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Sometimes_ count data explicitly omit Teto counte Examples include the numbers of days patients spend hospital (only patients who actually stay overnight in hospita...

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Sometimes_ count data explicitly omit Teto counte Examples include the numbers of days patients spend hospital (only patients who actually stay overnight in hospital are considered and the smallest possible count is one); the number of people per using rural road (the driver at least must be in the car); and suVer of the number of people living in each household (to respond the households must have at least one person) _ Using Poisson distribution is inadequate, the zero counts will be modeled t

Sometimes_ count data explicitly omit Teto counte Examples include the numbers of days patients spend hospital (only patients who actually stay overnight in hospital are considered and the smallest possible count is one); the number of people per using rural road (the driver at least must be in the car); and suVer of the number of people living in each household (to respond the households must have at least one person) _ Using Poisson distribution is inadequate, the zero counts will be modeled true zero counts In these situations, the zero-truncated Poisson distribution may be suitable, with probability function exp( AJAU f(y; A) {1 _ exp(_ A)}yl where 1,2,- and [3 marks] Show that the zero-truncated Poisson distribution is member of the exponential family by identifying p, a(0), 6(0) ; and cly; 0). b) [2 marks] Show that E[Y] = M/{1 exp(-A)} [3 marks] Find expression for the Deviance statistic for sample of independent zero-truncated Poisson observations with rates At What is distribution of the Deviance statistic? [2 marks_ Plot the zero-truncated Poisson distribution and the Poisson distribution for A = 2, and compare



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Another important discrete probability distribution is the Poisson distribution, named in honor of the French mathematician and physicist Simeon Poisson (1781-1840). This probability distribution is often used to model the frequency with which a specified event occurs during a particular period of time. The Poisson probability formula is $$P(X=x)=e^{-\lambda} \frac{\lambda^{x}}{x !},$$ where $X$ is the number of times the event occurs and $\lambda$ is a parameter equal to the mean of $X .$ The number $e$ is the base of natural logarithms and is approximately equal to 2.7183. To illustrate, consider the following problem: Desert Samaritan Hospital, located in Mesa, Arizona, keeps records of emergency room traffic. Those records reveal that the number of patients who arrive between 6: 00 P.M. and 7: 00 P.M. has a Poisson distribution with parameter $\lambda=6.9 .$ Determine the probability that, on a given day, the number of patients who arrive at the emergency room between 6: 00 PM. and 7: 00 PM. will be a. exactly 4. b. at most 2. c. between 4 and $10,$ inclusive.

Part one. The measure of skinnies for income is about 1.86. After taking a log of income. The skin is measure is three, Sorry .36. So there is much less skin ist in the log of income, which means income is this variable before taking luck is less likely to be normally distributed. Part two. The skin is for B W G H. T. It's about -1. When we use the lock of this variable, the skinnies measure is minus two points 95. And you can see that schools has increased in absolute terms. This is totally the opposite of what we observe in part one with variable income. So in part two there is much more skinnies. After taking look. The results of part one and 2 show that The statement we need to evaluate in part three is not always true. Taking log and increase or decrease. The skin is of the distribution. This statement is not quite correct. When you take the lock, um you force the variable to follow a different distribution and that distribution can have larger or smaller skinniest. So the new distribution can be more like a normal distribution or not. For many economic variables, particularly the one those measured in dollar values, for example, GDP or export import values taking the lock often helped to reduce or eliminate skin is, but it is not always the case. March four. In regression analysis, we should study the conditional distribution of Y. N, log Y. Additional distribution is the distribution of these variables conditional too. The explanatory variables. If we think that the means of why and log Y. Is linear Under Assumptions. one and 3 in this chapter, studying the mean is similar to you studying the distribution of the population error. The skin is measure in this question is often applied to the residuals from all L. S. Regression.

