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When the mutuallstlc, defensive ant populatlons decllne and CCacn tree and Its resident ant populatlon roplaced by competing change? The relationshlp changes (rom m...

Question

When the mutuallstlc, defensive ant populatlons decllne and CCacn tree and Its resident ant populatlon roplaced by competing change? The relationshlp changes (rom mutualism paraslism The relationshlp changes from mutualism predation: The relationship changes from mutualism competition . The relationshlp changes from mutuallsm commensallsm The relatlonshlp remalns mutualisticspcces;, how 'does tha reladdonshlp betweenBy simply comparing the condition of acacla trees Inslde and outslde ot the

When the mutuallstlc, defensive ant populatlons decllne and CCacn tree and Its resident ant populatlon roplaced by competing change? The relationshlp changes (rom mutualism paraslism The relationshlp changes from mutualism predation: The relationship changes from mutualism competition . The relationshlp changes from mutuallsm commensallsm The relatlonshlp remalns mutualistic spcces;, how 'does tha reladdonshlp between By simply comparing the condition of acacla trees Inslde and outslde ot the enclosure can the Teseanchen actually Detennine Prescnce non-mutuallstic anr caused the observed Incrovst borng bcetle Infestatlon? The reseancher wolldeneec stan Over an monltor ant colonization and beetle Infestatlon on como etcly new Sets Fcnceo and unfenced acacl rees Yes. Tne correlation between non mutualese ants and beetles strong enough [0 sho causatlon: No. The rescarchers would need compare Deetle Infestation levels trets where the non-mutualistc ants had been physically removed beetle Infestation levels trees where the ants had not been removed Yes: Tine dliference the number beetles between trees Inslde and outsde the encosurc sufncicnt [0 {ho causation Tner othar possibla explanabon for the Increase banng bccte Infestabon that ias posenicd acacia tree; (nhablted bY non-mutubllistc ants



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In a hypothetical population of beetles, there is a wide variety of color, matching the range of coloration of the tree trunks on which the beetles hide from predators. The graphs below illustrate four possible changes in the beetle population as a result of a change in the environment due to pollution that darkened the tree trunks. What would be the most likely change in the coloration of the beetle population after pollution and why?
a. The coloration range shifted toward more lightcolored beetles, as in diagram I. The pollution helped the predators find the darkened tree trunks.
b. The coloration in the population split into two extremes, as in diagram II. Both the lightcolored and the dark-colored beetles were able to hide on the darker tree trunks.
c. The coloration range became narrower, as in diagram III. The predators selected beetles at the
color extremes.
d. The coloration in the population shifted toward more dark-colored beetles, as in diagram IV. The
light-colored beetles were found more easily by the predators than were the dark-colored beetles.

