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Slep Dy slepFor each problem; please stale the Jort alternative hypotheses; wrile declsion rule rejecting the null hypothesis; (€) calculate the appropriate ...

Question

Slep Dy slepFor each problem; please stale the Jort alternative hypotheses; wrile declsion rule rejecting the null hypothesis; (€) calculate the appropriate statistic Ise in deciding whether reject the null hypothesis (show all calculattons} (d) state Yowr decision a5 uuhetler= reject the null hypothesis; and (e) state conclurion that the investigator entitled drdw: Use alpha level of .05 for relebant calculatlons First ycar college students New York reqfiged Uake an intelligence test; Iean L

Slep Dy slep For each problem; please stale the Jort alternative hypotheses; wrile declsion rule rejecting the null hypothesis; (€) calculate the appropriate statistic Ise in deciding whether reject the null hypothesis (show all calculattons} (d) state Yowr decision a5 uuhetler= reject the null hypothesis; and (e) state conclurion that the investigator entitled drdw: Use alpha level of .05 for relebant calculatlons First ycar college students New York reqfiged Uake an intelligence test; Iean L.Q. score for all students Lking the test is 108. The mean of a smple 0f 45 sludente from New York University 3. The standard deviation of the population the mean of college students fromn New York University signifieantly different from the Iean of all first year college sturdents New Yotk? The tean number of how many curs one has wned for the population of people living the U.S.is 5.7. An invesligator predicls that the mean number of people who own Cars cities will be significantly Iess than Ie In Fennl population. The mean number people wto own cars cities samnic Oi people 6.5. The standard deviation of scores in this sample = Do Ihese dulu $upport the investigator $ prediction? In the state of Califomia; third graders get Wvcrlec SCOIl 0f 20 relding test (higher corcs Fcnee higher levels Perlormance) Iccher USIng * ncw mcthod 5 (ejch reading She predicls that by the end of the third grade, students getting her new mcthod significantly highcr Scofes reading than those tnc populalon mcan oftthe sludeni her class and dhe standard devialion ot the students the cluss It is predicted that overull job sulisliction for employees Iuec compinies will be 'significantly lower than the mean job suistiction the entire population cmplovecs , Thc mcan 0f this populalion 0n 4 mcasure of joh uisluction 48,55 (higher scofes rellect higher Ievels xusuction) The mean of 4 <uple of 38 employees from Ltpc comnanics 51,23 The stundard devintion ofuhis $mple Do the data provide supPOrt for the: rescearcher'$ prediction?



Answers

(i) What is the level of significance? State the null and alternate hypotheses. (ii) Check Requirements What sampling distribution will you use? What assumptions are you making? What is the value of the sample test statistic? Compute the corresponding $z$ or $t$ value as appropriate. (iii) Find (or estimate) the $P$ -value. Sketch the sampling distribution and show the area corresponding to the $P$ -value. (iv) Based on your answers in parts (i)-(iii), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level a? (v) Interpret your conclusion in the context of the application.Note: For degrees of freedom $d . f$. not in the Student's $t$ table, use the closest $d . f$. that is smaller. In some situations, this choice of $d . f .$ may increase the $P$ -value a small amount, and therefore produce a slightly more "conservative" answer.Answers may vary due to rounding. wanagement: Intimidators and Stressors This problem is based on information regarding productivity in leading Silicon Valley companies (see reference in Problem 21 ). In large corporations, an "intimidator" is an employee who tries to stop communication, sometimes sabotages others, and, above all, likes to listen to him-or herself talk. Let $x_{1}$ be a random variable representing productive hours per week lost by peer employees of an intimidator.$$\begin{array}{llllllll}x_{1}: & 8 & 3 & 6 & 2 & 2 & 5 & 2\end{array}$$. A "stressor" is an employee with a hot temper that leads to unproductive tantrums in corporate society. Let $x_{2}$ be a random variable representing productive hours per week lost by peer employees of a stressor.Use a calculator with mean and standard deviation keys to verify that $\bar{x}_{1}=4.00, s_{1} \approx 2.38$ $\bar{x}_{1}=5.5,$ and $s_{2} \approx 2.78$ (a) Assuming that the variables $x_{1}$ and $x_{2}$ are independent, do the data indicate that the population mean time lost due to stressors is greater than the population mean time lost due to intimidators? Use a $5 \%$ level of significance. (Assume that the population distributions of time lost due to intimidators and time lost due to stressors are each mound-shaped and symmetrical.) (b) Find a $90 \%$ confidence interval for $\mu_{1}-\mu_{2}$. Explain the meaning of the confidence interval in the context of the problem.

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.

What we want to conduct A p D. T. A pair differences tests at the alpha equals 1% confidence level. Testing the claim that the population means A bar next pr are not equal. We have the data for A and B. Given below assuming amount shapes, not your distribution on the right. I've already calculated D. Bar noted that an equal seven and calculated SD as 70.47 So we proceeded to do the five steps listed below to solve this first. We evaluate the requirements and hypotheses. So the requirements to use the students distribution have been met because the distribution shape we have degree of freedom and minus one equals six. Are null hypothesis is mute equals zero. Or alternative is beauty does not equal zero. And we're testing at alpha equals 00.1 confidence Next will compute the test statistic and P value. So the statistic is T equals D. Bar divided by SD over. Uh huh. 2.083 from a tea table. This puts p between .1.05. So we can conclude that P is greater than alpha, which means we fail to reject the null hypothesis, which means that we lack evidence that beauty does not equal to about.

