3

TCCHOiogy tO write on the .paf file and save that file to submit ) For questions YOU may choose from the follow1ng tests for part (). Proportion Test Proportion Tes...

Question

TCCHOiogy tO write on the .paf file and save that file to submit ) For questions YOU may choose from the follow1ng tests for part (). Proportion Test Proportion Test, Sample t-Test Sample Test Goodtess of Fit (Chi-Square) Test Test of Independence (Chi-Square) ANOVA (6 pts) random Sample of neonle invnlued

tCCHOiogy tO write on the .paf file and save that file to submit ) For questions YOU may choose from the follow1ng tests for part (). Proportion Test Proportion Test, Sample t-Test Sample Test Goodtess of Fit (Chi-Square) Test Test of Independence (Chi-Square) ANOVA (6 pts) random Sample of neonle invnlued



Answers

Construct the histograms and answer the given questions. Use the frequency distribution from Exercise 15 in Section $2-1$ on page 49 to construct a histogram. Does the histogram appear to be skewed? If so, identify the type of skewness.

Okay, the following is the solution to number eight. And this looks at the browse uh normal distribution of um natural distribution of browse and seeing if it fits the deer feeding pattern. And so they give us the observed frequencies for these different feeding patterns. 100 and 225, 43, and 23. And then I found the expected. So I did a little preliminary work here. And the way to find expected is you just take The percent times the sample size and there were % 320 deer that were um sampled. And so I took 32% of 320.32 times 32 or 320. And I got 100 and 2.4. Likewise, I took 0.387 times 3 20. I got that 1, 23.84 and so on and so forth. That's what I did here and with that I what I usually like to do is I look at the observed and they expect, and I kind of compare them and see how close they are. If they're really, really close. I'm thinking that these are pretty good matches two distributions. If there if they seem to be pretty far away, then then I start to think that I'm probably gonna get up a small p value and it looks like these are pretty close together, so I have a hunch that we're going to get a pretty big P value. Okay, so let's actually answer some questions now. So in part a says, what's the significance level? And that's just your alpha value. So it's .5. And we're also meant to state what our hypotheses are, and this always follows the same pattern. The null hypothesis is basically stating that the observed and the expected are the same way that these two distributions are the same. And then the alternative is that they're not the same. So I'm gonna kind of paraphrase you can write this however you want, but I'm gonna say the natural distribution of brows fits the deer feeding pattern. Okay, So that's the knoll that there, that it fits these two are roughly the same, and then I'll just say not for this one. So the natural distribution of browse does not fit the deer feeding pattern. Okay, So the second part, we're asked to find the test statistic, the chi square value. We're gonna use software to find that you can use the formula, you can use any sort of software you want, but you'll get a chi square value. And then the second thing we need to do is check the expected values. All the expected values are greater than five, and I already did that over here, and those are all bigger than five. So we can go ahead and continue with our test knowing that the expected values are greater than five. And then it also asks us what distribution are we using? Well we're going to use a chi square distribution obviously because it's a chi square good, it's a fit test but we also need to say how many degrees of freedom there And in this case there are 4° of freedom. It's always the number of categories minus one. So I have five categories minus one which is four. So four degrees of freedom. Okay, so from here on that we're gonna use technology now. I'm gonna use the T- 84. I think it works well. Whenever you have a small data set like this for elementary stats it does pretty much everything for you. But you can use Excel, you can use SPS s or SAS or or whatever software program you like or you can just do it old school and use the formula. So I took the liberty and go ahead and punched in these numbers. So I put L. One is my observed and then L. Two is my expected. So you can see that those are already pre filled and if you go back to stat and then air over two tests and it's one of the last test. So I went up first but it's the chi square G. O. F. Test that stands for goodness of fit test. And I went there and L one was my observed. L. Two is my expected degrees of freedom was four. And then I went in and calculated. And that gave me a chi square value of I'm gonna go in and around but I don't think it's gonna matter. But 1.084, let's say it's a 1.084. 1.084. That's my chi square value. And then I don't know if you looked at the p value, but that's also shown on the screen. So this p. Is the P. Value and it's about 0.9. So we have a pretty sizable P value there, so it's 0.9 which is greater than any alpha. We're gonna have. So part D. Says make a decision and our decision if we compare the P value explicitly with the Alpha, if the P value is greater than alpha, then we're going to fail to reject. H not especially if you have a P value as big as 0.9. That means these are pretty darn close to each other. So my hunch was correct. I mean one oh 21 oh 2.4. Those are pretty close. 1 25 to 1 23.84 Pretty close. So these are pretty good fit. So then the last part party we just conclude and sort of bring it back to the hypothesis. Bring it back to this problem. So whenever we failed to reject, here's kind of the wording we we can use but you can fluctuated however you want. We say there is not sufficient evidence to suggest that the natural distribution of brows is different and that dear feeding pattern. Okay, so something to that effect, basically saying that the natural distribution of browse is a nice fit for the natural feeding pattern for these deer.

