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Biomedical researchers are studying skin wound healing rates for developing technologies to improve healing and tissue egeneration technologies Healing rates are me...

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Biomedical researchers are studying skin wound healing rates for developing technologies to improve healing and tissue egeneration technologies Healing rates are measured by determining the rate at which cells closed wcund made with sterile razor blade on anesthetized newts. The data for the hea ling - ates for = samp of 18 newts below microns nr Examine the data graphically using boxplot to decide whetner the data pproximate norma symmetric distribution: Showvour boxplot and explain your decisi

Biomedical researchers are studying skin wound healing rates for developing technologies to improve healing and tissue egeneration technologies Healing rates are measured by determining the rate at which cells closed wcund made with sterile razor blade on anesthetized newts. The data for the hea ling - ates for = samp of 18 newts below microns nr Examine the data graphically using boxplot to decide whetner the data pproximate norma symmetric distribution: Showvour boxplot and explain your decision_ Assuming these data do approximate symmetric distribution, ycu will need decide if vou will calculate the confidence interval based on either the Standard Normal distribution curve oron the distribution curves. Which will you choose? Explain ycur answer- Calculate the 95% Confidence interval for estimating the true mean healing rate for newts Be sure show work: What your interpretation conclusion regarding the skin healing rate of newts? Healing rate of Newts (um/br) Below are the lengths in feet of 44 great white sharks Examine the data graphical using boxplot to decide whether the data pproximate norma symmetric distribution Showvour boxplot and explain your decision_ Assuming these data do pproximate symmetric distributicn, ycu will need to decide if vou will Iculate the confidence interval based on either the Standard Normal distribution curve or on the distribution curves. Which will you choose? Explain ycur answer Calculate the 95% Confidence interval for estimating the true mean length of Ereat" white sharks_ sure t0 show work: It is popularly claimed that that great white sharks average 20 feet in length: Does your sample of the lengths of great white sharks provide evidence to support these claims? Explain ycur answer_ there any additional information You may want to know bcut the sample of great white sharks that may be important to ycur conclusion Length of Great White Sharks (feet) 17.6 16.8



Answers

SCIENTIFIC INQUIRY INTERPRET THE DATA Researchers
studied genetic variation in the marine mussel Mytilus edulis
around Long Island, New York. They measured the frequency of
a particular allele (lap ") for an enzyme involved in regulating the
mussel's internal saltwater balance. The researchers presented
their data as a series of pie charts linked to sampling sites within
Long Island Sound, where the salinity is highly variable, and
along the coast of the open ocean, where salinity is constant.
(a) Create a data table for the 11 sampling sites by estimating
the frequency of $l a p^{94}$ from the pie charts. (Hint: Think of each
piechart as a clock face to help you estimate the proportion
of the shaded area.) (b) Graph the frequencies for sites $1-8$ to
show how the frequency of this allele changes with increasing
salinity in Long Island Sound (from southwest to northeast).
Evaluate how the data from sites $9-11$ compare with the data
from the sites within the Sound. (c) Considering the various
mechanisms that can alter allele frequency, construct a
hypothesis that explains the patterns you observe in the data
and that accounts for the following observations: $(1)$ The $\operatorname{lap}^{94}$
allele helps mussels maintain osmotic balance in water with a
high salt concentration but is costly to use in less salty water;
and $(2)$ mussels produce larvae that can disperse long distances
before they settle on rocks and grow into adults.

