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(20) 5 Chris} has eumed created what would ` rcal cstate Every month be the equivalent = Portfolio with SS0,000 _ noral = alter - that; of a 5% anual Jin &sscts...

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(20) 5 Chris} has eumed created what would ` rcal cstate Every month be the equivalent = Portfolio with SS0,000 _ noral = alter - that; of a 5% anual Jin &sscts The first month bc distribution = annual rctum" retum relums with would Inovc (compoundcd for 30 [can monthly } ycars (360 of 0 . and cithet up r down on the #14 months). standard deviation based on & worksheet Now do the simulation of 1%. Simulate bis 1000 Limc_ Do this simulation Change Annual Return Probability 0% 590 15%

(20) 5 Chris} has eumed created what would ` rcal cstate Every month be the equivalent = Portfolio with SS0,000 _ noral = alter - that; of a 5% anual Jin &sscts The first month bc distribution = annual rctum" retum relums with would Inovc (compoundcd for 30 [can monthly } ycars (360 of 0 . and cithet up r down on the #14 months). standard deviation based on & worksheet Now do the simulation of 1%. Simulate bis 1000 Limc_ Do this simulation Change Annual Return Probability 0% 590 15% 25% 40% 0% What Was his average final balance? What was the maximum fina) What was the balance? minimun What final balance? % of the time did he have more than SIB by the cnd of 30 ycars? 15. (15) Using the information from # 14,add this piece of complexity. Thirty percent of the time he Expects extra income at the cnd of the month; &nd he will add S5,500 to the portfolio but this will occur randomly: Do the simulation 000 times on the #15 worksheel What was the average final balance? What % of thc time did he have more than SIB by thc end of 30 ycars? What was the 30t percentile of the final balances?



Answers

A not-so-clever employee decided to fake his monthly expense report. He believed that the first digits of his expense amounts should be equally likely to be any of the numbers from 1 to 9. In that case, the first digit Y of a randomly selected expense amount would have the probability distribution shown in the histogram.

(a) Explain why the mean of the random variable $Y$ is located at the solid red line in the figure.
(b) The first digits of randomly selected expense amounts actually follow Benford's law (Exercise 5 )
What's the expected value of the first digit? Explain how this information could be used to detect a fake expense report.
(c) What's $P(Y>6) ?$ According to Benford's law, what proportion of first digits in the employee's
expense amounts should be greater than 6$?$ How could this information be used to detect a fake
expense report?

So we can see the table here. That is the age in the prophet. Oh, so the probability So, Adri, How 21 22 23 24. And if I, when you call a lot driven 26. So for the fourth year, do some one by 21 Rick Crawford. Well, bury my notes of 100 Sultan's plus 250 that ihsaa money that person paid for the first year. And also for the second year, we need to plus another 220 again, not there to hang greed and 30 year olds. You rode your role the other one so that it's a fright. And if the person rather than by for those years the profit will be to Hendry and comes well, 300 for a coupon fraud and the probability here we got all those probability. And for this one, the little P here thus equals one miners The rope on zero to your old 183 rhinos 0.186 rhinos point 00189 Miners 0.1 night one 193 Reach Yukos who point Sinai zero for out eight. And those offered exactly. While we know that equals who each profit hums each probability off it. So we can God, uh, Miyu here, just you call. So 303 0.35

We want to conduct a pair differences test at alpha equals 5% significance testing. The claim that the population means X. A and X. B are not equal. The data is given below. With mount shapes and at your distribution on the right. I've already computed D. Bar 2.25 The sample size N equals eight and the standard deviation S. D equals 7.78 We proceeded to five steps listed below to solve first. We'll check the requirements and evaluate hypotheses. So because the distribution shape requires have been met to use the student's T distribution, the degree of freedom is at minus 27. We have no hypothesis. H. And R equals zero alternative mu D. Does not equal zero Alpha equals 00.5 significance. Archer statistic is T equals D. Bar over SD over route and 4.8181 from the tea table R. P value is 0.5 is less than ph 7.25 So we have enough information to conclude our test. Now, since P is greater than Apple, we fail to reject the null hypothesis, which means that we lack evidence that beauty does not equal zero.

We have data sets A and B listed here. We want to test the claim that there is no difference in the underlying distributions for populations A and B at the alpha equals 0.1 significance level. We proceed through steps A through D listed below this question is testing an understanding of non parametric tests in particular how to implement the ranks on test. So first we stayed alpha and hypotheses, this is alpha equals 0.1 has stated H. And distribution of the same H. A distributions are different and be we can take the test that we also see the sampling distribution which is normal and check the requirements which are met. So the ranks are as follows. We compute N one and two as 12 and 12. That's when you are is 1 50 single. Rs 17.32 are this summer brings from A is 1 66. Z is not unless you're over signal articles 660.92 Thus from a normal distribution we compete the P value as two PZ greater than absolute values equals 20.28 Thus, we can conclude, indeed, that we fail to reject nation on, since P is greater than alpha, which means we lack evidence to support the alternative hypothesis H. A.