Okay to solve this problem. First, we need to start by finding the palladium so apart, eh? Eh? So let's start with the first equation. Because this is only an equation of one variable. So we'll set that equal to zero. So here, see, one is equal to five. So we have five p one, uh, times one minus p one minus m one. Em one is equal to three. So three P one is equal to zero. So let's, um, work this out. So we have five p one minus five p one squared, minus three P one is equal to zero. So we have, uh, two p one minus five p one squared is equal to zero. Let's factor Rdp one. So we get p one times two minus five p one equal zero. Let's set each factory with zero. Then we get either p one equals zero or P one is equal to 2/5. All right, now, let's look at equation too. All right. We're gonna try to simplify equation to first as much as possible. Yes, we have 30 p too. Um, minus four times one minus P. One minus p too. And then minus I am, too. Is equal to three. So three p, too. And then minus soc one is five. So minus five p. One p. Too busy with zero. Now, we can factor out a P two from all of it. So we get 30 minus 30 p one minus 30 p too. Okay, then. Just minus three p too. Minus five p. One equal zero. So again, let's keep simplifying. We have P too. Times 30 minus 35 p one. 35 p one minus 33 p two equals zero. Whoops. Sorry. I, um Ms. Calculated here. So that should be minus three. Just minus three and then minus three minus five p one, like so is equal to zero. So then now we're left with this is gonna be 27 minus 35 p one minus just 30 p too. Okay, so we can break that into our two factors. So we have p two equals zero, or, um, we can say pee too, You know, 30 p, too is equal to 35 p one minus 27 minus 27. Like so. Okay, so now we plug in these two solutions from the first equation into both of our, um, so called this 13 This 14 rights, we plug in three and four into both of these and for so first, we have p two equals zero and p one equal zero. That's fine. Then we have p two equals zero and p two equals 2/5. So, uh, we have that. Our equilibrium points. First, we have 00 uh, 00 hips and then 0 2/5 or sorry, um, tooth ifs comma zero like so. Okay, so those aren't too. Now, let's do the, uh, second half second set. So we have P too is going to be equal to 35 times. 1st 0 minus, um zero minus 27. Divided by 30. So we just have negative 27/30 there. So we have that pee, too, is equal to 9/10. Now, if we plug in for the next one, p two is equal to 35 times 2/5 minus 27. Divided by 30. We get that pee too. P two is equal to and here we get 13/30. Okay? We get 13/30. So are two equilibrium points. are, uh, our other two. So we have four total. We have zero comma, 9/10 and then we also have to fifth comma, 13/30. So these are our four equilibrium points for part A. All right, so now for the Jacoby in, it'll be easiest to write out. Um, the equation in, um, you got that equation? Expanded form. So we have. For the first equation, we have five p one. Okay, minus five p. One squared and then minus three p one. So this is equal to F one. Um, now for the second equation, let's expand that out. I'll, um, So we can use this aversion here, Okay? Except we just multiply a P two throughout. So we have 27 p two minus 35 p one times p too, and then minus 30. Key to square. Okay. And this is able to r f, too. So to find the f one the, uh, p one, this is equal to five minus 10 p one minus three er 10 p one, minus three. So this is equal to two minus 10 p. One now. 40 f ah. One D p too. This is just going to be equal to zero. Now, let's look at the f. Uh huh. I mean, actually, put that, um, up here instead. Okay? Now for D F too. The P one. There's only one term with P one here. So is gonna be minus 35 p too. Okay. Now, for D F to the P two, this is equal to 27 minus 35 uh, 35 p one minus 60 p, too. So for part B, our answer the Jacoby in P one p two is equal to, uh so to minus 10 p one 10 p. One zero negative, 35 p too. And then 27 minus 35 p one minus 60 p too. So that's our answer for part B. Now, for part C, we need to plug in all of our equilibrium points. So I see. So we needed to take J of 00 was our first equilibrium. Point zero comma zero. So this is equal to to minus 10 times zero. Okay. Uh, zero negative. 35 times zero men 27 minus 30 was a 35 p. Once a 35 times zero, minus 20 right or minus 60 times zero times zero. Hey, this simplifies to two zero 0 27 Okay, so since this is triangular matrix, all of them will be trying to the matrices. Actually, we actually only care about these two numbers here to find the stability. Okay, so these are our, um, Aiken values here. Since they're both greater than zero, we have ah, um, unstable. We have an unstable equilibrium at 00 Okay, So unstable at, um 00 Okay, Now, for the next equilibrium point, that was 2/5 comma, zero. So to over five comma zero. Now, this is going to be equal to so to minus 10 times to fifth over. Five here. Zero negative. 35 times zero. Um, who's 27? Minus 35 times to fifth. So two divided by, uh, five and then minus 60 times zero. Simplify that. We have. So to minus 24 5 um, to minus 20/5. So to minus four. That's gonna be negative, too. Zero. Oh, see here, zero 200 And then again, um, so too minus 27 we have. Okay, So here are Aiken. Values are too and negative. 27. So, again, this is gonna be unstable, Dad. Zero common. 9/10. Okay, now our last equilibrium point. So that's J ads, Um, to fifth 2/5, comma 13/30. Okay, so here we have. So to minus, uh, 10 times. 2/5 Negative. 35 times 13/30. 27. Minus 35 times two over five. Minus 60 times searching over 30 zero. So 13/30. 13/30. Plug in our 2/5. Two very five. Right. So this gives us a kn answer of, um, here. This gives us a negative, too. Zero negative. 91/6, and then negative 13. So here are Aiken. Values are negative, too. And negative. 13. Therefore, um, this is actually going to be stable. Uh, so stable, Dad. 2/5. Um, 2/5. 13/30. So, for part B, yes, we will have, um, a part where or there will be able to exist at a stable equilibrium at, um, that last one. That last equilibrium

In this exercise, we are given the equation for a parameter data which is given here. And we can also estimate theater by drawing samples from each of populations 123 and four. In that case, our estimator would be given by this equation. Now, we're also told that when and one and two and three and four are large, then our estimator is normally approximately normally distributed and so, with that in mind, were asked to derive a confidence interval for theta. Now, since data are estimated for data is normally distributed and our confidence interval can be of this form and so are critical value is found in the in the usual method and our estimated for data is given by this equation. And so what we have to do is find the standard deviation for our estimator for data. Now, we can do this by using our standard rules for periods. I hope you can see this here. That is the variance of X sub four bar. The standard deviation would be the square root of the variance. Another variance of X sub one bar is is going to be s sub one squared, divided by an sub one. So that is the standard deviation of sample one. The sample standard deviation of sample one divided by the size of sample one and the same can be done for the variance of except to bar the sample standard deviation from sample to divided by the size of sample too. So we take the square root of all of this. And so now, returning to our equation for a confidence interval, we can see it is equal to you. So I'm just adding the estimator for theater and then our critical value. So are critical value in terms of Alpha can be written like this. And now I just have to multiply this by the standard deviation of our estimator for theater. And so this answer is the first part of the question. This is hour derivation of the confidence interval in terms of Alpha. And then, we're asked, is the data that is presented in the question to find a 95% confidence interval for data now for 95% confidence interval are critical. Value is 1.96 and now we really just use this equation and plug in all of the information given in the question. So this is the average for sample one and so on. And then we just have to calculate this. And this gives us an interval ranging from minus 0.837 two minus 20.163 And that answers the second part of the question. This is hour 95% confidence interval for data.

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.


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