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Suppose you perform the hypothesis test Ho: p 70 versus Hi: p F 70_ The population variance, 02 is known: The sample size is n 39. Assume the significance level is ...

Question

Suppose you perform the hypothesis test Ho: p 70 versus Hi: p F 70_ The population variance, 02 is known: The sample size is n 39. Assume the significance level is 0.03.1) Should you use 2 or t to find the critical value?Oz2) Choose the correct critical region:Reject Ho if 2 > Za Reject Ho if z > Za Reject Ho if 2 > 24 Reject Ho if 2 Z 24 Reject Ho if 2 < Za Reject Ho if 2 < Za Reject Ho if 2 < _ 24 Reject Ho if z < ~ 24 Reject Ho if 2 < ~ Za Or 2 > Za Reject Ho if 2 &

Suppose you perform the hypothesis test Ho: p 70 versus Hi: p F 70_ The population variance, 02 is known: The sample size is n 39. Assume the significance level is 0.03. 1) Should you use 2 or t to find the critical value? Oz 2) Choose the correct critical region: Reject Ho if 2 > Za Reject Ho if z > Za Reject Ho if 2 > 24 Reject Ho if 2 Z 24 Reject Ho if 2 < Za Reject Ho if 2 < Za Reject Ho if 2 < _ 24 Reject Ho if z < ~ 24 Reject Ho if 2 < ~ Za Or 2 > Za Reject Ho if 2 < Za or 2 > Za Reject Ho if z < 24 Or 2 > 24 Reject Ho if 2 < 29 or 2 > 24 Reject Ho if t > ta Reject Ho if t 2 ta Reject Ho if t > tg Reject Ho if t 2 t; Reject Ho if t < ta Reject Ho if t < ta Reject Ho if t < ~ t4 Reject Ho ift < - t; Reject Ho if t < = ta or t > ta Reject Ho if t < _ta or t 2 ta Reject Ho if t < t4 ort > t4 Reject Ho if t < _ tg ot 2 t4 3) Identify the critical value(s). If there are multiple critical values, separate them using commas. If using 2, round to 2 decimals; if using t, round to 3 decimals" _



Answers

Consider independent random samples from two populations that are normal or approximately normal, or the case in which both sample sizes are at least $30 .$ Then, if $\sigma_{1}$ and $\sigma_{2}$ are unknown but we have reason to believe that $\sigma_{1}=\sigma_{2},$ we can pool the standard deviations. Using sample sizes $n_{1}$ and $n_{2},$ the sample test statistic $\bar{x}_{1}-\bar{x}_{2}$ has a Student's $t$ distribution, where $$\begin{aligned}&\frac{\bar{x}_{1}-\bar{x}_{2}}{\sqrt{2}} \text { with degrees of freedom } d . f .=n_{1}+n_{2}-2\\&\sqrt[4]{\frac{1}{n+1}+\frac{1}{1}}\end{aligned}$$ and the pooled standard deviation $s$ is $$s=\sqrt{\frac{\left(n_{1}-1\right) s_{1}^{2}+\left(n_{2}-1\right) s_{2}^{2}}{n_{1}+n_{2}-2}}$$ Hypothesis tests: Use $H_{0}: \mu_{1}=\mu_{2}$ and an appropriate alternate hypothesis. The $t$ value of the sample test statistic is t=\frac{\bar{x}_{1}-\bar{x}_{2}}{s \sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}} \text { with } \mathbb{L}_{2}=\frac{n_{1}+m_{2}-2}{2} $$\text { A } c \text { confidence interval: }\left(\bar{x}_{1}-\bar{x}_{2}\right)-E<\mu_{1}-\mu_{2}<\left(\bar{x}_{1}-\bar{x}_{2}\right)+E$$,where $E=t_{c} s \sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}$,$\begin{aligned} t_{c}=& \text { critical value for confidence level } c \text { and degrees of } \\ & \text { freedom } d . f .=n_{1}+n_{2}-2 \end{aligned}$.Note: With statistical software, select the pooled variance or equal variance options. There are many situations in which we want to compare means from populations having standard deviations that are equal. This method applies even if the standard deviations are known to be only approximately equal. Consider Problem 19 regarding average incidence of fox rabies in two regions. For region $I, n_{1}=16, \bar{x}_{1}=4.75,$ and $s_{1} \approx 2.82,$ and for region II, $n_{2}=15, \bar{x}_{2} \approx 3.93,$ and $s_{2} \approx 2.43 .$ The two sample standard deviations are sufficiently close that we can assume $\sigma_{1}=\sigma_{2}$ (a) Use the method of pooled standard deviation to redo Problem $19(\underline{a})$ (b) Use the method of pooled standard deviation to redo Problem $19(\mathrm{b})$

