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Arborists are often interested in developing procedures that will allow them to accurately measure trees without subjecting themselves to danger. Suppose an arboris...

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Arborists are often interested in developing procedures that will allow them to accurately measure trees without subjecting themselves to danger. Suppose an arborist collects sample of trees and measures the height (ft ) and circumference (ft ) of each. The following data resulted: Height Circumference (ft) (ft) 1Assuming we meet all necessary assumptions, is our sample correlation coefficient of a sufficient magnitud: to indicate that; in the population of trees circumference and height are cor

Arborists are often interested in developing procedures that will allow them to accurately measure trees without subjecting themselves to danger. Suppose an arborist collects sample of trees and measures the height (ft ) and circumference (ft ) of each. The following data resulted: Height Circumference (ft) (ft) 1 Assuming we meet all necessary assumptions, is our sample correlation coefficient of a sufficient magnitud: to indicate that; in the population of trees circumference and height are correlated? Answer this question by conducting formal hypothesis test using an 0.05 significance level: Assume that the mean circumference is 5.429 the standard deviation for the circumference is 2.992, the mean height is 45.571 , and the standard deviation for the height is 19.823, and 0.9281 _ What is the null hypothesis for this test? Ho: p # 0 Ho: r = 0 Ho: r # 0 Ho: p = 0 Arborists are often interested in developing procedures that will allow them to accurately measure trees without subjecting themselves to danger: Suppose an arborist collects a sample of trees and measures the height (ft ) and circumference (ft ) of each: The following data resulted: Height Circumference (ft) (ft) 35 53 29 53 Assuming we meet all necessary assumptions, is our sample correlation coefficient of a sufficient magnitude to indicate that; in the population of trees, circumference ad height are correlated? Answer this question by conducting formal hypothesis test using an 0.05 significance level. Assume that the mean circumference is 5.429, the standard deviation for the circumference is 2.992, the mean height is 45.571, and the standard deviation for the height is 19.823, and 0.9281. Calculate the value of the test statistic for this test. a,t = 6.106 b.t = 8.635 t =5.574 d.t=-5.574



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Expand Your Knowledge: Logarithmic Transformations, Exponential Growth Model There are several extensions of linear regression that apply to exponential growth and power law models. Problems 22-25 will outline some of these extensions. First of all, recall that a variable grows linearly over time if it adds a fixed increment during each equal time period. Exponential growth occurs when a variable is multiplied by a fixed number during each time period. This means that exponential growth increases by a fixed multiple or percentage of the previous amount. College algebra can be used to show that if a variable grows exponentially, then its logarithm grows linearly. The exponential growth model is $y=\alpha \beta^{x}$, where $\alpha$ and $\beta$ are fixed constants to be estimated from data. How do we know when we are dealing with exponential growth, and how can we estimate $\alpha$ and $\beta$ ? Please read on. Populations of living things such as bacteria, locusts, fish, panda bears, and so on, tend to grow (or decline) exponentially. However, these populations can be restricted by outside limitations such as food, space, pollution, disease, hunting, and so on. Suppose we have data pairs $(x, y)$ for which there is reason to believe the scatter plot is not linear, but rather exponential, as described above. This means the increase in $y$ values begins rather slowly but then seems to explode. Note: For exponential growth models, we assume all $y>0$. $$ \begin{array}{|l|lrrrr} \hline x & 1 & 2 & 3 & 4 & 5 \\ \hline y & 3 & 12 & 22 & 51 & 145 \\ \hline \end{array} $$ Consider the following data, where $x=$ time in hours and $y=$ number of bacteria in a laboratory culture at the end of $x$ hours. (a) Look at the Excel graph of the scatter diagram of the $(x, y)$ data pairs. Do you think a straight line will be a good fit to these data? Do the $y$ values seem almost to explode as time goes on? (b) Now consider a transformation $y^{\prime}=\log y .$ We are using common logarithms of base 10 (however, natural logarithms of base $e$ would work just as well). Consider the following data, where $x=$ time in hours and $y=$ number of bacteria in a laboratory culture at the end of $x$ hours. (a) Look at the Excel graph of the scatter diagram of the $(x, y)$ data pairs. Do you think a straight line will be a good fit to these data? Do the $y$ values seem almost to explode as time goes on? (b) Now consider a transformation $y^{\prime}=\log y .$ We are using common logarithms of base 10 (however, natural logarithms of base $e$ would work just as well). $$ \begin{array}{l|lllll} \hline x & 1 & 2 & 3 & 4 & 5 \\ \hline y^{\prime}=\log y & 0.477 & 1.079 & 1.342 & 1.748 & 2.161 \\ \hline \end{array} $$ Look at the Excel graph of the scatter diagram of the $\left(x, y^{\prime}\right)$ data pairs and compare this diagram with the diagram in part (a). Which graph appears to better fit a straight line? (c) Use a calculator with regression keys to verify the linear regression equation for the $(x, y)$ data pairs, $\hat{y}=-50.3+32.3 x$, with sample correlation coefficient $r=0.882$. (d) Use a calculator with regression keys to verify the linear regression equation for the $\left(x, y^{\prime}\right)$ data pairs, $y^{\prime}=0.150+0.404 x$, with sample correlation coefficient $r=0.994$. The sample correlation coefficient $r=0.882$ for the $(x, y)$ pairs is not bad. But the sample correlation coefficient $r=0.994$ for the $\left(x, y^{\prime}\right)$ pairs is a lot better! (e) The exponential growth model is $y=\alpha \beta^{x}$. Let us use the results of part (d) to estimate $\alpha$ and $\beta$ for this strain of laboratory bacteria. The equation $y^{\prime}=a+b x$ is the same as $\log y=a+b x .$ If we raise both sides of this equation to the power 10 and use some college algebra, we get $y=10^{a}\left(10^{b}\right)^{x}$. Thus, $\alpha \approx 10^{a}$ and $\beta \approx 10^{b}$. Use these results to approximate $\alpha$ and $\beta$ and write the exponential growth equation for our strain of bacteria. Note: The TI-84Plus/TI-83Plus/TI-nspire calculators fully support the exponential growth 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 $\mathbf{0}$ : ExpReg. The output gives values for $\alpha, \beta$, and the sample correlation coefficient $r$.

