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You wIsh {0 [25t the following claim (IIalaca sieniticance levelof o 0,.05:Hap = m Han <pYou obtain 165 successes in & sample oi Si7e nl 499 from the first p...

Question

You wIsh {0 [25t the following claim (IIalaca sieniticance levelof o 0,.05:Hap = m Han <pYou obtain 165 successes in & sample oi Si7e nl 499 from the first population You ootain 274 successes In 4 samplc of size {12 528 from the second population For this testyou should NOT use {he continully correctlon; and you should use the normal distributlon J5 an approxlmation for the binomial distribution,What Is the test Stpllstc (or thls Sample? (Report answor accurate (0 three declmal places ) t

You wIsh {0 [25t the following claim (IIalaca sieniticance levelof o 0,.05: Hap = m Han <p You obtain 165 successes in & sample oi Si7e nl 499 from the first population You ootain 274 successes In 4 samplc of size {12 528 from the second population For this testyou should NOT use {he continully correctlon; and you should use the normal distributlon J5 an approxlmation for the binomial distribution, What Is the test Stpllstc (or thls Sample? (Report answor accurate (0 three declmal places ) test statistic = Whot /s the p valuc for thls samplc? (Rcport answer accurate t0 Iour decimal placcs ) Avalue The p-value Eue Oless (/un (0r uuudl (0) ( preater (hun € Ths (est stalislic Ieads (O 4 decision t0 eject Int null acceplthc null fall t0 reject the null Ns such; thc iInalconclusion Dmncro [a sutliclent ovidence to warrant rejcrtmn otiho clalm thattho tlrz: Popuiatnn



Answers

Conduct the hypothesis test and provide the test statistic and the P-value and/or critical value, and state the conclusion. In analyzing hits by V-1 buzz bombs in World War II, South London was subdivided into regions, each with an area of 0.25 $\mathrm{km}^{2} .$ Shown below is a table of actual frequencies of hits and the frequencies expected with the Poisson distribution. (The Poisson distribution is described in Section $5-3 .$ ) Use the values listed and a 0.05 significance level to test the claim that the actual frequencies fit a Poisson distribution. Does the result prove that the data conform to the Poisson distribution? $$\begin{array}{l|c|c|c|c|c} \hline \text { Number of Bomb Hits } & 0 & 1 & 2 & 3 & 4 \text { or more } \\ \hline \text { Actual Number of Regions } & 229 & 211 & 93 & 35 & 8 \\ \hline \begin{array}{l} \text { Expected Number of Regions } \\ \text { (from Poisson Distribution) } \end{array} & 227.5 & 211.4 & 97.9 & 30.5 & 8.7 \\ \hline \end{array}$$

No. Okay, so we know that the number of patients given sustained care is 198. Among that, 82 0.8%. They're no longer smoking after one month. So 198 times 82.8 percent. Well, give us 163.94 Okay, so we can round that 264. Yeah. Andi making, then Right down. Arnel Hypothesis, which is given that P is 1 80 80% since the test claims that 80% of patients stopped smoking after giving state care and our culture bosses will therefore be P is not equal to your right June at a given level of significance. Which is your prince? Your one. Okay. Me? Yeah. So we will get a Z statistic value off 0.99 on the critical value for the normal area. Table movies equals plus or minus +257 eight A p value. 20 above output is just your 200.32 Sure. Yes. On. Since this is greater than our little of significance, we can include that there's not sufficient evidence to support the rejection of the claim that 80% off patients stop smoking when sustained care. What about

