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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. $f^{\prime}(t)$

The following is a nova test based on the mean salaries for different metropolitan areas. So the alternative or the null hypothesis is that all the means are the same. So there are six metropolitan areas, I think it goes Chicago, Dallas Miami, Denver san Diego and Seattle. Uh So the null hypothesis is that all the means are the same. And then the alternative is that at least one of them is different. The second step is to find the critical value and you can do that using either software or a table, But they're essentially three things you need. The first thing is your alpha value, your significance level and that's usually given to you the problem and that's .05. Then you need the degrees of freedom for the numerator and the degrees of freedom for the denominator. And the way you find that Is the degrees of freedom for the numerator is the number of categories -1. So there were six cities that we looked at our metropolitan areas, so 6 -1° of freedom would be five for the numerator. And then for the denominators, the total number of data values minus the number of categories. So there were 36 data values minus the six metropolitan areas. So 30 is your degrees of freedom for the denominator. So that should be enough to use a table. But I use a calculator and I wrote a program in here called inverse. F. I'm not going to show you how to how to write the program. You can youtube it if you wish. Um But this is what I do. So um I put in my area which is my alpha value, my degrees of freedom is five and then my degrees of freedom for the denominator is 30 and That gives me my critical value. About 2.534 2534 is my critical value. I call f. star. So 2.534. Okay so anything greater than 2.534. We reject the annual hypothesis that all the means are the same And anything less than 2.534. We failed to reject meaning the h not is true. Okay so the second step is to find the F statistic and there's a formula but it's a bit of a mess. I always use software you know technology is a great thing. So if you go to stat and you can type in your data values. So these are the mean salaries um So again L1 I think was Chicago and then this is the mean salary for Dallas Miami Denver San Diego and Seattle. So there are six categories. And if you go to stat tests and then we're gonna go to the Unova test and then you just type in your columns separated by commas remember there were six columns, six data columns that we used and we need to make sure that all of them are in there and last one and then also you know make sure you separate those by commons, otherwise it's going to read it wrong. So then um that gives us everything we need. So the F. Is the F statistic, that's the third step. So we're looking at this it's about 2.281 as our F. Value. So two point 281 is our f statistic Which is actually barely in the non rejection region 2.281. So that means we fail to reject. Okay and also we can verify that with this p value here. So the p values 0.7 which is a pretty small p value, but it's still in this case greater than the alpha value. So the alpha value remembers point oh five, so it's barely greater than the alpha value. And whenever it's greater than the alpha value, uh we failed to reject, I should probably put H not there, so we failed to reject H not whenever the P values greater than the alpha. Okay. So then the last step is to summarize everything with actual words. So what does this all mean? It just means that there is not sufficient evidence, there is not sufficient. I guess you could say statistical evidence to suggest that the mean salaries from the different metropolitan areas are different. Okay. And that's the five step process for an Innova one way and over test

Following is the solution in number 14 at one way Innova test. Uh and this is about the mean sales prices for three cities. And the null hypothesis here is that the mean sale prices are the same for these certain houses. And then the alternative is that at least one of them is different. The second step is to find the critical value and you need three pieces of information to find the critical value. One is your alpha, Your significance level, that significance level. In this case, that's usually given to you is 10. They also needed the degrees of freedom for the numerator, which is the number of categories in this case, the number of cities minus one. So there were three cities that we looked at minus one is two, so degrees of freedom for the numerator is two degrees of freedom for the denominator is the total number of data values minus the number of categories. So in this case, if you counted up those data values, there were 31 -3 cities that we looked at. So 31 -3 is 28. So that's what we need. So from there you can use a table or you can use software. I'm gonna use software. So I wrote a program and I called it inverse. F. I'm not going to show you how to write this program. You can youtube it if you wish. But um it makes Makes it easier for me. So the area is the alpha value. So we'll put in .10 for that. Degrees of freedom from the numerator was too. And then degrees of freedom for the denominator was 28. And that's going to spit out my f. star my alpha value and my uh critical value which is about 2.503. Let's call it 2.503 is my F. stars 2.503. So anything greater than 2.503. We're going to reject h not anything less than 2.503. And we're gonna fail to reject the null. The next step is to find the f statistic and you can do that manually but especially with bigger data sets that can be really time consuming. So I went ahead and punch this into stat. Edit. And these are my data values. So these are the I think these are in thousands of dollars but these are the mean sale prices and if you go back to stat and then tests and the very last one in nova And you put in your columns just make sure you separate them. This is on the T. 84 by the way but make sure you separate them by commas otherwise it's not gonna read it right so nova for those three columns and that's gonna give us everything we need to. The f statistic is about 0.966 Let's go and write that down. So 0.9 66 which is somewhere over here. So that lands in the non rejection region. So that's actually gonna tell us why our fourth step which is the decision and we're going to fail to reject H not since the F statistic is less than the critical value. Now you can also use the P value method that's what this second piece of information is good for it. Now this other stuff doesn't really matter. Um You can just kind of ignore it because this is really what we need. We need the F statistic and we need the p value and the p values pretty large. It's about 0.39 And what you do is you explicitly compare the P value with your alpha value. So the p value in this case is greater than your alpha value. 0.39 is bigger than 0.10. And any time you're P values greater than alpha, you failed to reject H nine. If it's less than alpha then you you reject. And then the final step is to conclude this, you know, with actual words and bring it back to the question at hand. And so what we're going to say is that there is not enough evidence or there is not sufficient statistical evidence. So there's not sufficient evidence to suggest that the mean sales price prices of houses In the three cities are different. Okay that's the five step in Nova process.


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