Right. So this problem is involving catching fish in Pyramid Lake, in Nevada and you've been given some data, The likelihood of catching zero fish in a six hour window is 40 4%. The likelihood of catching one fish is 36%. The likelihood of catching two fish is 15%. three fish was four And four or more fish was 1%. So for part A we need to convert the percentages, two probabilities. So we are going to construct a probability distribution. I like my probability distributions to be a more vertical style. So we could catch no fish, one fish, two fish, three fish or four or more. And as a probability, since there is a 44 chance of catching no fish, that is translated into 44 out of 100. So our probability would be .44 For one fish, it would be .36 for two fish could be 20.15 Three, fish would be .04, and for four or more would be .01. Now, this problem also calls for you to construct a hissed a gram. So we are going to draw are vertical and horizontal axis and the horizontal axis is going to be the number of fish Caught in a six hour period. And the vertical axis is going to be our probability and we need to number our vertical axis. So I would number it as 5, 10, 15 2025, 30, 35, 40 and 45. So the second line is 10 0.20 0.30 and 0.40. So we're ready to create our towers. So for zero fish we have to get up 2.44 said somewhere like right here, So that be zero fish for one fish, we have to get to about 36. So that's about right there. for two fish we need 15, for three fish, we need to be at four And for four or more were at one. So it's just right above there. So that would be a history graham representing the fish caught um Or the probability of fish caught in a six hour period for part B. You are trying to determine the probability that a fisherman randomly selected will catch one or more. So we can define that as X is greater than or equal to one. So that would be these four. So we're going to take the probability of catching one fish plus the probability of catching too, Plus the probability of catching three and the probability of catching four And that will be a .56 probability for part C. We want the probability of catching two or more fish. So that means we're talking two or three or four. So we would take the probability of two, Add it to the probability of three and add that to the probability of catching four fish. And we would get a .20. Now for part D. You are asked to determine the mean or the expected value. And in order to find that we're going to apply the formula sigma of X times P of X. So what I'm going to do is I'm going to start our chart all over again. So we're going to have our possible Outcomes. So we could have zero fish, one fish, two fish, three fish or four plus. But it did say to round the four or more back to just plain old four and then we have our probabilities. So we had a .44 0.36 0.15 0.4 And .01. And if we look at our formula, it's telling us we need to create a column called X times P of X, which means we're taking the X and multiplying it by its corresponding probability. So if I take one times 36 two times 1, five, And I'm going to keep that going, so I'd get .12 and .04, and when I add up that column, I will get a .82. So therefore our average or are expected value Is going to be .82. So in a six hour time period at Pyramid Lake in Nevada, we would expect less than one fish caught from the shore in that six hour period. Now we're ready to tackle part E and in part E, you are trying to find the standard deviation and to find that standard deviation, we are going to apply the formula the square root of the sum of x minus mu squared times p of x. So we're gonna go back over to our chart and we're going to add on a couple more columns, so we're going to add on x minus mu And that means we're taking the x value and subtracting the average, and we're going to get a negative .82, then we're gonna take the one minus the 10.82 and we'll get a positive 0.18 and we'll keep that going to minus the 0.82 we're going to get 1.182 point 18 and 3.18 So the next thing you're going to want to do is you're going to want to square each of those because this formula over here calls for you to square that quantity, so we're going to create a column called x minus mu quantity squared. And I'm going to bring in my graphing calculator for that. So notice I have already placed my um ex values enlist one. I have placed my P. Of X values into list too. And I want to Put my next values in list three. So I'm going to come down here and I'm going to place them in their relatively quickly. I wanted to take every value Enlist one and subtract that average. And there are the values that you have in your um X minus mu column. So now I want to square each of those. So in order to square those I'm gonna say let's take everything in column three and square them. And in doing so you can see I'm going to get some values .6742. So I'm gonna keep coming back and forth here .6742. Then I get a .03-4. Then I get A13924, A 475- four. And then a 10.11- four. And the last thing we want to do now is we want to do one more column. And this column is going to be called X minus mu squared times the corresponding p of X. So that means you're taking everything in this column and multiplying it by everything in this column. So in our graphing calculator, we called this column list one, we called this column list too. This was list three. This was list four. So technically we're taking list for and multiplying it by list too. So here's my graphing calculator. And I'm going to clear out list five and Enlist five. I want to take list too and multiply it by list for And as you can see, we get these values. So I'm going to right those values down for you. So here we get point 295856 .011664 .20886 1900 96 And .101124. And the formula calls for us to add that list. So I'm going to now add up that list And when I add up that list, I'm going to get a sum of .8076. So therefore our Standard deviation is going to be the square root of 80 76. Or we are going to have a standard deviation of 89 eight seven. So just to recap real quickly here, we have taken our data and we constructed a hist a gram and a probability distribution. We found the probability of Catching one or more fish From the shore to be 56, The probability of catching two or more fish From the shore to be .20. We found the expected value, or the Average, or mean to be .8 to fish in a six hour period, and we just found our standard deviation To be .8987.

