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3Motl,tueal Mimuimim Muiniedtuimiii jetly |u M Oiniuu Miuiniui Mdt utae, iinimiiiu mii Nud Oruot Wiatlar (") Ihiu aull wa; ch tt ( dl "m |eu WtW Miiimaui Wue4 € ''YII eald ,1445 4"40"1147 1,270"M6" 1 1,| 71 0,6'1 1,41 1.1 (1|117 1,57 3 166 2'Io valuns 0 Il Jumple uuui Jampla etondouu devialioi, WWd (astimated) standard error 0l me (nean 'W 6uWiu { T [auvuly Isabbr: "Wi *ilii swuti Mat te |rue Avorage purcentage of mmiuim Jiuuniu or Mm

3 Motl,tueal Mimuimim Muinied tuimiii jetly |u M Oiniuu Miuiniui Mdt utae, iinimiiiu mii Nud Oruot Wiatlar (") Ihiu aull wa; ch tt ( dl "m |eu WtW Miiimaui Wue4 € ''YII eald , 1445 4" 40"1 147 1,2 70" M6" 1 1,| 71 0,6'1 1,41 1.1 (1| 117 1,57 3 166 2' Io valuns 0 Il Jumple uuui Jampla etondouu devialioi, WWd (astimated) standard error 0l me (nean 'W 6uWiu { T [auvuly Isabbr: "Wi *ilii swuti Mat te |rue Avorage purcentage of mmiuim Jiuuniu or Mmn: "l iwlii idui1u Iuiiii'ut nt y Out 10st 0/ the appropriate hyootheses at "uu mine leve I, | Noto m minu [2 Ouobimy Vldt otho datu shaws acceptable pattetr In lignt of Ulm renaonably |oueje edmbu s. | 'unioday #IiIdo Idelu Ui #iuis, 16 < Calculate Ule tcut statistlc Ont detcrming tha P-valuc_ (Round vaur test statistc to twa declmal places and Oniea-4 Jnoa three decdmal piacen. ) P-valuc Whal cun Vou conclude? Reject te null hypothests_ There Is sufficlent evldence t0 conclude that the true average percentuge oranic matter thi; type Dr S0il /s something ather thon 39_ Reject the null hypothesls_ There not sufricient evldence t0 conclude that the true average percentage o/ organlc maltcr thls type 0f soil somethlng otner tnan 3"0_ Do not refect the null hypothesls. There aufilcient &vidence conclude that the true uverage percentage of organlc matter In thle type ot soll (z sornethlng other than 39 Do not reject the rull hypothesle. There not sufficlant evidance t conclude that the true averaqe percentage organlc matter thls type of soll something other than 390 Would vour conclusion be different 0.05 had been Used? Reject the null hypothesis There sufficient evldence ta conclude that the true averuge percentage of organlc matter Mlos jo admi 5/41 something ather than 3%. Reject the null hypothesis, There not sufficlent evldence ta conclude that the true average percentage af organlc matter this type of soll sornething other than 3%. Da not refect the null hypothesis_ There I5 sufficlent evldence to conclude that the true average percentage of organlc matter In this type or soll Is somethlng other than 3%. Do not reject the null hypothesis, There not sufticlent evidence to conclude that the true average percentaga pf organlc matter thls type of soll Is somethlng other than 39.



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Expand Your Knowledge: Logarithmic Transformations, Power Law Model When we take measurements of the same general type, a power law of the form $y=\alpha x^{\beta}$ often gives an excellent fit to the data. A lot of research has been conducted as to why power laws work so well in business, economics, biology, ecology, medicine, engineering, social science, and so on. Let us just say that if you do not have a good straight-line fit to data pairs $(x, y)$, and the scatter plot does not rise dramatically (as in exponential growth), then a power law is often a good choice. College algebra can be used to show that power law models become linear when we apply logarithmic transformations to both variables. To see how this is done, please read on. Note: For power law models, we assume all $x>0$ and all $y>0$.
Suppose we have data pairs $(x, y)$ and we want to find constants $\alpha$ and $\beta$ such that $y=\alpha x^{\beta}$ is a good fit to the data. First, make the logarithmic transformations $x^{\prime}=\log x$ and $y^{\prime}=\log y .$ Next, use the $\left(x^{\prime}, y^{\prime}\right)$ data pairs and a calculator with linear regression keys to obtain the least-squares equation $y^{\prime}=a+b x^{\prime}$. Note that the equation $y^{\prime}=a+b x^{\prime}$ is the same as $\log y=a+b(\log x)$. If we raise both sides of this equation to the power 10 and use some college algebra, we get $y=10^{a}(x)^{b} .$ In other words, for the power law model, we have $\alpha \approx 10^{a}$ and $\beta \approx b$. In the electronic design of a cell phone circuit, the buildup of electric current (Amps) is an important function of time (microseconds). Let $x=$ time in microseconds and let $y=$ Amps built up in the circuit at time $x$. $$ \begin{array}{l|lllll} \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y & 1.81 & 2.90 & 3.20 & 3.68 & 4.11 \\ \hline \end{array} $$ (a) Make the logarithmic transformations $x^{\prime}=\log x$ and $y^{\prime}=\log y .$ Then make a scatter plot of the $\left(x^{\prime}, y^{\prime}\right)$ values. Does a linear equation seem to be a good fit to this plot? (b) Use the $\left(x^{\prime}, y^{\prime}\right)$ data points and a calculator with regression keys to find the leastsquares equation $y^{\prime}=a+b x^{\prime} .$ What is the sample correlation coefficient? (c) Use the results of part (b) to find estimates for $\alpha$ and $\beta$ in the power law $y=\alpha x^{\beta} .$ Write the power law giving the relationship between time and Amp buildup.
Note: The TI-84Plus/TI-83Plus/TI- $n$ spire calculators fully support the power law model. Place the original $x$ data in list $\mathrm{L} 1$ and the corresponding $y$ data in list L2. Then press STAT, followed by $\mathbf{C A L C}$, and scroll down to option A: PwrReg. The output gives values for $\alpha, \beta$, and the sample correlation coefficient $r$.