Now for Part C, we are told that a random variable is simulated 10,000 times and the sample mean is 41.63 and the sample standard deviation is 8.5 and we are asked to find and interpret and 95% lower confidence bound for the true expected value of this random variable. Now the lower conference bound is given by this formula, and here we can use a critical value from the standard normal distribution because and is so big and this comes out to 41.5. So this means that we are 95% confident that the true expected value of this random variable is at least 41.5.

In this case, we have to calculate the critical values for the information given to us. We are given the sample sizes and when and do, and the hypotheses null and alternative hypothesis. Over here, I can see that in the B option. This is going to be a two tailed test. This one is a two tailed tests, and all the remaining are single tail tests. For example, in the last one, this is going to be a left tail test because we're checking. If Sigma One is Sigma one by Sigma two is less than one. So this is a left tail test. We're here. We're checking for greater than one. So both of these tests are going to be right. Tail tests these air, right? All right. Now, how do we calculate the critical values? We can either use the tables or we can use some statistical Softwares like many tap or Excel or python, or are in this case in this demo or other. I am going to use an online software at online tool and try to calculate one of these critical values. For example, let's say I am taking the third example in C option I have. Anyone is 10 and and two is also 10. So the degree off freedom are going to be nine and al 50.1 So we let me just put nine as a degree of freedom in both of these, and probability level is 0.0 one. I'll just calculate this and the critical value that I'm getting is 5.35 approximately, and this is what I've put here. Similarly, we can calculate all the remaining values and these are the values that I'm going to get.

Hi. This question. We need to use many tab, so I'm gonna show you how to do most of the steps. So, first of all that you have to go on many tap. I use many top 17. So you go to college, then just goto Rhonda Data on Click Normal. You're gonna get something like that. Okay, then just put number of frogs 2000 as, ah, pacified. And then the question and just put them and see one, which is the first column. Um, we're gonna use the standard normal distribution, which is, um, a normal distribution off me and zero and standard deviation one, then head. Okay, then you're gonna see something with ah, uh, the The date is gonna be generated in and see one, so go back toe graph. Ah, you're gonna see ah, Choice called instagram. Then click on that with fit. Don't choose the simple one. Just go ahead and click with fit and and choose. The data said, that's we already generated, which is on C one. I happen toe called 1 51 for the question. So and as you can see here, it's absolutely normal because we generate from normal distribution and we get the the mean very close to zero and standard deviation off one. And the sample size was 2000. After you finish this one, you have to generate now at least samples from whatever we generated. So we are not generated from from a distribution that so But from the from the column that we just drink. So you have to go to Calcutta clearer random data and go Ah, to this option sample from columns. And after that, you're gonna generate Ah, turn off them. So see one tool. See, Tim, After you finish this one, you have to go to Stat Rose to the sticks here, and you're gonna be navigated to this one. So you go ahead and choose standard deviation here and choose all the columns. As you can see here from t one, I just rename it. Do you want to teach in or basically from sea to to see 11. So the 1st 1 was generated from normal distribution on from C two to C 11 is generated from the first column. And just click on this one center deviation and just choose them variables that we were just generated from the sample on store the the values off standard deviation and in any count that you would like, but usually but it here. So I haven't to To put it and sit well on, I rename it to standard deviation, then Hecht. Okay, After that, you have to do a comparison between the two values. Something like that. Um so let me show one and you're gonna continue. So here's the number biz. It's 12 and so on on, then you're gonna get something like that. The probability off it, Andi any of it, The area to the left. So for this probability, you can we know that we have 200 samples, so that will be won over 200. This one is going to over 200 and so on. So, basically, whatever ends in over so, honey and hear from the defense that we were gonna fund that as going Oh, the 93 on you gonna find this one is 1.1 all three and so on. So I'm gonna give you justice numbers for 10. It is 18.3. So after you calculate this one, you gonna find that Onley probability and in one is matching. But the other one actually are higher than calculated. Hopefully that Wasim full anti