All right, So before we get started, we're going to review a few key concepts. The first one we're going to talk about is illegals. Just what are they if you look at every individual in the species and you look at all of those individuals genome, whatever form it's in if that genome has been sequenced and identified on, you know how to find the same locations within each genome. For example, on a human, if you looked at the same chromosome, the same spot on that chromosome, you know, barring genetic defect, um, there's going to be the same gene at the same location in each person. The difference between person to person is little variations in the genetic code and the DNA of each gene and those little variations when they're significant enough to be identified are called the wheels. And when we refer to someone having, you know, a certain illegal were saying they have a particular variation in that gene or in their genome, the reason that's important is because if those jeans go on to encode a protein that does an important job and then an individual acquires a variation in that protein through a varied illegal, then that protein will still probably go on to do the same job. But it may do it a little bit better, or it may do it a little bit worse and the individual may die. But there's always a little chance that just at random little tweak is going to give them an advantage. And whether it gives them an advantage in living longer in reproducing more effectively, it doesn't matter. Um, it's if it leads to them having more opportunity to reproduce or simply in whatever fashion they create progeny. If it gives them or opportunity to pass their own DNA onto that progeny than there will be more, there will be higher prevalence of their particular Lille in the future generations as long as it creates an advantage. And this leads to an increased prevalence of that. A little of in this case, we're talking about the Lap 94 legal. We're told in this instance that there have been some observations made about a population of muscles in a particular body of water, and now we're talking about mussels like oysters, you know, not muscles, like go to the gym and flex mussels like oysters. We're told that the last 94 Alil encodes a protein that is really good at maintaining Oz morality. So it's really good at maintaining. You know, I on balance and concentration within the muscle, but it does this at a great cost. Um, you know, humans. We don't really think about an energy cost because, generally speaking, we don't have to worry about it. We have so much energy stored up, and we have such a ready access to food that, you know, we don't really experience the perils oven energy deficit. Really? Unfortunately, it still does happen, but we would refer to this as starving to death. Um, the last 94 Lille significantly increases this risk in the muscle because it is so energy expensive. And a muscle doesn't always know when it where he's getting his next meal from, um, in the last 94. Little kind of exacerbates that problem. But if it's in an environment where it needs it, then that energy cost is worth it. Because the muscle otherwise couldn't survive, it couldn't maintain its appropriate toe. I On balance, it's salt balance. We're also told that muscle larva are able to disperse broadly upon repair on hatching, and this one just imply that they don't have to stay where they are. They just have the opportunity. Thio float around and spread and, um, you know, if they end up somewhere, that they're able to thrive and out compete the surrounding population. Then there's a very good chance that they're going to get Thio more or less. Take over. Um, the more scientific way to say this would be the illegal frequency would increase in the area where they established. If they're able to out compete the other day, why would I say this? The organisms with other Leal's? In this case, we're talking about the last 94 wheel, and the way we talk about it is Prevalence were given pie charts, and it's convenient to analyze pie charts as a clock face just because we're all used to looking at those. And we're looking at the prevalence of Lap 94 in different populations of muscles at different sites. We're told that there's 11 sites site number one being within the sound, and we're told that site number one Why did that happen? There we go. We're told that site Number one is the lowest salinity, the least salty water of all of the sites. And at this site, the pie chart shows us about 1/12 of the muscles are gonna have that lap 94 Leo. Um, site number two, A little bit saltier, but still one out of 12. Site number three, salty or still and still one out of 12. Now. Site number four is where we see our first change. Site number four. We see about three out of 12, about 1/4 site number five saltier still and slightly more lap 94 wheels. Every four out of every 12 muscles will have last 94. Site number five Same insight six. Another tick up its site number seven. Another tick up a larger one at number eight and then at number nine, we see eight out of 12 muscles are expected to have the last 94 Lille, and we see that same number as sites number 10 and 11. And we're told that sites 9, 10 and 11 all have the same solidity. Eso we increase it cites one through nine and then it's constant. On the last three. There's a visual representation of that. This graph over here, the X axis is just site numbers. Now, this is kind of a surrogate for celebrities, you know, We're told that site number one is the least salty. And then the solemnity increases through site nine and then cites 9, 10 and 11 of the saltiest. We don't have those solemnity values, but we don't need them. We can still see this trend. And then on the y axis, we just have prevalence instead of these fractions instead of 1, 12, 3, 12, 4, 12, 5, 12, 7, 12 and 8. 12. We just labeled it as the numerator. So we have one. This represents 1 12, 23456 all the way up to 12. 12 would be at the top, but we don't end up using that part of that. So what we see is the same thing we saw in that data table site 1 1/12 or just one site to same thing. Site three Corresponds still one out of 12. Site four. Remember? Over here, Site four is where we saw the first increase. So I number four. We saw 3 12 and then so on site 567 Now I've added this red line here. Really, this is pretty arbitrary, but it need to think about this. Red Line is at 6 12. Otherwise, you know you could think of it is halfway at this point and above. This would be where the last 94 Leo is more common than the field, and that's kind of a gambling term. But the last 94 Lille is not only the most common one you would find, but it's It's more common to find a muscle that has the last 94 wheel above this solemnity than it is to find a muscle with any other combined. So if you either ask, does it or does it not have the lap 94 Leo above site seven salinity? It's more likely that yes, it does have it. Then we do see a plateau, you know, kind of as we would expect, given the observations that we knew. Um, sites 9, 10 and 11. Same salinity, same prevalence of of lap 94. So wrap this up in a nice package, come up with a hypothesis to potentially explain, um, why we see this relationship and it makes sense if we look back at the observations were told that the Lap 94 Lille it's expensive, but it's good in the right environment. And what we've seen here is kind of what exactly is the right environment? How much solemnity do you need now? If you wanted to be very precise, you could go back to these locations and you could sample the salinity of the water and each of these locations. And you could say, Well, site number seven, you know, that was the last time we saw less than half of the muscles, the last 94. So if the solemnity is sufficiently high, then in the last 94 Lille will compete all other. Leo is becoming the most common illegal, and not only that, but occupying more than half of its of the population. And for now, we just have this conceptual