What we want to conduct A p D. T. A pair differences tests at the alpha equals 1% confidence level. Testing the claim that the population means A bar next pr are not equal. We have the data for A and B. Given below assuming amount shapes, not your distribution on the right. I've already calculated D. Bar noted that an equal seven and calculated SD as 70.47 So we proceeded to do the five steps listed below to solve this first. We evaluate the requirements and hypotheses. So the requirements to use the students distribution have been met because the distribution shape we have degree of freedom and minus one equals six. Are null hypothesis is mute equals zero. Or alternative is beauty does not equal zero. And we're testing at alpha equals 00.1 confidence Next will compute the test statistic and P value. So the statistic is T equals D. Bar divided by SD over. Uh huh. 2.083 from a tea table. This puts p between .1.05. So we can conclude that P is greater than alpha, which means we fail to reject the null hypothesis, which means that we lack evidence that beauty does not equal to about.

We want to conduct a pair differences test at the alpha equals 5% level testing the claim that population mean X bar A is greater than population X barbie. We have data a be given here, we assume amounts to mr distribution as you can see on the right. I've already calculated the mean difference D bar 6.125 The sample size and eight and the sample standard deviation of differences SD and 8.7 We complete the five steps us to blow to solve this problem first, let's evaluate the requirements to use a student's T distribution of the hypotheses because of the distribution shape it is appropriate to use a student distribution your degree of freedom and minus 27. No hypothesis mute equals zero. Alternative media is greater than zero and alpha equals 00.5 for confidence nexus, complete the test at and the P value our test that is T equals D. Bar over SDR. Again this gives 2.14 U. T. Table. This gives us a P value between 0.5 point 025 That means we can include that P is less than equal to alpha. So we reject the null hypothesis H not which means that we have evidence and you D. Is greater than zero.

We want to conduct a pair differences test at the output equals 1% confidence level testing and claim that sample or rather population means that our A. Is greater than expire. Be given the data for A and B. Below here assuming the mound shaped in symmetric distribution. As you can see, I've already computed D. Bar the mean the difference is 12.6 and equals five. And a standard deviation differences 22.66 on the right. Using the appropriate formulas, we proceeded the five steps below to compute this PDT first we check the requirements, evaluate hypotheses because the distribution shape it's appropriate to the students distribution degree of freedom is n minus one equals four. Null hypothesis is not equal zero. Alternative nuclear than zero and alpha equals 00.1 significance. Next our test statistic is T equals D. Bar over SD over route and or 1.24 from the tea table. This gives p value between 0.25 point 125 Thus we can conclude That thing is greater than α.. So we fail to additional hypothesis, meaning we lack evidence, moody is greater than zero


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