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(6 points) You want 0 Icst thc following hypothesc s using Icvcl of significance of 1% Hc: A Mi z Hi: p 0.60 The sample dala arc: n = 10. 13481 0292 2.46,s"_ 0...

Question

(6 points) You want 0 Icst thc following hypothesc s using Icvcl of significance of 1% Hc: A Mi z Hi: p 0.60 The sample dala arc: n = 10. 13481 0292 2.46,s"_ 0320 You have reason to believe that 0 , but You do not know #hat cither is. Each sample comes from population with normal distnbulion: What is the pooled What thc allaincd Valuc of the Iest _Llistic? What is the Valuc for thix lest? Can wereject the null hypothesis? (6 points) You Mn( to tcst thc following hypotheses using levelon sig

(6 points) You want 0 Icst thc following hypothesc s using Icvcl of significance of 1% Hc: A Mi z Hi: p 0.60 The sample dala arc: n = 10. 13481 0292 2.46,s"_ 0320 You have reason to believe that 0 , but You do not know #hat cither is. Each sample comes from population with normal distnbulion: What is the pooled What thc allaincd Valuc of the Iest _Llistic? What is the Valuc for thix lest? Can wereject the null hypothesis? (6 points) You Mn( to tcst thc following hypotheses using levelon significance of 5%: Ho: H 30.0 Hi: p 0i 7 30.0 The sample data a: 2043 1,242 "=15. 2137.F 1,.605 . Ejch sample comes from population with nonal distribution_ but then is no feuson [0 believe that thcir population variance > are cqual: Whal the estimated variance ol &-n? What is thc ctimaled Aandard cnUr 0l X What i> the attained value of the tcst s utistic? How many degrees of freedom should you Usc with this test statistic? What is the valuc? Can Wc rcject the null [ hypothesis? (8 points) You want to lest the following hypothescs using Ievcl of significance of S%: He:T 01 20.25 Hi:# 0.25 Thc sumple data Jc: 200. 500. n = 240 What are the sample proportions P1 and pz? What is the estimated variance 0f pr ~ Pi Whal is thc cstimaled standard crTor ofp? What is thc altained value of the Iest suatistic? What i the valuc? Can we reject the null hypothesis?



Answers

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. Meteorology: Normal Distribution The following problem is based on information from the National Oceanic and Atmospheric Administration (NOAA) Environmental Data Service. Let $x$ be a random variable that represents the average daily temperature (in degrees Fahrenheit) in July in the town of Kit Carson, Colorado. The $x$ distribution has a mean $\mu$ of approximately $75^{\circ} \mathrm{F}$ and standard deviation $\sigma$ of approximately $8^{\circ} \mathrm{F}$. A 20 -year study $(620$ July days) gave the entries in the rightmost column of the following table. (i) Remember that $\mu=75$ and $\sigma=8 .$ Examine Figure $6-5$ in Chapter $6 .$ Write a bricf explanation for Columns I, II, and III in the context of this problem. (ii) Use a $1 \%$ level of significance to test the claim that the average daily July temperature follows a normal distribution with $\mu=75$ and $\sigma=8$

We have data sets A and B on the right. We want to utilize these data sets to test the claim that there is no difference in the underlying distributions for A and B it out the equals 0.1 significance. This question is testing an understanding of non parametric tests, particularly how to take perform the ranks on tests we proceed through steps A through G to solve. So first we see the outside of hypotheses. These are alpha equals 0.1 hypotheses. H. And distributions are saying hk distribution are different and be we can be the test at the sampling distribution which is normal and check the requirements which are met. So their ranks are for A and B. As follows That we have and one equals 11 and two equals 12. You are sigma are 1 30 to 16.2 from this, we have our equals and summer range from a 1 73 so Z equals ar minus. Moreover signal articles 2.53 thus received complete the P value. As for normal distribution two PZ greater than 0.114 Thus, we conclude, indeed, that we fail to reject nation on since P is greater than alpha, which means we lack evidence for AJ.

So in this example, we're looking at how the correlation coefficient relates to row And whether or not it are as significant or not significant. So looking back to the example about the Shetland ponies, we know that our was approximately .975 and we're looking at Alpha being 0.05 and the population was five. So, coming over here to my chart, Here's five for N At the 5% level. So I'm looking at 0.8, eight .975 Is greater than 0.88. And when the absolute value of r is greater than or equal to our table entry, we say that our is significant. So in this case we can conclude that the age and weight of the Shetland pony are correlated. So that's what are compared to P would tell us in this particular example. Now dealing with a cyclone, Our population is six. So I need to go to my chart where Sixes And I'm looking at the 1% significant level. So The table entries .92 and the absolute that's the absolute value of our so I'm going to use the positive. So .990 is indeed greater than .92, it's greater than our table entry, so therefore our is significant. And the way I can relate that according to this particular problem is to say, we conclude that the lowest barometric pressure reading and the maximum wind speed are correlated.

