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Fome MYUVUAssignments MG'4T--Hcmetor 7Inbax (1) _ wademisbaGe: Hemevcrk HeqPencohttps / Inewconnectmheducation com flowfconnecthtmlHomeworkDeHelp Save Exit SubmitCheck my Wotk16 After years rapid growtn; Illegal Immigration into the United States has declined, perhaps owing Ine recession and Increased border enforcement by the United States (Los Ang les |imes; September 2010]. While its share has ceclined, California still accounts for 219 ofthe nation estimated 14.3 million undocumented im

Fome MYUVU Assignments MG'4T-- Hcmetor 7 Inbax (1) _ wademisba Ge: Hemevcrk Heq Penco https / Inewconnectmheducation com flowfconnecthtml Homework De Help Save Exit Submit Check my Wotk 16 After years rapid growtn; Illegal Immigration into the United States has declined, perhaps owing Ine recession and Increased border enforcement by the United States (Los Ang les |imes; September 2010]. While its share has ceclined, California still accounts for 219 ofthe nation estimated 14.3 million undocumented immigrants [You may find useful tO reference the z lable-] ocints sample of 70 Illegal Immigrants, what the probability that more than 1622 Iive In California? (Round "2' value t0 decimal Places. and final answerto decimal Places ) cBco" Probability Rafetence= sanpe 0f120 Mlegal Immiarants Wnar the probabilizy that more than 1693 live California? (Round "> value [0 decimal places, and finol answe decimol loces; Probabilif Prev Next 3.39 PM 325/2019 Search the wveb and Windows



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Immigration The following graph shows how immigration (in thousands) to the United States has varied over the past century. Source: Homeland Security.
(a) Find the average annual rate of change in immigration for the first half of the century (from 1910 to $1960 ) .$
(b) Find the average annual rate of change in immigration for the second half of the century (from 1960 to 2010 ).
(c) Find the average annual rate of change in immigration for the entire century (from 1910 to $2010 ) .$
(d) Average your answers to parts (a) and (b), and compare the result with your answer from part (c). Will these always be equal for any two time periods?
(e) If the annual average rate of change for the last half-century continues, predict the number of immigrants in $2013 .$ Compare your answer to the actual number of
$990,553$ immigrants.

For the match pairs given on the right. We want to conduct a sign tested matched pairs testing P does not equal 0.5 at alpha equals 0.5 significance. This question is testing our understanding of non parametric tests in particular how to conduct a scientist of matched pairs. We proceed there steps A through D below to solve. So first, in a we stayed alpha hypotheses, this gives alpha equals 0.5 H. And r P does not equal five. H. A P does not equal five. So the no hypothesis P is equal to have, the alternative is not A and B. We compute the test out. So are signs for these matched pairs are as follows and equals 12 is the total number of matched pairs. So X equals the number of plus or the total number plus and minus which is 4/12. Thus we have seen it equals x minus 5/45 over and equals negative 1.155 Thus the p value is for a normal distribution, P equals two PZ greater than zero equals 248 Thus we conclude that we fail to reject asian off because he is greater than alpha, which means that we lack evidence to support the claim P does not equal five.

The following is a nova test based on the mean salaries for different metropolitan areas. So the alternative or the null hypothesis is that all the means are the same. So there are six metropolitan areas, I think it goes Chicago, Dallas Miami, Denver san Diego and Seattle. Uh So the null hypothesis is that all the means are the same. And then the alternative is that at least one of them is different. The second step is to find the critical value and you can do that using either software or a table, But they're essentially three things you need. The first thing is your alpha value, your significance level and that's usually given to you the problem and that's .05. Then you need the degrees of freedom for the numerator and the degrees of freedom for the denominator. And the way you find that Is the degrees of freedom for the numerator is the number of categories -1. So there were six cities that we looked at our metropolitan areas, so 6 -1° of freedom would be five for the numerator. And then for the denominators, the total number of data values minus the number of categories. So there were 36 data values minus the six metropolitan areas. So 30 is your degrees of freedom for the denominator. So that should be enough to use a table. But I use a calculator and I wrote a program in here called inverse. F. I'm not going to show you how to how to write the program. You can youtube it if you wish. Um But this is what I do. So um I put in my area which is my alpha value, my degrees of freedom is five and then my degrees of freedom for the denominator is 30 and That gives me my critical value. About 2.534 2534 is my critical value. I call f. star. So 2.534. Okay so anything greater than 2.534. We reject the annual hypothesis that all the means are the same And anything less than 2.534. We failed to reject meaning the h not is true. Okay so the second step is to find the F statistic and there's a formula but it's a bit of a mess. I always use software you know technology is a great thing. So if you go to stat and you can type in your data values. So these are the mean salaries um So again L1 I think was Chicago and then this is the mean salary for Dallas Miami Denver San Diego and Seattle. So there are six categories. And if you go to stat tests and then we're gonna go to the Unova test and then you just type in your columns separated by commas remember there were six columns, six data columns that we used and we need to make sure that all of them are in there and last one and then also you know make sure you separate those by commons, otherwise it's going to read it wrong. So then um that gives us everything we need. So the F. Is the F statistic, that's the third step. So we're looking at this it's about 2.281 as our F. Value. So two point 281 is our f statistic Which is actually barely in the non rejection region 2.281. So that means we fail to reject. Okay and also we can verify that with this p value here. So the p values 0.7 which is a pretty small p value, but it's still in this case greater than the alpha value. So the alpha value remembers point oh five, so it's barely greater than the alpha value. And whenever it's greater than the alpha value, uh we failed to reject, I should probably put H not there, so we failed to reject H not whenever the P values greater than the alpha. Okay. So then the last step is to summarize everything with actual words. So what does this all mean? It just means that there is not sufficient evidence, there is not sufficient. I guess you could say statistical evidence to suggest that the mean salaries from the different metropolitan areas are different. Okay. And that's the five step process for an Innova one way and over test