The following is a solution in number 18 which is the goodness of fit test for Poisson distribution. Uh This involves like bacteria in the mouth or something. And the lambda the mean is 2.8 or 2.80. And we're asked to find the probability that zero occur whatever it is, I think it's like bacteria colonies or something. Zero probability that one occurs 234 and then greater than or equal to five. So you can use the formula but I like to use the software so I'm gonna use the C. I A. T. Four. And if you go to second bars which is the distributions we can go to this possum pdf a sense for the probability density function. And then we're just gonna type in the mean which is mu in this case at lambda, they call it lambda for the present. And then the x. value is just what I'm trying to find. So this is the probability of zero given that the mean is 2.8 and that gives us .0608. So .0608. Okay I'll do that one more time and then the other ones I'll just copy down. So then the possum pdf 2.8 and then this time I'm going to make it one And then we press enter and then it's .17 03 Let's say. So .1703. Okay so that's what you're gonna do and then you're going to change it to 234 So I'll just go and copy these down whenever you do that. For the two should get 20.2384 for the three, you should get 30.22 to 5. And then for the probability of four you should get 40.1557 Now for greater than equal to five it's one minus the things that you just found. So what you can do is you can just add 0123 and four together. These these probabilities and you take one minus that and that will give you the probability that it's great and equal to five because we don't have enough time to do probability of five plus probability 6789 all the way up to infinity. You know, we don't have enough time. So Instead it's quicker if we just take the total which is one minus the probability is that we don't want which is zero through four. Now fortunately um there is a function of the calculator that adds it up for you. If you want to use it you don't have to but it's called the person C D f. C D F stands for the cumulative density function. So it's one minus the person C. D. F. So if we go second distribution and you may have seen it already it's the person CDF actually you know what when the second quit because I forgot to do one minus so one minus posen CDF. And remember the mean was 2.8. Now instead of the X. Value being I'm not gonna say 01234 I'm just gonna say four. So the calculator automatically knows that it's zero plus one plus two plus three plus four. The probabilities associated with that And we paste it in there and that gives us .15-3. Okay so let's write that down some point. So equals .15-3. So those are all the probabilities. The second part of the support be. It is finally expected value if the sample size N was 100. Now that's kind of nice because the math is super easy. We're just gonna take and I'll just show it for the first one we take the sample size which is 100 times each associated probability. And that's going to give you the expected value. So really all you gotta do is just move the decimal place over twice to the right and that's gonna give you 6.08 and then 17.03 when we multiply 23.84 22.25 15 57 And then greater than equal to five is 15.23. So those are all the expected values And that's going to lead us to our goodness of fit test. So they observed that was the chart that was given so zero that we observed 12 to have zero. We observed 15 to have one, we observed 29 to have two and so on, and so forth. So I wrote those down now. I got to write down the expectancy. Remember the expected For each of these. 6.08 17.03 23.84 22-5. So these are all the expected values from the previous part, Part B that I can plug in now. Okay, so now I need to find the chi square test statistic and I can go and do this now. The degrees of freedom is always in minus one or the number of categories. So k minus one. So 123456 That means there are five degrees of freedom, So 6 -1. Okay, so let's go back to the calculator Or you can use um you know, again, any software you want or you can certainly use the formula that they give you. But this is That's not right, this is a lot easier. So if you get a stat and edit you see already put those in. So L1 I put the observed And then L. two I put the expected. Now if you have something different or if you plug them in differently that's fine. But I just do L one observed L two is expected to go back to stat and then tests and it's the very close to the very bottom, that's the chi square G. O. F. Test which stands for goodness of fit test. And as you can see there I do L one for observed L. Two for expected. So if you have something different there, if you put observed in L. Three let's say you just change that to L three degrees of freedom. Number was five already said that. And then calculate and that gives us the chi square value of 13 point 138, let's say 13 .138. And that's that. Okay. The last part is we actually finish our test. So let's just look at this p value, that's gonna be the easiest route. So the P values .022. So let's go and write that down. P values 0- two. And we're going to explicitly compare that to the alpha value. In this case it's less than alpha, it's less than 0.5 point two, is less than 20.5 anytime the alpha value or the p values less than alpha. You reject. H not well, we didn't formally write down what the, you know, h not was but it's understood in a goodness of fit test, the H not is that these two distributions are the same, the observed and the expected are about the same. And we're rejecting that. So in this case, because we're saying that it follows a person, a person distribution, we can say this data, I think it's bacteria does not follow a person distribution because we're rejecting that meaning, we're accepting the the alternative hypothesis and the alternative hypothesis would be that this does not follow a Poisson distribution.


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