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The critical score(s) (the value(s) for t that separate(s) the tail(s) from the main body of the distribution, forming the critical region) is/areTo calculate the t...

Question

The critical score(s) (the value(s) for t that separate(s) the tail(s) from the main body of the distribution, forming the critical region) is/areTo calculate the test statistic, you first need to calculate the estimated standard error under the assumption that the null hypothesis is true Theestimated standard error isThe test statistic is t =Use the tool to evaluate the null hypothesis (Note: You can place the purple Iine on your statistic to tell whether it Iies within or outside the critical

The critical score(s) (the value(s) for t that separate(s) the tail(s) from the main body of the distribution, forming the critical region) is/are To calculate the test statistic, you first need to calculate the estimated standard error under the assumption that the null hypothesis is true The estimated standard error is The test statistic is t = Use the tool to evaluate the null hypothesis (Note: You can place the purple Iine on your statistic to tell whether it Iies within or outside the critical region It's possible that this line may not rest exactly on the statistic_ but vou will be able to tell whether it lies within or outside the critical region ) The statistic in the critical region for two-tailed hypothesis test: Therefore_ the null hypothesis Is The graduate student conclude that playing video games alters peripheral visual perception_ The graduate student repeats his study with another random sample of the same size. Thls time, Instead of the treatment belng playing the combat simulation video game, the treatment is playing soccer video game: Suppose the results are very similar After playing soccer video game, the mean score was still 0.04 higher, but this time the standard deviation of the difference was 0.10 (vs_ the original standard deviation of 0.08) This means has the more consistent treatment effect: This difference In the standard deviation also means that 95% confidence interval of the mean difference would be for the original study, when the treatment was playing the combat slmulation video game than the 95% confidence interval of the mean difference for the second study, when the treatment was playing soccer video game_ Finally, this difference In the standard deviation means that when the graduate student conducts hypothesis test testing whether the mean difference is zero for the second study, he will be likely to reject the null hypothesis than he was for the hypothesis test you completed previously for the original study.



Answers

Highway Accidents: Poisson Distribution A civil engineer has been studying the frequency of vehicle accidents on a certain stretch of interstate highway. Long-term history indicates that there has been an average of $1.72$ accidents per day on this section of the interstate. Let $r$ be a random variable that represents number of accidents per day. Let $O$ represent the number of observed accidents per day based on local highway patrol reports. A random sample of 90 days gave the following information. (a) The civil engineer wants to use a Poisson distribution to represent the probability of $r$, the number of accidents per day. The Poisson distribution is $$ P(r)=\frac{e^{-\lambda} \lambda^{r}}{r !} $$ where $\lambda=1.72$ is the average number of accidents per day. Compute $P(r)$ for $r=0,1,2,3$, and 4 or more. (b) Compute the expected number of accidents $E=90 P(r)$ for $r=0,1,2,3$, and 4 or more. (c) Compute the sample statistic $\chi^{2}=\Sigma \frac{(O-E)^{2}}{E}$ and the degrees of freedom. (d) Test the statement that the Poisson distribution fits the sample data. Use a $1 \%$ level of significance.

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.

Problem. 22 switch and a each node, which is m one smaller than our ableto m two. And each one, which is M one, is bigger than him too. So the degree of freedom, which is in one plus in little minus doors for 19 plus 20 to minus two, is 39. So the critical value corresponding to this degree of freedom with also equal to open toe or five one tail. So using table five teams different toe, 1.6 eight fine. So the pool the standard deviation, which is square root off anyone minus one. So 19 minus one times this one square, which is 7.6449 squared plus in two minus ones who is 20 to minus one times is to square with its 6.9 square over in one plus into a minus 2 19 plus 20 to minus to approximately equal to seven point 253 So the distance statistic is equal to X one bar. One. It's extra war, so 83 minus 79.49 or nine over full standard deviation, which is 7.253 times square boat off 1/19 plus 1/22 which is equal to 1.7 to 1, as this value is bigger than 1.685 So we reject the non high processes.

