babies phone that to women in poverty. And so that is what we're measuring it for. You could read that even though it's a little bit hard to right here. And so the normal tone of hypotheses we came up with right here and again. That's what mu is. This is just a wordy version of what Mu is were saying that the mean birth weight of babies born to women in poverty It's equal to 2800 always great in the 2300 when we're talking about specifying the hypothesis test criteria. What we're looking at is the conditions that we need in order for this and in order to start looking at a sample on the test. And so what we have to do is we have to assume no normality on what that means is we're assuming that the population off birth weights roughly follows a normal curve. And we can often assume this using something called the central limit their, um and so for use in central over there. Um, we have an is 36 on our standard deviation. Starting off the basics, reading the wrong news. Um, for this we have and is 25 Andi, Explore. Sorry. My expo, my dear Onda Ah, sigma is 500. And so we can use these to form standard error which helps us use the settlement there to issue normality. Now, today, sample evidence is the information we get from our sample. And so I thought, again, we given the any 25 I'm were given the Expo is 3750. So that's our sample evidence. And so when it's asking for their probably probability distribution information, that's what we calculate up here. That's said Star on RPI value values. And again, determining the results is what we did. Just hear. While he compared up, he value with a significance level and decided that the Coast P was less an Alfa. We can reject a channel, and so that's looking a bit closer at some of the specifics of the questions that arrest.

At this problem involves a study of babies born with very low birth weights. And in this study we have 275 Children that are given an I Q test at the age of eight. The results of that test yielded on average of 95.5 and a standard deviation of 16.0. Then there was a follow up study done where they gave a smaller portion of the 275 the test again. And this time it was 50 of those Children. And the question where the problem comes in two parts, we have part a and part the in part A has, ah couple questions within it and party is asking you when considering the distribution of the mean i Q scores for samples of the 50 Children. Should the standard deviation of the sample means be corrected by using the finite population correction factor, and you're going to use that finite correction factor if you can answer yes. 22 questions and the two questions are one is the sampling performed without replacement, and we're gonna answer these in a moment. And to Was the sample size greater than 5% of the finite population size, and we're going to answer it again. If we can answer yes to both, then yes, we need to utilize the finite population correction factor. So the first question was sampling performed without replacement. And the answer is yes, because when we chose thes 50 Children to do the follow up study, once we chose a child, we didn't put them back in the mix that they had the opportunity of being selected again. And the second question is the sample size greater than 5%. So our sample size was 50 and we want to test if it was greater than 5% of the finite population size and the finite population size was we started with the 275 Children. So we're asking ourselves is 50 greater than and when you perform that calculation, you get 13.75 and the answer that is yes. So because we've answered yes to both yes, we must use the finite population correction factor when determining the standard deviation of sample meets. Now there's still more to part a. The next part is why or why not. So we've answered the why or why not? Because we've answered those two questions, and the next part says to actually determine the value. So we need to find the value of the standard deviation of the sample means and the correction factor to find that you're going to take the populations standard deviation divided by the square root of N, which is the one we were using from the Central Limit Theorem in previous problems of this section. And here's where the correction factor comes in. Oops, we're going to multiply it by the square root of begin minus little n over Big n minus one. So in terms of the numbers involved in this problem, the standard deviation was 16. The sample size was 50. The big end represents the finite population size, which means we started with 275. Little end is the sample size on that secondary study and 2 75 minus one. So when we calculated out standard error of the mean or the standard deviation of the sample, means would equal approximately 2.504 584 I don't recommend rounding because you're going to use that then in probability uh, questions. And if you round in the middle of a problem, it tends to skew the overall end result. Now, let's answer part B in part B. We want to determine the probability that when we issued this week test again to the follow up sample, their average was between a 95 and a 105. So again, we are using a sample. Keep that in mind. So our sample size was 50. We are going to need the average of the sample means and we are going to need the standard deviation of the sample means So the average of the sample means, according to the central limit theorem is going to be equivalent to the average of the population, which was 95.5 and the standard deviation we had to make that correction. So we're going to utilize the 2.504 584 So we're going to draw our normal curve. We're gonna put the average in the center, and we're trying to determine the probability that we're between a 95 and 105. They were trying to figure out that probability, so we will need the scores. So we will need to use the formula. Z equals X bar minus the average of the sample means divided by the standard deviation of the sample means. And we're going to determine the Z score for 95.4 95. So we're gonna do Z equals 95 minus 95.5, and we're going to use the corrected value for the standard deviation of the sample means and we're going to get a Z score of negative 0.24 So we could put a negative 0.24 up here on our bell. And we need to find the Z score, associate it with 105 syringe a 105 minus 95.5. And again, the standard deviation of the sample means we're gonna use the corrected one 2.504 584 and we're going to get a 4.63 And again, we're gonna put it up on your bell. So when the problem is asking with these 50 Children, we did the follow up study. What's their average? Or what's the probability that their average I Q score was between a 95 and a 105 is no different than saying, What's the probability that your Z score falls between a negative 0.24 and a 4.63? So at this point, we're going to utilize your standard normal table in the back of the book, which is table aid to. And when you use that table, the areas and the probabilities extend into the left tail. So we will have to rewrite this as the probability that Z is less than 463 minus the probability that Z is less than negative. 0.24 and using your table 4.63 cannot be found in the table. But there's information that says anything greater than or equal to a 3.50 Z score will have a probability of 9999 And the probability that Z is less than negative 0.24 is 0.4052 So when we subtract, we get an overall probability of 0.59 4/7. So just to summarize our final answer, we started with 275 Children. We found their average I Q. We did a follow up study selecting 50 of them and we wanted to know what was the chances or the probability that the average score I Q score of the 50 students fell between a 95 and 105 and the answer would be 0.5947 after we've made the correction for the finite population correction.