The following is a solution to number 12 goodness of fit test. And this looks at in a library, the number of books that are checked out and comparing it to the distribution of uh students and classes are the subject taught or something like that. So here are the observed data values 268 114 115 115 76. And then the expected values they just give you percent. So the way you find the expected value, you take the percent times the total sample size and there are 888 in the sample. So 32% times 888 is 2 84 16, 25% times 8 88-22, You know, and so and so forth. So I did that for the data that we're going to use now we can start answering some questions. So it says that in part a what's the significance level and that's the alpha value. And they tell you that it's a 5% significance level. So alpha equals .05. And then the alternate or the Nolan alternative. The goal is always saying that these two distributions about the same and then the alternative saying that they're not the same. So in the context of this problem, h not will be the subject distribution. The subject distribution of books in the library fits the distribution of books checked out. Okay, that's the first um the null hypothesis and then the alternative, you know, just the same time, I'm just gonna say not, but you would just say the subject distribution of books in the library does not fit the distribution of books that are checked out. So the second part, we need to find the chi square value and we can use technology to find that. But you can certainly use the formula if you wish. And then we need to check to make sure the expected values. All the expected values need to be greater than five. That's one of those conditions for inference. And we can verify but those are all significantly greater than five. So we're good on that. And then because of that we can use the chi square distribution. Okay. But we also need to say how many degrees of freedom there are? There are four degrees of freedom. Okay. The reason why it's four is you just take the categories minus one. So there are five categories here and you subtract one? You get four. Okay so I think that's it now. We just need to go to our calculator. Now I use A. T I. T. For especially whenever the data sets pretty small like this, but you can use whatever software you like or you can use the formula. So if you go to Stapp and I took the liberty to go and just type these in already. But L1 is where I put the observed to 68 to 14 to 15 1 15 and 76. And then L two is where I put the expected so you can see there and then if you go back to stat and then tests it's one of the last one. So I go up first and I go to the Kaiser. This the D option chi square G. O. F. Test that means goodness of fit. That's what that means. And L one is the observed L two is the expected degrees of freedom remember was four. So I can calculate this thing and that gives me everything I need. So the chi square value. Is this this top thing right here? So 11 point 866 is my chi square value, so 11866. And then the p value is also given, it's about .018 .018. And then if you compare that with the alpha value, remember the alpha value is 0.5, that is less than alpha. And the rule is so you explicitly compare the P value with the alpha value, the significance level, and any time it's less than alpha, you're going to always reject the null hypothesis. So we reject H not. Okay. And then the final step is the conclusion. And you can write this however you want, basically just saying that these are not uh the same distribution. But in the context of this problem, I'm going to write there is sufficient evidence to suggest that the subject distribution of books in the library does not fit the distribution of books checked out by students. Okay, so that's the conclusion that I came up with. There is enough evidence to suggest that these two distributions are not the same.

Soon. Number 28. What's your name? The level of significance is mentioned is all for equal 0.1 on our hypothesis, which state that the population mean new is equal to the value mentioned in the claim, so it would two new equal 4.55 g. The alternative hypothesis stated the opposite off the novel hypothesis according to the clean, thus using lesson. So each one to mu is less than 4.55. Dreams is alternative hype with his old listen. Then the test is left. 15 is the alternative. Hypotheses uses Biggers in, so the desk is right field. The alternative hypothesis uses non equal, so the test is toe field, so the answer will be left field question. Number be given export equal 3.75 being equal 4.55 Zita equal 0.7 and unequal six. The sampling distribution off the sample mean export is normal because the population distribution X is assumed to be normal. The sampling distribution off the sample mean has mean new and stirred and standard deviation. Zita over square root and does it? Value is a sample mean decreased boy. The population mean divided by the standard division so that equal explore minus new over Zita over square root in equal 3.75 minus 4.55. Mhm 0.7 over square root six equal. Minus 2.8. Question number C Riddle Sport A and B. It snowed too. Mu equal 4.55 g each one to mu is less than 4.55 g and did equal minus 2.8. The B value is the population off obtaining a value more extreme or equal to the standard test static Zet that remind the probability using tepidly. So, probability equal probability off export is less than 3.75 Equal probability off. That is less than minus 2.8. Equal 0.26 Question number We? Yeah, given Alpha giving Goma equal 0.1 Result port. See, P equal points here is you 26 If the value is smaller than significance level Alfa, then then I'll Hypothesis is rejected. Mm. Is listening 0.1 So I reject h No. If we rejected them on hypothesis the data is the data easy, statically significant at level Alfa Question number e result, 40 each node to new equal, 4.55 g and each one to mu is listening four point 55 graham result for the project. Each node. There is sufficient evidence to support the claim that the mean, the mean weight off these periods in three sports off the Grand Canyon is less than 4.55 g.


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