Following is a solution video to number 24 and this is where we compare to means uh for the soil, water content for field a compared to feel b. And the first part is just to verify that the mean and the standard deviations are in fact these numbers, So the mean for field day is 12.53 and the standard deviation is 2.39 And then the mean for field B was 10.77 with the standard deviation of 2.4, and it says use a calculator something use this T I 84 I went ahead and typed in the means, or sorry, the data values L one and L two, so L one represents fuel day and then L two represents, you'll be and if you go to stat falcon and it's one of our stats and we're going to change that to L one and calculate and that gives us everything we need. So the X bar, is that 12.53 So that's verified. And then we're looking at the s the standard of the sample standard deviation is about 2.39 So for the field A. That is correct. And then let's just go and double check field be. So we're gonna change it to L. Two now because that's where field B. Is, and then that verifies its 10.77 for the mean, then about 2.40 for the standard deviation. So the first part is done, that's verified. And then the second part it says conduct not formally, but we're basically just going to conduct a uh uh two sample T. Test with an alpha value of point oh five. And we're gonna go back to the calculator because doing it by hand can be pretty cumbersome. So if you go to stat and then test now, since we already have, we're gonna go to two sample t. Test since we already have the data in there. I'm just gonna use the data instead of the summary stats. So list one is L. One that's the field A. List to is L. Two. That's field B. And then these frequencies just keep them the same and then we have the the alternative hypothesis and you kinda have to use your context clues here. But this one actually explicitly states is field A. Is the soil content or water content higher or greater than. So I'm going to change this to greater than So μ one is greater than you to warm you. A. is greater than YouTube pulled is usually zero or no unless they tell you otherwise. So we can go ahead and calculate and that's gonna give us everything we need. So the T. Value if you want you can put it down there. But really it's just this P. Value that I want to see. So 00.27 is the p value. So P value equals 0.27 And then we compare that with the alpha value and it's less than the alpha value. So any time it's less than the alpha value. That means we reject the no hypothesis. So we're rejecting the statement that these means are the same. And we are I guess you could say accepting the fact that these two means are in fact the water content and field A. Is greater than the water content and field be on average. So we can type this out as there is enough evidence to suggest that field A. Has on average a higher soil content soil. Sorry, water content. Banfield beat. Okay so this field A. Has on average a higher soil water content on field because since we are rejecting that null hypothesis and accepting the alternative hypothesis.

So in this particular problem were asked to do several things. So first of all, we need to identify Alpha, the significance level is 1% 0.01. And we know that the null hypothesis is that P is 0.301. And then the alternate hypothesis would be that P is less than now to figure out what kind of distribution it is. It's going to be standard, normal. And a quick check Is N. Times P. Greater than five. And indeed it is. So to 15 time zero point 301 is indeed greater than five because it equals 64.7 about And then his end times Q Greater than five. So if P is 301, Then 1 -301 will give us Q. So Q will be 0.699. So when I multiply those together To 15 times 0.699, I get approximately 150. So yes, that is indeed greater than five. Now I need to find the test statistic which in this case will be P hat and they're asking us to find the Z value. So I'm going to do this all at once with my calculator. Probably got it worked out but let me walk through it with you. So stat tests. This is a one proportion Z test. So number five. Now the probability of success is .301. We're told that our is 46. So in the calculator that's the x the total is population is 215. Were testing if it's actually a less than cursor down to calculate and the information they need. It's all right there. So I have this on the other screen. So P hat Is .21 depending on rounding here 214 and Z is -2.78. And in this particular situation I'm also given the p. value Which is this one. I know you got all these peas, you got the little P. For probability and you've got row and you've got your P. Value 27 So that's part C. Were asked for to find the P value. So if we're if we're actually shading this then we make our normal curve And here's negative 2.78. And I'm shading to the left. So that's if we had to shade it. Now I need to check this. So is this P value less than greater than or equal to my significance level? And I can see that my P value is less than or equal to. So that means I need to reject the no and then how do I write that out? How do I explain that? I would say something like At the 1% level of significance. The sample data indicate that the population proportion in the revenue is less Than 0.301. So this next part is asking for your opinion. And if you're doing a multiple choice question, the answer might be a little bit different than what I worded here. But This indicates the fact that P is in fact if he is in fact less than 0.31, Then that indicates that there are not enough numbers that start with one. So, yes. So what does that mean as a stockholder? Well, as a stockholder, that could mean that the value of your stock is inflated for the FBI. That might be a red flag to investigate. Because, according to Bedford's law, there should be a certain amount of Values that start with one. So for a stockholder it could mean that your stock is not worth as much as you think it is. And then for FBI this could be an indication that there's fraud. Now. Finally, just because we reject the null hypothesis doesn't mean that we have proved anything. So we did not prove H. Of zero, which was the fact that the probability should be this. Mhm. All we did was take some sample data and because the sample data let us to reject the null, then there could be too few numbers with leading digits of one. So you need to investigate more. So it's not an indication that this is actually false, but it's an indication that more investigation needs to be done. Maybe another sample or maybe a larger sample.


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