The following is a solution to number 10 and we're going to see if the days and I think it's january and Maui, there's a town in Maui or city in Maui and we're seeing if it follows a normal distribution, the average daily temperatures In January and there are 20 years worth. So it's like 600 20 or something days. They they looked at and there are two parts to this. The first part Um says, Okay, explain what these percentages mean, and essentially it just means that it's the empirical rule. Okay, so the, the reason why they did 34% and 34% is that makes up 68% and empirical rule says that 68% of the data lie within one standard deviation of the mean. So I'm kind of like looking at the middle of that chart right now and um that's where we get the minus sigma plus sigma. So we have a mean, or this is the at least the hypothesis, the mean was 68. I'm saying that in a race ago and so 68 and the standard deviations for that means 68% of the data is between 64 and 72°.. All right, if you do 68 minus 4, 64 68 plus 4, 72 then 95% lies within two standard deviations in the means. That's where we get the 13.5 and 13.5. If you add 13.5 plus 34 plus 34 plus 13.5, that equals 95%. So two standard deviations, if you take 68 minus two times four, that gives you 60 and then 68 plus two times four, that's 76. So 95% of the data is but then is between 60 and 76 degrees. And then finally, likewise, the 2.35 on either end, If you add all that together, you get basically 100%. So 99.7%, you know anything outside of that is definitely an outlier. So that means in this case Um essentially all the data will be between 56° and 80°.. Okay, so that's the first part, that's kind of the meaning behind that. Now let's look at the actual data. So we have observed days, so 14 data values were 14 days were, you know, within three standard deviations on the low side And then 86 within two standard deviations on the low side, and then 207 within one standard deviation and so on, and so forth. And then to 15, so that's all given to you. And then the expected, I just like we've been doing I did the 2.35 took the percent times the sample size, and that gives me the expected data values. Okay, So now let's take a look at The questions that says, what's the significance level? Well, that's given to you, it's the 1% significance level, and that's gonna be your office, your alphas .1. And then we need to write down our hypothesis. So the null hypothesis is that these are matching, these two distributions are matching. So, in other words, the average daily temp the average daily temperature in january for whatever this town is, I think something somewhere in mali follows a normal distribution with I mean 68° and then standard deviation for all right. And then the alternative is just essentially saying not so, you can pretty that up if you want, but the average daily temperature in january does not follow a normal distribution. Alright. Part B says, what's the test statistic? What's the chi square value? We're gonna use technology to find that. But the second part of that says um have the conditions of inference been met and we look to see to make sure that the expected value is greater are are all greater than five and they are. So if you look at these expected values, they're all greater than five, that's good. And then the last part of this, what type of distribution we're gonna use? We're gonna use the chi square distribution obviously because the chi square goodness of fit test, But we need to say how many degrees of freedom and we're gonna use 5° of freedom since there are 6 um observed values. So there are six different buckets I guess you could think of And -1 is five. Okay, so now let's look at some technology so I'm gonna use the IT4 but you can certainly use different form of software. You can certainly you can even use the the formula if you want. Although that might take a while. So I took the liberty to go ahead and type all this stuff in. If you had a cal can edit or Staten edit At L1 is where my observed values are and then L2, these are my expected values and it should be a mirror image. These data values should be the same once you reach the peak. Okay? So then you go to stat once you type those in and then tests and it's one of the last ones. So I went up first since the chi square G O F. That stands for goodness of fit test. L one is your observed L two is you're expected at least. It is in my case. Now, if you have those switched or if you have them in different columns, you'll need to change those. And then the degrees of freedom in this case is five, so then calculate and that gives us everything we need. So that first part, that's the chi square value and that is point 256 and 7.256, which is a super small one. And if you look at if you look at that p value, I mean, that's one of the biggest p values ever seen. So p values 10.998 I pretty much can't get much better than that. That is a very, very big P value and it is definitely greater than alpha, which means we fail to reject. That's two words, by the way, I fail to reject. H Not All right, So, this does follow, you know, almost identical. And you can kind of see that these data values are really close to the same thing. So, um yeah, so, this does follow a normal distribution and concluding it again, you don't need to write it like this, you can deviate from this, but I would say there is not sufficient evidence to suggest that the average daily temperature in january does not follow a normal distribution With means 68 and standard deviation four Kind of a double negative there. So, another way you could say that is um the average daily temperature in January does in fact follow a normal distribution with mean 68 and standard deviation for