Here were given a bunch of data for um as for median income as a function of age. I was asked to use a quadratic quadratic model, a quadratic fit to find that. So what we do is um I put all of my ex values in that were given all my Y values that were given. And then I have a column here for X squared because I want that square to be part of my fit. So we're doing what's called a multiple regression. So we're doing a regression on X squared also as well as X. And so you can think of this function as being linear and X square. Now what I did is that you can go back in problem 24 see all the details of how I did this. But is the is the linear estimate function on both X and X squared. And then use index function to extract things. And we can see here that we get our coefficients A, B and C in our expression. So this is a coefficient of X squared of coefficient of X. And this is our constant. And so if we plot the data when you get it looks pretty quadratic here and then I actually had it to make a quadratic trend line in the plot. And we can see here that it basically reproduced the values that we got here a trend line. And then they asked us for, you know, what would really expect for someone that was aged 28. No, we'll be right in here, right between these two, closer to 30. So we should be somewhere in between here probably. And we can put in 28 there. And then I used their equation. So I said, you know, a times X squared plus B, times X plus C. And so we plug those holes in and we can see that for 28 were next is 28. We get about about 45 45 700 dollars. I guess that would be per year for the median income of somebody that was roughly 28 years old.


Similar Solved Questions

5 answers
Studen= Deked Ju F elelea Icoiectineerat #pondmuIncuohruu conlinuous oiscrere qualkuie
studen= Deked Ju F elelea Icoiectine erat #pondmu Incuohruu conlinuous oiscrere qualkuie...
5 answers
CounpcnunGasohneoctenc (c Hw) Mtn Inqu d octane Wumnd Cakulote the nloles occnenrernd produce 0,055 Irol ol Carbon Dlonide Henicert dlgles-TcacttOxyocn (01) gs oroquco canoon dlotde wotrr ~utc Your Dnsttrmas unitenmbol K noccssary, andirouau 66o 2CP
counpcnun Gasohne octenc (c Hw) Mtn Inqu d octane Wumnd Cakulote the nloles occnenrernd produce 0,055 Irol ol Carbon Dlonide Henicert dlgles- Tcactt Oxyocn (01) gs oroquco canoon dlotde wotrr ~utc Your Dnsttrmas unitenmbol K noccssary, andirouau 66o 2 CP...
5 answers
A traveling wave Is graphed at the Instant t = 0 If It is moving to the right with velocity find an equation of the form Y(x, t) = sin(kx kvt) for this wave. (Round all numerical values to two decimal places:)Y(x, t) =Need Help?uctemn an
A traveling wave Is graphed at the Instant t = 0 If It is moving to the right with velocity find an equation of the form Y(x, t) = sin(kx kvt) for this wave. (Round all numerical values to two decimal places:) Y(x, t) = Need Help? uctemn an...
5 answers
Asaatue dl George Washinglon is also in Drud Hill Park: If agrl slands 15 f away hrom the slate and places a mirror 12 ft. Irom Ihe slalue (3 It Irom her leet) , she can see the Iop ol the statue in the minor: The girl $ height is 5.5 ft How tall is the statue? (draw a diagram and solve)10.A Giant Redwood tree casts & shadow 125 feet long; Jason is 6 feet lall and he casts & shadow that 2.25 f , long How tall Is the Giant Redwood tree? (draw tlagram and solve)
Asaatue dl George Washinglon is also in Drud Hill Park: If agrl slands 15 f away hrom the slate and places a mirror 12 ft. Irom Ihe slalue (3 It Irom her leet) , she can see the Iop ol the statue in the minor: The girl $ height is 5.5 ft How tall is the statue? (draw a diagram and solve) 10.A Giant...
5 answers
Shcw #hat if s jg &1 odd integer, tben tbere ig wnique % g1c.1 #tat; # iz tle gwog ci % ~ l wd *+46. Frove thal thee ig mO irtegec # &xch thab %? + m' = 34,Froz #ba: tlere exiots 3,W integergjcb tha; 48 JclIoD Ie yctr Proof conatrctive @r Donooneti icitive?
Shcw #hat if s jg &1 odd integer, tben tbere ig wnique % g1c.1 #tat; # iz tle gwog ci % ~ l wd *+4 6. Frove thal thee ig mO irtegec # &xch thab %? + m' = 34, Froz #ba: tlere exiots 3,W integer gjcb tha; 48 JclIoD Ie yctr Proof conatrctive @r Donooneti icitive?...
