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Flight formula for Indian spotted owlets The following table shows the average body mass $m(t)$ (in $g$ ) and average wing chord length $w(t)(\text {in } m m),$ along with the derivatives $m^{\prime}(t)$ and $w^{\prime}(t),$ of $t-$week-old Indian spotted owlets. The flight formula function $f(t)=w(t) / m(t)$ which is the ratio of wing chord length to mass, is used to predict when these fledglings are old enough to fly. The values of $f$ are less than $1,$ but approach 1 as t increases. When $f$ is close to $1,$ the fledglings are capable of flying, which is important for determining when rescued fledglings can be released back into the wild. (Source: ZooKeys, 132,2011 ) $$\begin{array}{lrrrr} t & m(t) & m^{\prime}(t) & w(t) & w^{\prime}(t) \\ \hline 1 & 23.32 & 41.45 & 10.14 & 14.5 \\ 1.5 & 50.59 & 64.94 & 20.13 & 26.17 \\ 2 & 82.83 & 57.95 & 36.7 & 39.86 \\ 2.5 & 105.13 & 31.08 & 58.92 & 47.11 \\ 3 & 115.48 & 12.48 & 81.55 & 41.38 \\ 3.5 & 119.4 & 4.44 & 98.99 & 27.94 \\ 4 & 120.76 & 1.51 & 109.75 & 15.74 \\ 4.5 & 121.22 & 0.51 & 115.5 & 7.99 \\ 5 & 121.37 & 0.17 & 118.34 & 3.85 \\ 5.5 & 121.42 & 0.06 & 119.69 & 1.8 \\ 6 & 121.44 & 0.02 & 120.32 & 0.84 \\ 6.5 & 121.45 & 0.01 & 120.61 & 0.38 \end{array}$$ State the units associated with the following derivatives and state the physical meaning of each derivative.

A. $m^{\prime}(t)$ b. $w^{\prime}(t)$
c. $f^{\prime}(t)$

.

Hello and welcome. We're looking at chapter eight, Section one Problem 39. So that just to this problem is we want to find out if circles overlap s So we have these two circles. I noticed that this first circle has a center of 23 in a radius one. The second circle here has a center of 32 in a radius of 1/2.

So 1/2 not 1/4. Remember whatever this is the square of the radius. So 1/2 square. This one. So we have these two circles.

We'll find out how they overlap so we can graft them. I'll show you the graphs in a second, but I want to show you how we could do this. Algebraic Lee. So if we have two circles, there's some distance between their centers. So these are the centers of the circles.

Call this center one in center too. So there's some distance between the circles. Those circles also have their own respective radio are one and are too. So if the distance between the centers is greater than the sum of those two Rady, I means the two radio can't reach can't reach each other there's too much distance to cover, so they would not overlap. However, if the distance is less than the sum of the radio, then they will overlap.

So that this is this right? Here is the logic we're gonna use. So for part A, we just need to find the sum of the radio. So, um, are one Let's do this. Um, here are one plus are too equals 1.5 and then we need to find the distance between the two points. The two centers, this will be two miles, three squared, plus three minus two squared that is the square root of two.

And so, um, square to two is about 1.4. So the distance is less than the sum of the radio, so it must overlap. All right. So part A, they have to overlap on the reasoning eyes from this logic up here. Essentially.

So we did the calculation for the distance here in the sum of the radio I their rights. That's part a duck. Um, if we go to part B, it's a very similar problem. Just in three dimensions. Uh, four.

Do part B. Let's just look at the graph. here s O. I grafted these two circles on Dismas and you can see there's just a sliver. They barely overlap, but they do overlap, right? We were We were exactly right with our work for part B, Part B.

We have 23 D circles, so Mrs X minus two squared starts the same. Then we're introducing dizzy value. So our first circle has a center of 231 with radius one in our second circle has a center 32 zero with radius 1/2. So definitely some similarities there. Um, now we're gonna use the same logic it worked in to Space and two D.

It's gonna work in three D if the sum of the Raid e I, um, is not enough. If it's not greater than the distance between the centers, then they're not gonna be able to overlap. So the sum of the radio I is 1.5 again. Let's check the distance between these two points so the distance will be two minus three squared plus three minus two squared, plus one minus zero squared. That's all going to simplify to be the square root of three squared of three is about 1.7.

So the distance, uh, between the center's is greater than some of the radio. That means the radio. I don't have enough, uh, enough juice to cover the distance between the centers. Um, so they don't overlap. All right, So, uh, part B is done.

We've proven that it doesn't overlap using most of the same logic we did with part. All right, if we want to see some visual proof of that, I got a Geo Deborah three graph for to graft These two and they are both, um they don't overlap, so we can see clearly that they do not overlap now for part C, it's pointing out that there are a lot of similarities between part in part B. Um, what's the relationship between the two s o? If we actually rotate our view for part B here, it becomes really clear. So I'm just if you just imagine rotating this, the red line is the X axis. The green line is the y axis, but we're just basically looking from the top down.

Essentially, what we're looking at is the projection of those two spheres onto the X y plane. You'll notice it looks very similar to our picture for the two D I. So that's the main. That's the relationship right there that if you do, ah, projection onto the X Y plane. Oh, part B.

It is part. So that is a B and C or problem. 39. We're done..

For the match pairs given on the right. We want to conduct a sign tested matched pairs testing P does not equal 0.5 at alpha equals 0.5 significance. This question is testing our understanding of non parametric tests in particular how to conduct a scientist of matched pairs. We proceed there steps A through D below to solve. So first, in a we stayed alpha hypotheses, this gives alpha equals 0.5 H.