Okay, Let us look at this question. Now, the fish and the game department stopped Lake Lulu with fish in the proportions that were given to us. Okay, so what we have here are different kinds of fish. So let me just make a column over here. So this is going to be kind off fish. Okay, so first we have catfish, we have catfish, then we have bus. Then we have blue kin. Then we have bluegill, and then we have bike. So this is going to be 1 20. This is these are the observed values that we have. 85. This is 2. 20. And this is 75. Okay, now they stopped them in different proportions. 13 30%. 15%. 40%. 15%. Again five years later. So this was the observed value. These are the observed values, the observed values. Okay, now, over here, we'll read the original proportions. Original proportions, original proportions. Now, the original proportions, who are 30% for catfish in this was 15. This was 40%. And the last one is 15 again. All right. Now it was found that 500 fish in the sample were distributed the way it has given the observed values. In the five year interval, did the distribution of the fish changed at 0.5 level? So for this question, we can say that our Alfa is 0.5 Okay, Now, what is going to be another hypothesis? Arnel hypothesis will be that the distribution, the distribution did not change did not change in the past five years in the past five years. All right. And what will be the alternative hypothesis? The alternative hypothesis will be that the distribution changed in the past five years in the past five years. All right, so now the tests that were conducting is the Chi Square test and what is the first step in order to calculate the value for the chi square statistic. First, we need to find the expected values, the expected values, the expected values for all the categories. Now it is given by sample size. The sample size for a question is 500 right sample size multiplied by the probability or the proportion probability Oh, are the proportion off each category off each off each category. Okay, off each category. All right, So what is my sample say's my sample size is 500. If I just add all of these up, I'm going to get 500 rate. So what will be my expected values? My expected values. My expected values are going to be if I just use my calculator over you. Just a moment. Yes. They are going to be 30% off. 500 now, which is going to be 1 50 Then we're going to have 15% which means 75. This is also going to be 75. And this one is 40%. So 40% is going to be 1 60 right? Really? This is going to be 1 60? Nope, This will not be 1 16. So this is going to be 200. All right. 1 1503 100. 200. 500. Okay, these are expected values. Now, what is the next step? The next step is finding the guy sequester distinct. How do we do that for all the categories we're going to perform? The operation observed value minus the expected value. Whole square. That is we are squaring the difference between the observed and the expected values we divide this by the expected value and in the end, with some all of the individual values up. So this will give me the chi Square statistic for the entire problem. So if I go up, these are going to be my individual chi square values for all the categories. Right? Individual high school values. Now, what exactly did we do now? We're going toe. Do the observed minus the expected and then hold square. So this is the difference between 1. 15 and 1. 20. Now we're going to square this and we'll to divide this by the expected value. Okay, so this is 1 50 minus 1 20. We square this. So this becomes 900 on DWI. Divide this by 1. 50. So this is going to be six. Okay, then. 85 minus 75. We square this and direct this by 75. This is 1.331 point 33 All right, then we have the difference between 202. 20. So the square is 400 divide this by 200. So this is to And then I think over here the difference would be Cedo, right? Oh, minus zero. Yeah, Now what I'm going to do is I'm with some all of these individual values. So this is six plus 1.33 plus two and this becomes 9.33 So I can say that my chi square statistic turns out to be 9.33 Now, what else do I need In order to find the p value? I need the degrees of freedom. Degrees of freedom is what DF it is given by the formula number off categories number off care degrees minus one. All right. How many categories do we have if we look away? There are four categories catfish based, Google and pipe. So this is going to be four minus one or I can say that this is equal to three. My degrees of freedom is equal to three. Now, my chi square value is 9.33 and my degrees off freedom is three. Now, we can either use a chi squared even for this or what we can do is we can use a statistical package or a software which will give us an exact P value. What is our Alfa for this question? Are Alfa for this question is 0.5 Okay, so I put in all the values and I hit Calculate. I get my P value as 0.25 to 1. So we're here. I will say that my p value is 0.25 to 1. All right, Our Alfa was 0.5 and we can clearly see that RPI value is less than Alfa hands. We can say that we will reject that we reject are null hypothesis. H not know what was on a hypothesis. The null hypothesis waas that the distribution is still the same after five years. So I can say now this is the most important line. I have enough statistical. I have enough statistical evidence Enough statistical evidence to suggest do suggest that the distribution, the distribution, the distribution off fish off fish off fish in the league in the league has changed. Yeah, in the league has changed over the past five years over the past five years and this is how we go about doing this question


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