Uh huh. Consider two binomial experiments the first with and one equals 210. And observe successes are one equal 65. The second one and two equals 1 52 and R two equals 18. We want to construct a confidence interval for the difference in population proportions. P one minus P. Two for which. We proceed to the following three steps A through C. To complete step A. We have to determine whether it's appropriate to the normal distribution. To approximate the hat one minus P had to. The answer is yes that one is just R. One over N one P had to is our two over M. Two. And you get obtaining P. One P. Two. We see that all of N one P hat one and one Q. Hot one and two P. Had to. And and if you have to a greater than five so it's appropriate to the normal distribution next. Now this validated let's construct a 99% confidence interval with that interval given by the formula for the right to be here. So we have to find the margin of error. That's the critical Z score times the square root expression. The left at 99% confidence are critical Z score is from a Z table on google or a textbook 2.33 plugging in RP hats, Q hats and Z is equal 0.106 Or plug into the confidence interval 0.85 or less than T one minus P two is less than 20.297 Finally, we can interpret these findings, since our confidence interval only contains positive values, we have 99% confidence that portion one P one is greater than proportion to P two.

Alrighty. So this is one of those questions where the data is really news relevant and it would be very interesting to see if you could find updated information on the same topic to see how things have changed. So despite the curiosity there, but let's focus on what they're asking us to do. The big thing for you to realize is that this is not a frequency distribution table. All right. This is a probability distribution table. That means every number in there is already a probability number. And you should be able to recognize that just by the fact that they're all less than one. All right. So, the probability is they're looking for are basically already calculated for us. Right, Okay. So all the way might, once we get to the Gibbons, we might have a little calculations. Let's see what they're asking for. The first one is asking for the probability Of the citizens that arrived were naturalised, whatever the story is in the year 2002. Right? So that's the column. The second column. And again, this is question about a single variable. This is not a joint event. So, we're going to go down to the bottom of that column for that marginal probability and just read it off the table because they've already calculated the probabilities for us. Right? So, the second question, it's a little bit more interesting, Right? This is the probability that a randomly selected person from this study did not come from El Salvador. All right. That means they want everybody else except the people from el Salvador. Right? So how are we, what's gonna be the easiest way to do it? So, again, we're talking about these marginal totals. So we can be adding up all the countries except el Salvador. I think the easiest way to calculate calculate that is just to say, well, let's start with everybody which represents 100%. And let's just subtract the probability that people that were from El Salvador. Okay, so that gets me at Mine, is that .417? Right. So I put my handy dandy calculator here Off to the size and I know you're doing the same thing. 1 -1 1 7. Right? So that .58, 3 Right, or 583. 10,000 thousands thousands. Yeah, that's the right. 1, 503,000. Let's work on our number, number skills, numeracy. All right. Next question is the probability um that the citizen is from Honduras, right? And That the citizen came in the year 2003. So now we are looking for a joint event. It's very simple. Just run along row five and column three there as marked And find that joint probability in there which is a much smaller numbers. Now we're just at 38,000 or 3.8 get that way. And finally we get to the conditional probabilities right? Which is the new skill this time around. So here we cannot do the conditional probabilities directly right? Like we did at the beginning of the lesson here, we have to use the conditional probability general formula. So this one is asking what's the um probability that the citizen is from or the person is from Guatemala Given that they came in the year 2001. Right? So remember if you can't calculate it directly, what you're doing is you're doing the joint probability divided by the givens probability. That's a key idea that the given goes on bottom Right? So joint over the given. So the joint one, we just look up off the table, we go to row four for Guatemala and 2000 a column one for 2001. And we pull off the number of 0.079. And then the probability of year one then, is that marginal probability at the bottom of the column? So that's your .3 age. What? All right. So do that division there real quick On their handy Dandy calculator and I'm getting .20 5 7 were around. I just dug it out a little bit and that's almost that's all for that. Nobody's asking me. Let's see what are they asking? The answer to be left in. It does say interpret your results in terms of percentages. We should have been doing this a long way. We'll do it at the end here. So last one there asking for is the probability Okay, it's a reverse of the last one. The probability that it came in the year 2001. Given that they came from Guatemala. Okay, I'm doing the right one there. All right. So that's gonna take the same numerator noticed because the formula is very similar. So the numerator is still the joint probability of C. One and C. Four and Y one. But the denominator is now the probability of C. Four right Given becomes the denominator. So my new monitor still pointing their 79 like it was before. But my denominator now has to go over here to Guatemala's marginal probability way out there for the total number. The total problems of people from being from Guatemala. Just 1.2 four. Run that one through Handy Dandy Calculator Real Quick. And you get point 387 for your probability for parts Or 387000s. Right. All right. So now we do need to give answers in percentages we should have all along the way. So just real quick. Right, you make a sentence that says Something like 33 0.8% or again you got to make the whole sentence. So you're gonna say something like if a person is random, if a person is selected at random from the naturalized persons in this subset, what's the best way to say that? Anyway? There's 33.8 chance that they came in the year 2002. And if a person from this study is randomly selected, There is a then going on to the next one. There's a 58.3 chance That they came from El Salvador. And if a person from this study is randomly selected, there's a 3.8 chance that they were both From Honduras and came in the year 2003. And then everybody. There's a 20 .57 chance that a person randomly selected from this study Was from Guatemala, given that they came in the year 2001. And finally, there was a 38.7 chance that a person randomly selected from this study was arrived in the year 2001, given that they were from Guatemala, all right, steven type of all those sentences, and that's all for this problem.


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