First one. The test for a unit root in series you rate unemployment rate using the usual dickey fuller test with a constant yeah. And the augmented dickey fuller with two legs of change of unemployment rate. I find that seven both times we are unable to reject the now hypothesis that unemployment rate series is a unit fruit. The legs are not significant. However, the significance of the legs matters. So the outcome of the unit root test, we will repeat what we have done in part one two series vacancy rate and report the result in part two. I guess similar result. So the rate is a unit root. Well part one and two. I use package the R. Package A. T. S. A. And the function is a D. F. Dot test. R. Three. We assuming that unemployment rate and vacation re rate are both integrated of level one. We test for co integration using the angle grandeur test with no legs. So the step the steps are as follow. We first regress, you read on the rate then we yet the residual and we run the key fuller has on the residual to see whether the residuals our unit root. I find that you're right and we rate Arco integrated at the 5% level. Yeah Heart Forest. I get the leads and lacks estimator of the change in vacancy rate and I did note that uh CB rates up minus one. This is for the lack and plus one is for the lead. This is a regression result. So the usual centered errors are in green and in round brackets, the robots that Iran's are in blue and in square brackets you can see that the main estimate on vacancy rate is highly significant. This one is not correct. So the centered errol the usual one for the estimate of the first lack of change in vacancy rate is 164 In all cases except for the estimate of the lead of C. V. Right. The robust standard Iran's are larger than the usual standard errors. This is usually the case it happens but rare that the robot standard errors are smaller than the usual standard errors. The r square of this regression is 0.77 So for the rate, because the robot standard error is larger than the usual standard error. So we will get a wider confidence interval if we use a robot standard error and for confidence interval you will run this function in our count in and you impose the name of the regression. It was spits all the 95% confidence intervals for all explanatory variables. The default version is the 95% interval. But because the standard barrel of this estimate is are very close, two versions are very close to each other so the confidence intervals should be roughly equal. Yeah. Last part. What you could say about real business of the claim that you rate and the rate are co integrated. Yeah. When I run the test and good grandeur, the results are not consistent across alternative types of process. In one case I can reject the notion that the residuals are united and for all the cases I cannot reject. So I conclude that the claim that you rate and be rate our co integrated is not robust.

This question covers the concept of a confidence interval and whether a sample mean is inside the confidence interval to a certain degree of um confidence and how that how we can use that information to determine how confident the population or the actual mean is. So we're asked to do several things. First, we must construct construct a um normal probability distribution patient. Then we're asked to see whether that normal probability distribution is a normal, Is it a normal distribution? Then we're asked to find to 90 five 0.44% confidence interval. And finally were asked what um C means, in other words, were interpreting it. So the general overview of what we're going to do is for a we're going to construct a normal probability distribution which has the relative standard deviations and the plants, and if it's linear or normal distributed then we should expect a straight line or else if it's anything else, then we should not consider this normally distributed. So that's the first thing. We also want a 95.44 confidence interval, which means um we need information on the standard deviation which is helpful but were only given the population standard deviation, so we can use the formula, that's um the actual standard deviation is our guest standard deviation divided by the square root of n. Where n. In this case it's 15 since we have 15. Um We have 15 entries over here, so we take this and we times it by two from the mean. Which we guess is this because um this is a random sample, it's our best bet. Um Then then we interpret what that means, so what confidence in row actually means. So to get started, let's consider doing the first part of the normal distribution. So in general we need to find um we need to organize the lamps in increasing order, right? So um sorting this out in increasing order, we started by 15.7, finding the next highest one, 16.4 and so on. So after you find the correct order of things, you can generally constructed table where you have the lengths and the corresponding um Table three values. Okay, so table three is found at the back of the textbook and in this case where we have 15 values, So you would look for the 15th value of this right? Um Table three begins with negative one point negative 1.740 Um And it goes to 1.74 So this is the vertical axis, the horizontal access is the lamps. So once you have the graph it graphed out, you can then determine if it's normally distributed or not. So recall that normally distributed means that we tend to see a linear trend. So in this case we do see it, so yes, Yeah, it is normally uh it's roughly linear and it has no outliers. So yes, the cleaner with no else players, then we want to determine the 95 point for 4% confidence internal. So if you recall there is a 68 95 and 99% confidence interval pattern that should memorize and that basically represents the different levels of standard deviations. So 68% of the distribution Within one standard deviation 95 is within two and 99 0.7 or so is within three standard deviations. But in order to determine that we first need to know what the standard deviation is. So to find a standard deviation, approximate it with our current data provided by um n which in this case is 15. Okay, so it looks like I made a small mistake. Okay, large mistake. Um It's actually the other way around. Yeah, but the calculation should be same. And we're actually given this value to be um 1.76 and we know this is 15, so this is actually the population standard deviation that's given. Um Once we plug this in we can find that the value is points 45 for roughly. And if we have are mean approximation plus or minus two times our standard deviation then we should, because this is the 95.44% confidence interval within um Two standard deviations right. Two standard deviations. Yes, we should expect um We should expect the interval to be um Mhm I think but we also need to mean I guess so the mean you guys know how to, how to find the mean? It um turns out to be 18 point for two, So plus or minus two standard deviations, which is about 20.9 both sides should get you 19 points 34 two, 17.52 confidence interval, So 95.44%. It's within this interval, 95%. Um for 95% confidence that the mean carapace length of all adult male, Brazilian giant um Tony red trench loses somewhere between 17.52 millimeters to 19 points 34 millimeters. Finally, for the last part, we were asked what this really means um for if we want to find the mean, the mean is within two standard deviations, we're 95% dirt, 95.44 percent confidence. That's why it's called a confidence interval. That's our true mean is within this confidence interval and that's what it means, but we're not 100% sure. So it's not necessarily true that our population mean is within the interval. We're just 95.44% confident that it is


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