And r P does not equal five. H. A P does not equal five. So the no hypothesis P is equal to have, the alternative is not A and B. We compute the test out.

So are signs for these matched pairs are as follows and equals 12 is the total number of matched pairs. So X equals the number of plus or the total number plus and minus which is 4/12. Thus we have seen it equals x minus 5/45 over and equals negative 1.155 Thus the p value is for a normal distribution, P equals two PZ greater than zero equals 248 Thus we conclude that we fail to reject asian off because he is greater than alpha, which means that we lack evidence to support the claim P does not equal five..

All right, So in this question, we are given data regarding the costs of deliveries and we want to find a range in the sample standard deviation for this dataset. Um and what I've done here, I was actually just organize the data from least to greatest, which is helpful because when we want to find the range, that's actually just going to be the biggest minus the smallest, so it's going to be 34, 70 -16 84. So if you organize it from lowest to highest, the highest to lowest, um it's very clear which ones those are, especially when we're dealing with a lot of numbers or what big numbers, it's good to sort that out. Um so this difference is going to be 1,786, that's going to be in dollars. So the range is $1786 to find the example standard deviation.

Um Well, that actually relies on the square differences from the mean, so we actually have to figure out what the mean is and well the mean is just going to be the sum of all the terms. You add them all up and then divide by the total number of terms, which in our case is If you add them all up and divide by eight, that's going to give you a mean of $2,589.4 um and that gives you the average cost of a delivery. Well, the important reason we need to know the mean aside from giving us useful information about the data set, is that, like I mentioned, the standard deviation relies on the square differences from the mean, namely right for each data point, we are going to calculate the square difference from the mean. So we're going to subtract, so we're going to start with the first data point, subtract the means to 5 8 9.4. And we are going to find that square different.

So we're going to square it. And in this standard deviation expression in the numerator is the sum of the square differences from the mean. So we're going to some Across all of our terms. So this is our second term -2 means this is our third term minus the mean squared minus mean tree. 94 Weird.

Right? And you do this for all of your terms, right? Our denominator is simply going to be N -1. So and in this case is eight. So our denominator is seven. All right. Well, our setup is done.

All we have to do now is political calculator. And be careful. That's also what computers are for their very helpful with this. But when you add all of that up and do your squares, you'll get three million and 84,000, 890 approximately. That's going to be divided by seven and that all simplifies to be $663.9 as extended deviation.

And they accessed around 21 more data point or one more decimal than is provided. And that is precisely done. Yeah..

Now this problem were asked to investigate the difference between means for different bird types on whether the mean length of the coup eggs is different for different species. So what you're gonna want to do is plug all of this data into your statistical program or your calculator to get some basic statistics here. So we want our sample sizes. So I'm gonna say N s is for Sparrow. Well, Dio Robin will be in our on dwa AG tail will be in w.

So I have here 15 for my sample set 17 and 16 were also going to need the mean value for those eggs. So the mean for the sparrow was about 23.2 The mean for the robin was about 22.53 on the mean for the wag tail came to about C 22.8 four. We're then going to need the standard deviation. Cistern deviation came to 0.979 Standard for the robin came to 0.686 and the standard for the wag tail deviation came Teoh 1.6 five. All right, so from here, if you have statistical software, you can go ahead and run analysis and get your confidence interval.

If you need to know how to do it by hand, I will do it for one example. So if we wanted to compare the Sparrow to the robin with a 95% confidence interval, we're going to use the formula the mean of the first sample minus the mean of a second sample to get that difference, plus or minus R T statistic based on degrees of freedom times the standard era. So a T statistic based on degrees of freedom. We're going to look up in a tea table on this degrees of freedom. You need to use your computer to find you're gonna plug in this basic information that we've already found and we get for the first example that degrees of freedom for Spiro versus Robin comes to 20 four.

If you look up in a tea table Alfa of 0.5 for two tailed T test, This is coming from our 95% confidence, right? So one minus 10.95 gives us at 0.5 Alfa with degrees of freedom of 24 R t statistic is 2.63 nine. So we've gone ahead and found this first part of our formula. We also need to find the standard error. You could do this by hand by taking the square root of the standard deviation of your first sample squared, divided by and for your first sample, plus the standard deviation for your second sample square divided by the end for that second sample for us. If I was to plug in these values, it would look like 0.979 squared, divided by 15 plus 0.686 squared, divided by 17.

This value altogether gives you 0.302 six. Now we're able to actually construct are 95% confidence interval. So we get our first mean 23.0 to minus our second me and 22.53 plus or minus that t statistic 2.639 times 0.30 to 6. And if we were to solve all this out, get plus or minus, we wind up with a confidence interval of negative 0.565 20 Oh, I'm sorry. That is the in Kratch Interval.

I'm getting ahead of myself here. You would wind up with negative 0.114 to to 1.962 So the really important part of this confidence interval, whether you found it by hand or you ran a test in your program, is that we're ranging from a negative to a positive. So what that means is that if we were to place this on a number line, our range of possibilities includes zero meaning with 95% confidence. There might not be any difference between these two egg sizes. The same is true if you compare the sparrow doing the exact same process the sparrow to the wag tail.

You mind up whether you do this in your program where if you do it by hand like we just did with an interval of negative 0.565 to 0.936 one. All right, so that again we have zero included in that confidence interval. So what that really means is that here we go. I ran out of room. There is that there might not be any difference at all.

So is 95% confidence intervals. We say that these do not have any statistically significant difference between them.

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