Solution number 17. And this is a, an interesting problem with the Poisson distribution. And the Poisson distribution is the type of discrete distribution where the events are random and rare. And uh, this, in this case it's a traffic accidents, Daily traffic accidents, and there is an average, that's what we use this lambda for. So λ,, which is the average of the mean is 1.72 accidents per day. Were asked to find the probability that zero accidents occur and the probability that one occurs To occur three occurs and then greater than four. Curse. Now you can use the formula, but again, I like to use the uh, software. So what I'm gonna do is I'm gonna go to second distribution and then I'm gonna go down to the present pdf and then here it asks from you, or sometimes it lasts for lambda, and that's 1.72 And then the X value, I'm just going to find the probability that zero occurs, And that gives me .1791. I'm going to go in round here some point 1791. So you can do this with any type of software or you can just use the formula. Although the formula can take awhile .1791. And I'm gonna do it one more time. Just show you whether they're the second bars for distribution. And then I went to the Plaza pdf, right? That's the probability density function. So pdf. And the mu the mean is that's the land of 1.72. And this time we're gonna find the distribution or the probability that one occurs and it's about Point Let's say .308 point 308. And that's what you're gonna do for, you know, basically all the rest of them. Until you get to the greater than so zero point 2649, I'll go ahead and give you these answers here and then 0.15 one night. So then uh to get something that's greater than what you're gonna do is you're gonna take one go all the way up to infinity and save some time. We're just going to take one minus the four that we've already found. 0123 So there's another function in the calculator we can use is called the plaza CDF. The cumulative density function. And we're going to go up to three. So the CDF calculates the probability of zero plus the probability one plus probability two plus probability of three whenever you do CDF of three, so one minus that. Or you could just do one minus these four numbers here whichever you like. So one minus. And then 2nd distribution. And I'm going to go to the present CDF. And the main remember was 172 and then the X value. Now I'm not gonna put 0123 I'm just gonna put the three And it automatically calculates 0 1, 2 and three combined. And whenever you do that that should be your answer. So .0962 0962. You might get 61 if you just use these numbers here but you get the same thing if you do one minus and these all added together. Next up we find the expected value and any time you find the expected value just take the sample size times the probabilities. So for zero, remember there were 90 days that we looked at. So 90 times the probability of zero was remember that .1791? That should give you 16.119. And then all of these are gonna be the same 90 times something. And those some things are the probabilities. So .308 that's going to give you 27.72. So that means we can expect about 27.72 days where there are there's one wreck And then 90 times the point 2649. That should give us 23 841 And then 90 times .1519. He was this 13671. And the 90 times 0.0 962 Gives you 8658. Okay, so those are the probabilities. Those are the I'm sorry those are the expected values. And then the part c we find we're gonna find our use the good as fit test. So they observed that was the chart that was given and the expected we actually just found. So we're just gonna copy those expected values down 27.72 23-841 13.671 and then eight 658. So those are the expected values where we take the probabilities times the sample sizes. Okay? So now we can go back to our calculator and do the goodness of fit test. So if you go to stat and then edit you can see that here in L. One. I put the observed values and then L. Two I put the expected values. And then if you go to stat tests it's the chi square goodness of fit test. And the observed is L. One the expected sell to or you can use the formula or any other software. And degrees of freedom was four. So the degrees of freedom, that's actually another answer. The degrees of freedom of four. Since there are five categories there of 0123 and then greater than equal to four. So we can calculate that And that gives us a chi square value of about 12 509 So chi square equals 12.509 And then it also asked for degrees of freedom. The degrees of freedom, like I said, was equal to four because it's five minus one. Okay, so the p value let's look back at the P value, it's about .01. Let's go at some .014 and that is greater than the alpha, barely, but it is greater than the alpha. So whenever the p value is greater than the alpha than we fail to reject to reject H not. Which means in this case, actually, in all these goodness of fit test cases, the null hypothesis is that the distribution does in fact follow whatever we're talking about in this case, we're failing to reject that. So this follows a person distribution so the traffic accidents, and per day in this particular area, it does in fact follow a Poisson distribution.

So we're gonna let d stand for the time of no nap minus the time where they take that four hour nap after having only four hours of sleep. And so are non hypothesis is going to be that that mean difference is Less than a report to zero. And alternately that this no nap time is going to be higher than the nap time. So that D. Is greater than zero and you can go the opposite way as well and have this these signs being reversed. So that's up to you. But I draw my little picture so that's part A Part B. We're gonna end up having our for part B. We have to find that critical value. So we're assuming that we're at zero and there's a cut off point where we have that significance level of point to one and are degrees of freedom is nine. So we want to find that T. Star value that's going to have the degrees of freedom of nine and have the upper tail be 1%. When we look that up, that comes out to be 2.8 to 1, which means our rejection region Will be if the T value we get for our test statistic is greater than 2.8-1. So no one part C. I subtracted all the values and I found that the mean. D the uh d bar value came out to be negative 0.0. Excuse me? Positive .097. Yeah, let me erase that. And we got the standard deviation of those values. So I just I truly put the one value into list one and the other value into list too. And then I have the calculator to list one minus list too. And then I did one variable stats to find the mean and the standard deviation. And this comes out to be about that. So now we need our test statistic, our test statistical have 9° of freedom and we will take that .097 and we will subtract away the mean and we will divide it by the standard deviation Over the square root of 10. And when we do we get the test statistic comes out to be very large 7.14. So we can see that for part E. R. Decision, we can see that this test statistic is way in the way, high, its way into this rejection reason region. So we have evidence to reject the null. And so we would conclude that it does appear that the nap does seem to reduce the time to reduce the runners times run or time in that 20 m dash, I believe.


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