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Chesterfield ALICAUEAa aan Probability 1 and id Statistics 1 CP Cenira P Fall endenc; 2020 1 1 1 ejabuv) Cannon)...

Question

Chesterfield ALICAUEAa aan Probability 1 and id Statistics 1 CP Cenira P Fall endenc; 2020 1 1 1 ejabuv) Cannon)

Chesterfield ALICAUEAa aan Probability 1 and id Statistics 1 CP Cenira P Fall endenc; 2020 1 1 1 ejabuv) Cannon)



Answers

Let $P\left(E_{1}\right)=\frac{1}{4}$ and $P\left(E_{2}\right)=\frac{1}{2}$ and find the probability of the event. Probability of $E_{1}$ not occurring.

All right. So for these problems were told that two events are happening at once, 1/4 probability happening event to is one half opportunity of happening. Um So for this one really care about E two, and that was one half. So not happening would be doing the complement of that, so one minus one half, which is equal to one.

The following is a solution for number six and this is a two sample Z test testing the two population means the difference to population means being 45 or greater than 45. And we're gonna test at the 450.1 level of significance or 0.10% level significant. So here you have the summary stats, we're gonna use this formula, we're gonna do the critical region method. Now the reason why we're able to do a two sample Z test instead of a T, Is because the sample sizes are large enough. As long as the sample size is greater than 30, the Z and the T essentially become the same thing. There may be a slight difference here and there, you know, with rounding and whatnot. But but with these small with these large sample sizes that usually takes care of any discrepancy. So we're essentially just gonna be plug in some numbers here into our formula. So we're going to find the test statistics. So it's the difference of the sample means, So that's 13, -12, 56 minus the hypothesized value. Without hypothesized value was 45 as the difference. And then that's divided by the square root of the sample variances. So 35 squared Divided by that sample size which is 200 and then plus Um this should be asked to actually, so plus 28 squared And then that's divided by 2 25. Okay, so whenever you plug that in you should get three point decided 3.548. So that's our test statistic which is kind of a big test test value especially with the sample size. So I think we're probably gonna be rejecting here. But let's just go to verify. So we're going to find the critical value. That's the star is what I use. So C. V. Or Z. Star. And the way we do that. So let's look at uh you can use a table or T. I. T. For any calculators. So wish but I'm gonna use inverse norm here And the area is going to be .001 And the mean and standard deviation will always be 01 in this case. And then this is a right tail test since it's greater than so we're gonna pace that rascal in there and we get about 3.090. So that's our critical value. three 090. So I like to draw a picture with the critical region method. Anything to the right of that critical value we're gonna reject. And this test value does lie in this rejection region. So we're gonna reject The null hypothesis saying that it's equal to 45. Now to find the p value. That's the second part of this of the p value. We should get the same answer as rejecting but let's just verify. So on the p value method we just look at the probability which means we need to use normal CDF. The lower bound will be since this is the right tail test will be that test value of 3.548. The upper bound is just any large number, I like to use the 99 because that's the biggest number calcula can handle. But you can use anything, you know, as long as it's probably greater than uh than 15 or 20, it should be fine. The mean and standard deviation of zero and 1 and go out and paste and you get pretty small number there, you know, by rounding this is four zeros with 19 at the end. So I'm gonna just go in and say it's About zero to which is less than alpha. Which either way we're gonna reject. H not Okay, so that's the first one. The second one is a lot like it. But this time we have a two tailed tests, because the whole the alternative hypothesis is not equal to And we're testing that is not equal to negative 12 at the 10% level of significance. And so here we have the summary stats and we're going to find that test value using that same formula. So it's the difference of the sample, means 1 21 minus 1, 35 minus the hypothesized value of negative 12. And then that's divided by the square root of the standard deviation squared or the variances of the sample divided by n. And then divided plus seven squared divided by that end was 40. Okay, and whenever you do that you should get about negative 1.332. So that's our test statistic. And then to find the critical value that Z. Star, we're gonna go back to the calculator or table, whichever you prefer. If you go to second distribution and go to that inverse norm. So this time since it's a two tailed tests, you're just gonna have to cut that alpha and 0.5.5 and it doesn't really matter which one you do left or right and I did left last time. So all good do right this time. And if we paste 1.645 and then it's also negative 1.645 that equals Positive, And I like to draw a picture here. So negative six 1.645 And positive 1.645. So anything less than the negative or greater than the positive, we would reject this negative 1.3 lands in the non rejection region. So we would fail to reject here or inject. It's not. Okay, so let's just verify with the p value. So whenever we find the p values, you remember on two tailed tests whatever you get on the p value. Just need to remember to multiply it by two. So let's just use this negative 1.332. If we go to second distribution and again we're finding the probability. So normal CDF is what we want. Lower bound is just some negative large number. The upper bound, like I said, is going to be our test value since it's negative, so negative 1.332. And then the mean and the standard deviation just to keep that a zero and one and then we paste and so this is not the actual P value. Remember that's just one side of it. We always have to account for the second side whenever it's a two tails. So it's about 20.182, which again, so 0.182, Which is greater than that alpha of .10. So had we not multiplied by two, it would actually be less than and we would reject. But but so don't forget you gotta multiply by two greater than alpha, which means we would fail to reject. Which is the exact same conclusion with that critical region method.

Alrighty. So for number 10, we are talking about probability of someone being 25 years old, are older with a bachelor's degree is 250.279 in the United States. And then we were told that the probability of someone being that age with a bachelor's degree, given that they live in D. C. Washington, D. C. Is 0.485 So the question is, are they independent events of each other? And because they're not equal to each other, they have to be mutually exclusive with each other. Then, um, that's the case where they would be the same probabilities. Since that's not the case here, the answer is no, that they're not independent.

Problem. 23. We have three events. See one, C two, C three which are independent events with probabilities. Uh huh. One third. One quarter expected, We want to compute the probability of C. One, Union C two, Union C three. Let's use the inclusion exclusion property. It equals the probability of C one plus the probability of C two plus the probability of C three minus a submission of the probability of the intersection between each pair. Let's see, one intersects C two. Close the probability of C one intersect C three plus the probability of C two intersect the C suite plus the probability for the intersection between the three events. As long as C one, C two and C three are independent events. We can change the intersection to be multiplication, then it equals one half plus one third plus one quarter minus half, multiplied by one third plus half, multiplied by one quarter plus one third, multiplied by one quarter plus the multiplication of the three on half. One quarter. One third equals three quarters


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