So we would be assuming that the mean number of the minutes for Disney movies and the mean number of minutes of non Disney or other movies, Children's movies showing tobacco, that those two are equal and alternately that they're just different. So again, we're looking at how many minutes or how many seconds actually we're seeing tobacco shown. So we're going to assume that the difference between these two groups is zero and we're doing a two tailed test. And so let's find what our test statistics are. Test statistics is not going to be and are smaller sample sizes 17, So we'll say we have 16 degrees of freedom. And so I take the difference 61.6 seconds for Disney minus 49.3 seconds, or the other Children movies. Okay. And then we have very, very large standard deviations. And that is something to really pay attention to notice that those standard deviations look how much larger they are than the actual even mean. And so normally, if you look at these two times you're gonna think, well they're going to be significantly different because, you know, they're over 10 seconds apart. However, with those huge standard deviations that's going to make that be suspicious, and when we get that we get 0.46 So we're getting a test statistic here, that is a T. Value, that is 0.46 and this area plus going symmetrically on the other side, those two together are going to be our P. Value. So we need to find two times the probability of that being greater than or equal 2.46 And I used my T. C. D. F. On my calculator and found out that that p value is about 65%. So if you have two distributions that their difference is actually zero, the likelihood of getting what we got, or more extreme in each direction happened 65% of the time, and this is definitely not less than 5%. Therefore we have we failed to reject now. Yeah. Okay. So it appears as though the means are equal, so they mean amount of time that you see tobacco means are equal even and again, because that's standard deviation is so large, That's what's causing that to happen now. Since we did a two tailed test, we would be finding a 95% confidence interval of our significance level was 5%. And so I can look up the 95% number for 16 degrees of freedom and for 16 degrees of freedom, I get that T. Star value being 2.1 to 0, so we can take that difference. And this difference looks like that is 61.6 minus the 49.3 comes out to be 12.3 plus or minus R 2.12 zero. And then the square root up. And let's get those big standard deviations here square divided by the sample size of 33 and that's 69.3 square divided by the sample size of 17. So let's find that margin of air 2.12 point 12 times the square root of 118.8 squared, divided by the sample size of 33 plus the 69.3 squared divided by the sample size of 17. And I get a margin of air of a whopping 56 basically 560.5, which I'm going to store that as X. And then we're going to take that 12.3 and subtract away X. And I get my low value as a negative value, negative 44 point roughly. Two. And then when I changed that and I add those values together, I get the upper limit as 68.8 and notice that it does include zero. So it includes zero where 95% confident that the difference between the two means is somewhere in here and it includes zero. So we would not have evidence to reject the not we would fail to reject it. And uh and I think you had a part, see that stated something about looking at those values and when you look at those, it appears to me that there are lots of values that are less than uh 25. About half of the values are less than 25. Less than 25. Almost half of them. Therefore you are going to have skewed data so it won't be normal. So we would have to be be careful be careful about using these results because we are assuming that the two distributions are normal and the test is pretty robust. So, um, so, but you know, we if we had some more samples, that would be better for us.