The following is solution video to number 26 which is two sample T tests comparing meantime loss in the workforce with stressors and intimidators. And you just got to be really careful here. The only kind of weird thing is They kind of switch up the order on you so X one represents the intimidators. In the next two represents the stressors. And we want to see if the stressors, so I'm actually gonna write that down. So this is intimidators and this is stressors. Okay, So it says is there enough evidence to suggest that the stressor is greater than the intimidator? So keeping that in mind, we're going to write less than some you won Is less than you two, Even though it says greater than but the orders just switched, if that makes sense? So the greater than in this case means less than for our alternative, since the order is opposite. Okay, so the first part of this is just verifying that these means. And standard deviations are four and 2.38 for intimidators, and then 5.5 and 2.784 stressors. So I went ahead and it says to use a calculator, so I'm using a T 84 if you go to stat and then edit here are the list, so L one is the intimidator and then L two is the stressor. So if you go back to stat over to couch and then it's one of our stats, we can change that to L one and this will be the four and the 238 that was over here. So four and 238 And then go back to stat couch. One bar stats, change that to L two calculate, and that's where we get to 55 and the 27855 to 78 So we verified that that is the way to do it. So then the second part of it is to kind of sort of not formally conduct a hypothesis test but essentially come up with the conclusion here and it is gonna be a two sample T test. We're gonna use the calculator since we already have the data input anyway. So it's two sample T test for the alpha is point oh five. And the alternative we already talked about, the alternative is less than not greater than but less than so. Um if you go back to stat and then we're gonna air over two tests and it's the fourth one down is the T. Test. Two sample T test because we don't know the standard deviation for the population, We don't know sigma, we only know as um so sincerity of the data, I'm just going to keep it as data instead of doing the summary stats. L one is the first list. L two is the second list and you can keep those frequencies as one and then we're gonna change this alternative to less than you two. And then pulled is usually no unless they tell you otherwise and then we're gonna go and calculate and you can have that test statistic there if you want but you really don't need to because the p value is all you need. So it's 0.14. So let's write that down. So the p values 0.14 and what we do is we explicitly compare the P value with the alpha. In this case 0.14 is greater than point oh five. So that means whenever the p values greater than alpha, we fail to reject H not so we failed to reject h. Not whenever the p values greater than alpha at the p values less than alpha, then we will reject the null hypothesis. So keep in mind that the null hypothesis is generally always saying that these two means are equal. We can we're failing to reject that, so we're accepting that to be the truth. So let's go ahead and complete this hypothesis test. And we're gonna say there is not sufficient evidence to suggest that the mean time lost due to stressors is greater than the meantime lost due to intimidators. That's spelled right? I have no idea. Looks like it's not so intimidators. Okay, that's why math teachers aren't spelling enthusiast. So intimidators um now you could also say there is not sufficient evidence to suggest that the meantime lost due to intimidators is less than the meantime loss due to stressors. If you want to keep to that less than thing. I just wrote it like this because that's the way the book had it. So there is not enough evidence to say that these two means are different or that one means more than the other. So we're accepting the truth that these two means are probably about the same.

Following is a solution to number 25 comparing to means for the amount of time lost due to hot temperatures Compared to the mean number need an amount of time lost due to disputes from superiors attitudes in the workforce. And the first part of this, all we're gonna do is just verify that the mean and the standard deviations for the two datasets are in fact 4.86 and 3.18 for the temperatures And then 65 and 288 for the attitudes. And it says use a calculator, so I'm using the T I T four and if you go to stat and then edit, you can see already put those numbers in. So L one represents, I think that's the temperature column and then L two is the attitude problem. And if you go back to stat and then cowpoke and then one bar stats, one variable stats the list, I'm just doing L one first and that is where we get that 4.86. And the standard deviation where this s is that's up 3.18. So that's where I get these numbers here, and then we'll do the same thing, cal one of our stats, but this time we're gonna do second to for L. Two, and we calculate that, and that's where we get the mean as 6.5, this X bar here, and then the standard deviation, we're not looking at signal, we're looking at essence is a sample sample standard deviation about 2.88 So that's where we get this 2.88 So we have verified that those are in fact the numbers, and now we're gonna do the two sample t test with the significance level of point oh five. Before we do that, we need to figure out what the um alternative hypothesis would be, because we already have the data. We can just punch that in after that and it's going to be a two tail not equal to test because it says one way or the other is just as are the two means different and whenever it doesn't specify which one is greater, you just assume that it's a two tail meaning not equal to so not equal to is going to be our alternative. Okay. And then there are no I didn't write it down but the Noles that they're equal to. Okay. So if we go to stat and then go over to tests and it's the two sample T test and since we already have the data, I'm just going to use the data instead of the summary stats because it's more accurate that way list one will be L one list too will be L to the frequencies can just stay as one and then the alternative the new one, let's change that to not equal to the pool is going to be no unless it's otherwise stated. And then we're going to calculate and this gives us everything. So the t. You know if you want you can put that in there. But really all I care about is this P value the P value is about 0.317 So I'm gonna write that down so the p value is equal to 2.317 And what we do is we explicitly compare the P value with our alpha value and this time the p value 0.317 is greater than our alpha value. Which means we fail to reject the null hypothesis. So any time the peabody is greater than alpha you fail to reject. If it's less than alpha than you actually reject. H not so if we fail to reject this null hypothesis, that means there's not enough evidence to say that these two means are different, or in other words, these two means appear to be the same, so I'm gonna write that out, or I'm gonna type it out because it's a little quicker. That way, I'm gonna, right, there is not sufficient evidence to suggest that the meantime lost due to hot temperatures is different from the meantime lost due to disputes from superiors attitudes in the workforce. Okay, so that is our two sample T test for these two population means.


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