So in part A you need to grab. Hey kissed a gram of that data with that is a decimal. And so you have values that go from 01234 or more. And we'll let four stand for that. And then here is 0.1 0.2 0.3 24. And our first bar is .44. So we'll do that in red. And so we have our first bar here for zero and now we have 0.36 So we'll just say it's right here. 0.3 point 15 for the next one About here and for this is one, And then the bar for three is .4. And then I'm just going to let for wanted five. We could let this go indefinitely higher because you have the number of fish but we'll just say that's approximately that and you can ask your instructor how they want you to deal this if this is a column that is up to an indefinite number. Alright. So not for part B. We want to find what's the likelihood that we have one or more fish caught. And that means all of these need to be added or 1 -1 zero which was .44. And that gives us .56. See What's the likelihood that two or more and two or more can be adding up all these bars? Well, I should say adding up all of these three bars or taking one minus these two. And probably just as easy just to add up these three. They're relatively small numbers and that gives us point to for part D. We want to find the mean or expected value. And so we're going to need to take that zero times .44 Plus the one times .36 Plus, continue all the way out to the last 14 times .01. And when we do we end up getting .82. So that's what the mean is. And we can see it's going to be relatively close to zero because this is nearly half of them are zeros and ones, well nearly half are zeros in the sum of these two is 80% of them are zeros and ones excuse me for part E. We want to find the standard deviation and that's going to be if we're showing a workout, we're going to have to take each element minus the main find the deviation squared times its corresponding probabilities. We want to wait those deviations squared And we continue to do that all the way up to this last one and told us to use for for that and whips and put foreign here for minus the pointing to squared times its corresponding probability. And when we some all that up, we end up getting a standard deviation of .898665. I'll just say seven or around .9 one. Little subtle hint if you go in and you put your data of zero through four into list one and put the corresponding probabilities Enlist to and do one variable. Staff With list one as the data element and list to as the frequency. You very nicely will get this for your x bar and you'll get this for your sigma. So this feature is nice and slick.

In the exercise. We're told that when you have a large sample of Poisson random variables for hypothesis testing we can use. This has our test statistic where X bar is the mean of the sample, and you not is the hypothesis is the no hypothesized mean. So we are given a sample and were asked to test the hypothesis as follows. The no hypothesis is the meanest for, and we want to check if the sample provides evidence to suggest that the mean is greater than for. And we are asked to test this at a significance level of 0.2 from what's given in the question, we can calculate our sample mean that will be 160 divided by 36. So that's 160 arrivals over 36 weeks, so the weekly mean is 4.44 So now we can calculate our test statistic using this formula, and this comes out to 1.33 And since this is an upper tail test, our P value will be the probability that's that is greater than or equal to 1.33 And this probability comes out 2.918 which is bigger than 0.2 which is our significance level and therefore we would fail to reject the null hypothesis.


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