Okay, So what do we have in this question? Now? Ben Ford's law states that the first non zero digits off numbers Ronald random from a large, complex data file have a certain type of a probability distribution, and it has been given to us. Oh, kit. So that is his draw. The stable. This is very important. So this is going to be first non zero digit first, non zero did it All right. Now, the probability according to Ben Ford's law. So this is going to be the probability. As for Ben Fords law, Okay. And then we're going to have the sample frequency. Let us call these the observed values. The observed values. Okay, so, non zero digits, these rains all the way from 1 to 9. So this is going to be a long table. So this is 1234 567 eat and nine. Okay. Now, what are the probabilities that are given to us According to this law, zero point 301 0.301 Then this is 0.176 Then this is 0.125 Then this is 0.97 This is 0.7 nine. 0.670 point 058 0.51 and then we have 0.46 Yes. Okay. This thing is disturbing as well. Again, It's not pay attention to that. Because if I refresh the page, it'll go. I left to right. All of this again. So what were the observed values? The argument was 83 49 30 to 22 83 49 32 20 to 25 18, 13, 25 18, 13. Then we have 17 and 16, 17 and 16. Okay, these are the observed values. Now, this is all the data that is given to us. And this total sample size is 2 75. Mr. Oral, sample size is 2 75. All right, now, what is the question they're saying? Use a 1% level of significance to test the claim that the distribution, our first non zero visits in this accounting file, follows the band Ford's Law. Okay, So the first point is that our Alfa 0.1 our Alfa is 0.1 What is our null hypothesis are null hypothesis is going to be that the distribution, the distribution off first, non zero desserts off first, non zero. Okay, digit in the accounting I am file follows the band. Ford's law doesn't mean that the distributions are basically the same. Follows the Ben Fords a lot. What was what is going to be the alternative hypothesis? The alternative hypothesis will be that the distribution off first non zero digits in the accounting file in the accounting file doesn't follow the band. Ford's law. Okay, now, in order to as this plane, we're going to perform ah Chi Square test and one of the first step to performing a chi Square test, it is finding the expected values finding the expected values for all the categories in the formula. For that is the formula to find this between values is sample size, the sample size that we have multiplied by the probability of the proportion off each category multiplied by the probability or the proportion off each category off each category. So Okay, so let me just put this formula interaction over you. This table will give us the expected values. This table is going to give us the expected values. Okay. All right. What is our sample size? It is to so many faith. So what is going to be the expected value for one? The expected value in this case will be to 75 multiplied by 0.301 82.775 82.775 Then for the next one is to 75 multiplied by 0.176 This is 48.4 48.4. Then it is to 75 multiplied by 0.125 34 point 375 Then we have 2. 75 multiplied by 0.97 This is 26.675 26.67 Faith. Then this is to 75 multiple of a 0.79 This is 21.725 21.72 Faith. Okay, then it is to 75 multiplied by 0.67 This is 18.4 to 5. 18.4 to five. Then there's 2 75 multiplied by 0.58 This is 15.95 15.95 Yeah. Then this is 0.51 multiplied by who's 35 that has 14.25 14.25 And then we have 0.46 multiplied by 2. 75. This is 12 0.65 These are the expected values. What is the next step? We're going to calculate the individual chi square values for all of these categories. Okay, so how is this given? Well, the formula to find the individual keister values is find the difference between the observed and the expected values square it divided by the expected value. Now, once you find all the individual values, you will some them all up, and it will give you the overall Caires questions. Stick for the problem. Let us look at this formula in action. So we're here for the first category. The difference is going to be between 83 on 82.775 We're going to square this and divide this by the expected value that is 82.775 So this is 0.6 So let me just write this zero. Then we have difference between 49 48.4. We square this and divide this by 48.4. This is 0.7 0.7 Then the difference between 34.375 and 32. We square this and divide this by 34 point 375 This is 0.16 0.16 This is the difference between 26.675 and 22. We square this 4.675 and divide this by 26 point six. Certain faith. So this is almost 0.82 Then it is the difference between 25 21 point 7 to 5. We square this and divide this by 21.75 21 0.7 to fight. This is 0.493 0.493 Okay, then we have a difference between 18 point 4 to 5 and 18. So this is 180.4 25 We square there's and divided by 18.4 to 5. So this is 0.90 point 0098 So, Yeah, this is 0.98 Okay, then the difference between 15.95 and 13, we square this 2.95 and divide this by 15.95 This is 0.0 point 545 Then we have the difference between 17 minus 14.25 We square this and divide this by the expected well do 14 0.0 to fight 0.63 0.63 Then we have the difference. Between 16 on 12.65 We square this and divided by 12.65 and we get over here 0.89 Okay, Now we're going to add all of these up. So this is zero plus 0.7 plus 0.16 plus 0.82 plus 0.493 plus 0.98 plus 0.5 for five. 0.63 plus 0.89 This is the total edition comes out to 3.553 point 55 So I can say that my chi square for this entire question is 3.55 Now what else do I need in order to find the P value? Well, I need the degrees of freedom which is given by the formula number off categories, number off categories minus one. How many categories do I have? 123456789 So this is going to be nine minus one, or I can write this as eight. All right, So I have my chi square statistic, and I also have my degrees of freedom. Now what I'm going to use is I'm going to use an online too, to get my p value. So my chi square statistic is 3.55 My visa freedom is eight. My level of significant if I look at the question is 1%. So this is 0.1 I have calculated and I find that my P value is 0.89 My people live is 0.895 Okay, it is 0.895 What was my Alfa? My Alfa was 0.1 So I can see that my P value is much greater than my Alfa hence, I will say that I fail to reject my null hypothesis. H not. This means What do I say? I say that I do not have enough statistical evidence. Studies tickle evidence to suggest that to suggest that what was the wording? To suggest that the distribution off first, non zero digits, that the distribution off first non zero digits in this accounting file in this accounting fine in this accounting file does not follow does not follow Ben Ford's lock When Ford's No right, what was the alternative hypothesis? It waas That doesn't follow the went for lawyer. So I do not have enough statistical evidence to suggest that the distribution of first non zero deserts in this accounting file does not follow Ben Folds Law, and that's how we go about doing this question.

We have a sample with N equals 83. And of that 83 members of the sample 64 satisfy a certain condition. Want to meet our equal 64. Now we want to use the sample data to test the claim that p does not equal 2.75 for the population at a confidence level of 5% or alpha equals 0.5 Now that we've identified the confidence level, we can go to the following procedural steps to conduct this hypothesis test first. Is the normal distribution appropriate to use? Yes, it is. Because both N. P and Q. P. R greater than five. B. One of the hypotheses were testing we're testing whether or not H not P equals 10.75 or H a p does not equal 0.75 Ak We're conducting a two tailed test. Next compute P hot. And the test statistic P hot is all over. N equals 20.77 Z is given by the formula on the right, taking in as input P hat P Q n N producing down to Z equals 0.42 Next we compute the P value. We use the table to see the area outside Z equals plus and minus 0.42 as we shaded in yellow and the graph on the right. This P value corresponds to 0.6745 Next we reject H not, no, we do not. P is greater than alpha and we interpret this signing to suggest that we lack evidence that P does not equal .75.


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