Similar Solved Questions

5 answers
2) The rate of change ofa chemical C) with respect t0 time is proportional tO the difference between the amount of the chemical and the current time () Find differential equation to model this situation;b) Can the differential equation from part &) be solved using separation of variables? Why or why not? #00n
2) The rate of change ofa chemical C) with respect t0 time is proportional tO the difference between the amount of the chemical and the current time () Find differential equation to model this situation; b) Can the differential equation from part &) be solved using separation of variables? Why o...
5 answers
2-Sy" + 12y' + 20y = 100 sin x
2- Sy" + 12y' + 20y = 100 sin x...
5 answers
Hypothesis Question The below population j j statement (5 36 9990 (15 H: points) mean points) confidence 100 Determine between an an interval whether 01 example Tliio 38 the estimate of of of the Lnull statement population hypothesis true and mean Or false using the dala
hypothesis Question The below population j j statement (5 36 9990 (15 H: points) mean points) confidence 100 Determine between an an interval whether 01 example Tliio 38 the estimate of of of the Lnull statement population hypothesis true and mean Or false using the dala...
5 answers
Question: Diagonalize the following matrix M by finding D, P and P-1 so that M PDP-1 _[ 1 3 3 M = -3 _5 -3 3 3
Question: Diagonalize the following matrix M by finding D, P and P-1 so that M PDP-1 _ [ 1 3 3 M = -3 _5 -3 3 3...
4 answers
QuestionYour answer partially correct: Try again_Let~13A =and2303Given thatdiagonalizesthen compute the matrix2303
Question Your answer partially correct: Try again_ Let ~13 A = and 2303 Given that diagonalizes then compute the matrix 2303...
5 answers
Area of Composite Figures PracticeCalculate the perimeter and area of the shaded region: Round toEmPerimeterArea
Area of Composite Figures Practice Calculate the perimeter and area of the shaded region: Round to Em Perimeter Area...
5 answers
Direcllons: Find the zero(s) and y-intercept of each function algebraically: 7 "()=r - IOx' + 9 8. f(x) = Vx+4-3 zerols}:zerols}:r-Intercept:r-Intercept:
Direcllons: Find the zero(s) and y-intercept of each function algebraically: 7 "()=r - IOx' + 9 8. f(x) = Vx+4-3 zerols}: zerols}: r-Intercept: r-Intercept:...
5 answers
A company has an issue of executive stock options outstanding. Should dilution be taken into account when the options are valued? Explain your answer.
A company has an issue of executive stock options outstanding. Should dilution be taken into account when the options are valued? Explain your answer....
3 answers
[18 points] (a use the Cauchy-Schwarz inequality to find the maximum value of x + y + given that x2 + y2 + 4z2 = 1; (b) tanx > 5>x>y> 0 tany
[18 points] (a use the Cauchy-Schwarz inequality to find the maximum value of x + y + given that x2 + y2 + 4z2 = 1; (b) tanx > 5>x>y> 0 tany...
5 answers
4J1 Points]DETAILSLARCALCTI 8.2.019.mi;Find the Indefinite Integral using Integration y parts With the glven cholces of u and dv. (Usa € for the constant of integration ) *Cos 4X &*; cos 4x dx
4J1 Points] DETAILS LARCALCTI 8.2.019.mi; Find the Indefinite Integral using Integration y parts With the glven cholces of u and dv. (Usa € for the constant of integration ) *Cos 4X &*; cos 4x dx...
5 answers
Find f.f"(x) 2Ox? + 12x2 + 6, f(o) = 5, f(1) = 1
Find f. f"(x) 2Ox? + 12x2 + 6, f(o) = 5, f(1) = 1...
5 answers
Ourrtlon Ne-LCLQ Lllomw (A)Refer to the ciruit below to detenmine if the given statements are TRUE or FALSERs And R, are connected in parallelThe circuit has & nodes= Ri Ri, + Raia + RyizRi, _Ri,1i(B) In the circuit below. find the curents h and E Then find node bvoltage6 V
ourrtlon Ne-LCLQ Lllomw (A)Refer to the ciruit below to detenmine if the given statements are TRUE or FALSE Rs And R, are connected in parallel The circuit has & nodes = Ri Ri, + Raia + Ryiz Ri, _Ri, 1i (B) In the circuit below. find the curents h and E Then find node bvoltage 6 V...
5 answers
Cox's bazar
Cox's bazar...
5 answers
18 Bendonly selected students took the calculus final . If the sample mean 15n5 03 and the standard deviation was 15.2 LBALA 33% confidence interval for the mean score of all students.selert onc: 57-5 t0 68.5b. 67-5 t0 78587-5 [0 98.5 d775 t0 885
18 Bendonly selected students took the calculus final . If the sample mean 15n5 03 and the standard deviation was 15.2 LBALA 33% confidence interval for the mean score of all students. selert onc: 57-5 t0 68.5 b. 67-5 t0 785 87-5 [0 98.5 d775 t0 885...
5 answers
Question 104pthe reaction shown, how many moles of Hz are produced when 4.6 moles of Alare reacted? AI (s) 6 HCI (ag) 7> 2AIClz (aq) 3Hz (g)949.2 molesOl5 mples02 3mmolesD bamolesJommoles
Question 10 4p the reaction shown, how many moles of Hz are produced when 4.6 moles of Alare reacted? AI (s) 6 HCI (ag) 7> 2AIClz (aq) 3Hz (g) 949.2 moles Ol5 mples 02 3mmoles D bamoles Jommoles...

-- 0.021834--