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Tl cont iuUou equivalent Othe Gometric distribution the Erponentidl ditributiou Tou irst suw in the Fractical Homenork ~ction Ll paurt C Suppact [ICauring tle titue btuutu consecutive @i_htning - ` Con dav Lut > dctote bi" 4 suppunc has an expote Itial distributiot O @00O _uupk fotn the LuIdotn ariabk . with tate 0.1 (#hichi Jitralki Mldee lightning 10 ~tom-) Waut tl tlury valuo bx thc Iican oh di-tibtion! Fatiuuua t cpitical uia 04 Flottk di-uibution o tlue uupko 4 Now AUppEA Juu aeauy

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Consider testing hypotheses $H_{0} : p=p_{0}$ versus $H_{a} : p<p_{0} .$ Suppose that, in fact, the true value of the parameter $p$ is $p^{\prime},$ where $p^{\prime}<p_{0}\left($ so $H_{\text { a }}$ is true). \right.(a) Show that the expected value and variance of the test statistic $Z$ in the one-proportion
$z$ test are
$E(Z)=\frac{p^{\prime}-p_{0}}{\sqrt{p_{0}\left(1-p_{0}\right) / n}}$
$\operatorname{Var}(Z)=\frac{p^{\prime}\left(1-p^{\prime}\right) / n}{p_{0}\left(1-p_{0}\right) / n}$
(b) It can be shown that $P$ -value $\leq \alpha$ iff $Z \leq-z_{\alpha},$ where $-z_{\alpha}$ denotes the $\alpha$ quantile of the standard normal distribution (i.e., $\Phi\left(-z_{\alpha}\right)=\alpha ) .$ Show that the power of the lower-tailed one-sample $z$ test when $p=p^{\prime}$ is given by
$\Phi\left(\frac{p_{0}-p^{\prime}-z_{\alpha} \sqrt{p_{0}\left(1-p_{0}\right) / n} {\sqrt{p^{\prime}\left(1-p^{\prime}\right) / n}}\right)$
(c) A package-delivery service advertises that at least 90$\%$ of all packages brought to its office
by 9 a.m. for delivery in the same city are delivered by noon that day. Let $p$ denote the true
proportion of such packages that are delivered as advertised and consider the null
hypotheses $H_{0} : p=.9$ versus the alternative $H_{\mathrm{a}} : p<9 .$ If only 80$\%$ of the packages are delivered as advertised, how likely is it that a level. 01 test based on $n=225$ packages will detect such a departure from $H_{0} ?$

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

Who can't move till question 1 75 Show part A were asked The weather, The American off the auto for inspect is less than one need a meter. So obviously the danger is going to be Yes, because all the data point we have all the merriment are less now. Want me demeanor? So we're confidence enough, Teoh, to say yes. For this question, all data points are less. Then one point narrow you a meter. Okay. And when will to move to part B? We're trying to find the 95% confidence interval. So we start with the simple sized and here it is equal to 36. And after the calculation, we will have the average off the sample to be karaoke wanks 254 So this is our sample size X bar. All right. And according Teoh, the question we have our sick of up, which is a center deviation, is going to be there a 0.1 all right. And our Alfa is equal to their oh, buying their own five. So the critical values that off Alfa over two is going to be 1.96 According Teoh Table for table for beat. Yeah. Okay. Now we're be able to calculate the maximum era. So e is equal to that off Alfa over two times Sigma over squared Rudolph and show we plugging numbers through it. We have 1.96 times there a 0.1 divided by a squared grove 36 we will have the maximum error to be there. Oh, point Narrow 33 How our read it again? Here he had a sequel to their applying their of 33 So our confidence interval is going to be equal to our ex far plus or minus maximum terror. So it is going to me they're open to 54 plus or minus. There are quite narrow 33 and we will get a interval. The lower bound off it is going to pee. Narrow point 2 to 1 in the upper bound is going to be there. Oh, bind! Teoh 87 So this is going to be our 95% confidence in trouble

This example explores rectangular distribution, also known as continuous uniform distribution. And after a lengthy explanation, were given a couple of formulas where basically we have a rectangle and a rectangle is on the closed interval alpha beta, and then we have another interval inside there of A and B. And we want to calculate the probability that something chosen at random, some value chosen at random from the interval of A to B will actually be in alpha beta. So this particular problem, alpha and beta already set for us 0.150 point 065 and then A and B are determined by the problem itself. So in the first example, were asked, what's the probability of getting a pellet that is Greater than or equal to 0.50 mm. So that means a being a smaller value will be 0.0 50 and be can't be any bigger than beta. Now I need to use my formula B minus A over beta minus alpha. And when I evaluate those, I end up with about 0.30. Now, in the next situation, I'm looking for the probability that X is less than or equal to, I need zero point 040 So in this case A will be the smallest value possible on my interval which is alpha and be will be the value in question. And then plugging the information in to my formula and evaluating that expression, I get 0.50 and then and see, I actually want to know the probability that X will be between these two measurements. So now A and B are both different from alpha and beta. And it's just a case again of substituting the values in the right place. In my formula B minus A divided by peter minus alpha, calculating that because these are equal signs. And then the next question I'm asked to find the mean, or mu. So I'm going to use this formula alpha plus beta divided by two. So the mean was Alpha Plus Beta Divided by two. So in our case zero 015 plus zero point 065 divided by two. And then the standard deviation. And that formula was beta minus alpha, divided by The Square Root of 12, Divided by Radical 12. And that is approximately .0144.

This example explores rectangular distribution, also known as continuous uniform distribution. And after a lengthy explanation, were given a couple of formulas where basically we have a rectangle and a rectangle is on the closed interval alpha beta, and then we have another interval inside there of A and B. And we want to calculate the probability that something chosen at random, some value chosen at random from the interval of A to B will actually be in alpha beta. So this particular problem, alpha and beta already set for us 0.150 point 065 and then A and B are determined by the problem itself. So in the first example, were asked, what's the probability of getting a pellet that is Greater than or equal to 0.50 mm. So that means a being a smaller value will be 0.0 50 and be can't be any bigger than beta. Now I need to use my formula B minus A over beta minus alpha. And when I evaluate those, I end up with about 0.30. Now, in the next situation, I'm looking for the probability that X is less than or equal to, I need zero point 040 So in this case A will be the smallest value possible on my interval which is alpha and be will be the value in question. And then plugging the information in to my formula and evaluating that expression, I get 0.50 and then and see, I actually want to know the probability that X will be between these two measurements. So now A and B are both different from alpha and beta. And it's just a case again of substituting the values in the right place. In my formula B minus A divided by peter minus alpha, calculating that because these are equal signs. And then the next question I'm asked to find the mean, or mu. So I'm going to use this formula alpha plus beta divided by two. So the mean was Alpha Plus Beta Divided by two. So in our case zero 015 plus zero point 065 divided by two. And then the standard deviation. And that formula was beta minus alpha, divided by The Square Root of 12, Divided by Radical 12. And that is approximately .0144.


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