2 answers
A+"totJeeel12[tui m {GMpoeocn: (ie4irplix - % ^Hannieea Heeenee Wieet
A+"tot Jeeel12 [tui m {GMpoeocn: (ie4ir plix - % ^ Hannieea Heeenee Wieet...
5 answers
9.22 beauty parlor finds that in n = 60 randomly selected days day with it had, 7 = 72.3 patrons per standard deviation of average, 8.5 patrons; (3) Construct 90% confidence interval for the actual daily beauty parlor patrons average number of Using the data of this exercise; but changing F = 72.3 t0 F = 82.3 , recalculate the 90% confidence interval for the actual daily average number of beauly parlor patrons_
9.22 beauty parlor finds that in n = 60 randomly selected days day with it had, 7 = 72.3 patrons per standard deviation of average, 8.5 patrons; (3) Construct 90% confidence interval for the actual daily beauty parlor patrons average number of Using the data of this exercise; but changing F = 72.3 ...
5 answers
PROBLEMS1 In an experiment involving smoke detectors, an alarm was set off at college dormitory at 3 a.m: Out of 216 residents of the dormitory, 128 slept through the alarm If one of the residents is randomly_cho: sen what is the probability that this person did not sleep through the alarm?
PROBLEMS 1 In an experiment involving smoke detectors, an alarm was set off at college dormitory at 3 a.m: Out of 216 residents of the dormitory, 128 slept through the alarm If one of the residents is randomly_cho: sen what is the probability that this person did not sleep through the alarm?...
5 answers
Hi, need help with these 3 herea. log4 X + logx16 = 3b. 3 log6X + 2 logx 6 = 5c. log2 X = 4 logx 2
Hi, need help with these 3 here a. log4 X + logx16 = 3 b. 3 log6X + 2 logx 6 = 5 c. log2 X = 4 logx 2...
5 answers
Question 16 of 16Compute the matrix of partial derivatives of the function f R3 and select the correct answer from the given choices _ Axy Dflx,Y, z) = 4ez 2x2R2, f (x,Y,2) = (3x + 4ez + 2y, 2yx? )Dflx,Y,2) = [5 + 4ez Axy- +2x2 ] Axy Df(x,Y, 2) = 2x2 4ez 4ez Df(x,Y, 2) = 4xy 2x2 4ez Dflx,Y, 2) = 4xy 2x2
Question 16 of 16 Compute the matrix of partial derivatives of the function f R3 and select the correct answer from the given choices _ Axy Dflx,Y, z) = 4ez 2x2 R2, f (x,Y,2) = (3x + 4ez + 2y, 2yx? ) Dflx,Y,2) = [5 + 4ez Axy- +2x2 ] Axy Df(x,Y, 2) = 2x2 4ez 4ez Df(x,Y, 2) = 4xy 2x2 4ez Dflx,Y, 2) = ...
5 answers
Q16. What is the product of the following reaction? (2 points)1) PCCii) Excess CH;OH, H+OH
Q16. What is the product of the following reaction? (2 points) 1) PCC ii) Excess CH;OH, H+ OH...
5 answers
Eight people participated in Professor Jackson’s experiment.Each person (a) answered a questionnaire about his or her emotionalstate, then (b) watched a very sad 15-minute video clip from themovie Old Yeller, then (c) answered the emotions questionnaireagain. Higher emotion scores indicate greater positive emotion. Thedata are presented below. Test the null hypothesis that µ Before =µ After . Subject 1 2 3 4 5 6 7 8
Eight people participated in Professor Jackson’s experiment. Each person (a) answered a questionnaire about his or her emotional state, then (b) watched a very sad 15-minute video clip from the movie Old Yeller, then (c) answered the emotions questionnaire again. Higher emotion scores indicate gre...
5 answers
21. Given Anxn, Unx] and Vnxl, use the Cauchy-Schwarz inequality to prove (vT A Au)? < (uT AT Au)(vT A" Av)_
21. Given Anxn, Unx] and Vnxl, use the Cauchy-Schwarz inequality to prove (vT A Au)? < (uT AT Au)(vT A" Av)_...
5 answers
X = (4Jazz is six years older than Rascal. Next year Jazz will be twice as old as Rascal will be then. How old are they now?
X = (4 Jazz is six years older than Rascal. Next year Jazz will be twice as old as Rascal will be then. How old are they now?...
5 answers
The major productof the treatment of 3-methyl-1-butene with dilute aqueous HzSO4a) 2-methyl-2-butene:b) 3-methyl-1-butanolc) 3-methyl-2-butanol:d) 2-methyl-2-butanol:e) 3-methyl-1-butanesulfonic acid
The major productof the treatment of 3-methyl-1-butene with dilute aqueous HzSO4 a) 2-methyl-2-butene: b) 3-methyl-1-butanol c) 3-methyl-2-butanol: d) 2-methyl-2-butanol: e) 3-methyl-1-butanesulfonic acid...

-- 0.023712--