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Question 120 ptsFrom the textbook data in StatCrunch, open data 9 ct_8. It'Il be toward the bottom of the chapter 9 data_ A tennis enthusiast wants to estimate...

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Question 120 ptsFrom the textbook data in StatCrunch, open data 9 ct_8. It'Il be toward the bottom of the chapter 9 data_ A tennis enthusiast wants to estimate the mean length of men'$ singles matches held during the Wimbledon tennis tournament From the Wimbledon history archives, he randomly selects 40 matches played during the tournament since the year 1968. The lengths of the 40 selected matches, in minutes, are in your spreadsheet:a) Calculate a 95% confidence interval for the popu

Question 1 20 pts From the textbook data in StatCrunch, open data 9 ct_8. It'Il be toward the bottom of the chapter 9 data_ A tennis enthusiast wants to estimate the mean length of men'$ singles matches held during the Wimbledon tennis tournament From the Wimbledon history archives, he randomly selects 40 matches played during the tournament since the year 1968. The lengths of the 40 selected matches, in minutes, are in your spreadsheet: a) Calculate a 95% confidence interval for the population mean length of all men's singles matches at Wimbledon: b) What function in StatCrunch did you use to obtain your answer in part a? (This question is insurance for YOU) c) Interpret your confidence interval, rounding the bounds to one decimal place and referring back to the problem: Label your answers a, b, Edit View Insert Format Too;j Table 12pt Paragraph



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Please provide the following information. (a) What is the level of significance? State the null and alternate hypotheses. (b) Find the value of the chi-square statistic for the sample. Are all the expected frequencies greater than 5 ? What sampling distribution will you use? What are the degrees of freedom? (c) Find or estimate the $P$ -value of the sample test statistic. (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis that the population fits the specified distribution of categories? (e) Interpret your conclusion in the context of the application. Archaeology: Stone Tools The types of raw materials used to construct stone tools found at the archaeological site Casa del Rito are shown below (Bandelier Archaeological Excavation Project, edited by Kohler and Root). A random sample of 1486 stone tools was obtained from a current excavation site. Use a $1 \%$ level of significance to test the claim that the regional distribution of raw materials fits the distribution at the current excavation site.

The following is a solution to number nine and this is a study to see if this is a normal distribution. It's a number of days is 20 year study. Number of days in july for I think it's like kit Carson colorado and see if those temperatures, the average daily temperature to see if there it's approximately normal and there's two parts, the first part I actually just kind of typed out because it be easier this way, But um the way that they structure the percentages that represents the empirical rule, So if this actually matches a Normal distribution, that means it's just basically follow the empirical role, which means 68% of the data lies within one standard deviation of the mean. That's why they had 38, and 34 that were minus one standard deviation plus one standard deviation from the mean, That's why it was between 67 and 83. In fact, I'll write that down. So 68 of data between 67° and 83°.. Okay, so this is because if you take 65 or sorry, 75 minus eight, you get 67 then 75 plus eight you get the 83 and then likewise, 95% of the data lies within two standard deviations of the means that that's where we get kind of think we're starting in the middle of the hump and then we just kind of work work out that way and 13.5 plus 34 plus 34 plus 13.5, that equals 95%. So 95 of the data is between 59 degrees And 91°.. And then finally, basically every day to value at least should be Otherwise, it's probably an outlier. But anything else? About 99.7% of the data, which is the rest of it. So 2.35 plus 13.5 plus 34 then back down the hill that equals 99.7. So 99.7 percent of the data is within three standard deviations of the mean and that three standard deviations. If you just tracked eight and add eight You get 51°.. So basically all these data values should be between 51 and 99. So that's the empirical rule. That's the first part. Okay, so the second part is just like normal. So we have these observed values. I was given two and then the expected like normal. I just took the percent times the sample size and I got these expected. And what's nice about this is you just really just have to do half of it because the other half should mirror image. So that's where I get those numbers. So let's go and answer these questions. The first part it says um what's the Significance level? And that's the alpha, and that's given to you as 1%. Okay, And then we do the um the knoll and the alternative, and again, you can probably write this however you want, but basically the knoll is going to say that these daily temperatures is approximately normal, and then the alternative is that they're not normal. So in in prettier words, let's go and say the average daily temperature, I'm going to abbreviate here. This because the room in july follows, and this is kit Carson colorado follows, a normal distribution, with mean 75 And standard deviation eight, and I should probably put their degrees. Okay, and then the alternative, I'm just gonna put not now, you may want to write that out. Um the average daily temperature in july does not follow a normal distribution. Okay, So now we have this next part and says, what's the chi square value will come back to that, I'm gonna use technology to find that. But the second part of that 2nd part of the question is to make sure that the expected values are all greater than five and they are. So I just verified their these numbers are all greater than five. That's one of those conditions for inference. And we're just gonna make sure that the expected values are greater than five. And then it asks what distribution we're gonna use. Well we're gonna use the chi square distribution since this is a chi square goodness of fit test. But we also need to say how many degrees of freedom and they're gonna be 5° of freedom. Okay. The reason why it's five is you just take The total number of values -1. So 1234566 values there -1 is five. Okay, so from here we can go to our calculator uh or Excel or SAs or or whatever um software you like and I went ahead and type these in already. If you go to stat and edit you can type L one is your observed values and then L two is your expected values or vice versa just as long as you change it and then your stat and go to tests and it's one of the last ones. So I go up and it's the chi square geo F that stands for goodness of fit test. Okay, observed as L one expected as L two degrees of freedoms five and I can go ahead and calculate and that gives me a chi square value of about four point We'll say 895. So four. Mhm. 4.895. And then the P value, I don't know if you saw it, but if we go back this P value is about 43 Okay, so that's the third part. And what we do is we explicitly compare the P value with the often in this case the p values pretty big 0.43 is a pretty sizable P value that's greater than alpha. That means we fail to reject, failed to reject. H not so we're accepting H not to be true. So that last part, we just sort of summarize that. And a lot of ways you can say this, I'm gonna say there is not I'm just going to type this because it's easier not sufficient evidence to suggest. Okay, that the average daily temperature in july does not follow a normal distribution With mean 75 and standard Deviation eight. So that's basically another way of saying that this Kit Carson does follow a normal distribution with mean of 75 and a standard deviation of eight.

We need to state what type of testes is given the options A through F, and the scenario is that we want to know if Company A and Company B pay their mid level professionals differently on average. So what happens is we're taking a sample from each of these companies, finding the averages of these samples and comparing these averages to see if there is a difference in the average is. So a sample from Company A is independent from a company sample from Company B because there's no relationship between these two companies. So we're talking about independent group averages or independent group means, and the other thing we're told from the question is that we know the standard deviation of the salary for all mid level professionals. So the entire population of mid level professionals we know the standard deviation. So the answer is a. It's independent group means with a population standard deviation is known

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


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