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To determine if “A” students tend to sit in a particular part ofthe classroom, a teacher recorded the locations of the students whoreceived grades of A....

Question

To determine if “A” students tend to sit in a particular part ofthe classroom, a teacher recorded the locations of the students whoreceived grades of A. He found that 17 sat in front, 9 sat in themiddle, and 5 sat in the back of the classroom. When testing theassumption that the “A” students are distributed evenly throughoutthe room, he obtained the test statisticof χ2 = 7.226. If using a 0.05significance level, is there sufficient evidence to support theclaim that the “A” students

To determine if “A” students tend to sit in a particular part of the classroom, a teacher recorded the locations of the students who received grades of A. He found that 17 sat in front, 9 sat in the middle, and 5 sat in the back of the classroom. When testing the assumption that the “A” students are distributed evenly throughout the room, he obtained the test statistic of χ2 = 7.226. If using a 0.05 significance level, is there sufficient evidence to support the claim that the “A” students are not evenly distributed throughout the classroom? Then state the conclusion. Fail to reject the null hypothesis; there is sufficient evidence to support the claim that the “A” students are not evenly distributed throughout the classroom. Fail to reject the null hypothesis; there is sufficient evidence to support the claim that the “A” students are evenly distributed throughout the classroom. Reject the null hypothesis; there is sufficient evidence to support the claim that the “A” students are not evenly distributed throughout the classroom. Reject the null hypothesis; there is sufficient evidence to support the claim that the “A” students are evenly distributed throughout the classroom.



Answers

Conduct the hypothesis test and provide the test statistic and the P-value and/or critical value, and state the conclusion. A classic story involves four carpooling students who missed a test and gave as an excuse a flat tire. On the makeup test, the instructor asked the students to identify the particular tire that went flat. If they really didn't have a flat tire, would they be able to identify the same tire? The author asked 41 other students to identify the tire they would select. The results are listed in the following table (except for one student who selected the spare). Use a 0.05 significance level to test the author's claim that the results fit a uniform distribution. What does the result suggest about the likelihood of four students identifying the same tire when they really didn't have a flat? $$\begin{array}{lc|c|c|c} \hline \text { Tire } & \text { Left Front } & \text { Right Front } & \text { Left Rear } & \text { Right Rear } \\ \hline \text { Number Selected } & 11 & 15 & 8 & 6 \\ \hline \end{array}$$

In this problem, we're going to be testing the claim that oxygen treatment is effective for people who have very painful cluster headaches. We have two samples. One sample received oxygen treatment on the other sample received possible treatments, so 150 patients were treated with oxygen, and out of the 150 people, 116 are free from headaches, so the proportion is 116 divided by 115. And for those who are given the possible, you have 29 free from from the headaches 15 minutes of the treatment out off a total of 148 patients. So we're going to test the claim that the oxygen treatment is effective and we're going to use two approaches. The fast approach will be the hypothesis hypothesis test, and the second one will be. The confidence interval meant better now for the hypothesis test method. We're going to have the Nahal hypotheses as P one equals p two, which is to say that the proportions are equal and to prove that them oxygen treatment is effective. It would be that the proportion off those who received oxygen treatment and war free from headaches is greater than the proportion off those who received the placebo. And we're free from, um, headaches. Now we can work out the test statistics by substituting the values we obtained into the formula for that. And when we do so, the calculator value of that is nine point 96 Next, we can get the critical value for this, uh, one tales test at the 0.1 significance level. On the calculated value of that is the critical value of set for that is going to be 2.33 So we can compare the conclusive body of that in the critical money upset. And in this case, we have 2.33 can shape that critical region, and we see that 9.96 is within the critical region. And for that reason we reject the national hypothesis. This'll means that there is sufficient evidence to support the claim that the cure rate with oxygen treatment is higher than the cure it for those given a placebo. So it appears that the oxygen treatment is effective from the hypothesis test. Let's see what happens when we conduct the confidence interval test So for the confidence interval, we need to substitute the value, Uh, the values that we have obtained the into the formula and to get the imagine of error e And when we do so, imagine of error is 0.1101 And when we can also substitute the values, uh, into the expression for the confidence interval you obtained, the interval limits US zero 0.4669 less than p one in a speed too less than 0.6871 And when you look at the intervals at the interval limits, we do not have zero in between. So the internal limits do not include zero, and that means that the two curates are not equal. In other words, we would have to weigh would have to reject the null hypothesis. So this confidence interval test is in agreement with the hypothesis test. And so it appears it appears that the cure it with oxygen treatment is higher than the cure it for those given a placebo, and that means that the oxygen treatment is effective. Next, we're going to to give a reason as's, too, as whether or not the oxygen treatment is effective based on the two tests. So based on the two results the conflict, the hype of the serious test and also the confidence interval test, there's an agreement that there is a difference between those two proportions. So the results. Both results suggest that the oxygen treatment is effective in curing cluster headaches.

This exercise, we're going to be considering a trial that was conducted with 75 women in China were given, uh, 100 you one bill while another 75 women in China for giving 100. You want in the form off smaller bills. So we told to test the clean that when a single large bill is given a smaller proportion off, women in China spend some or all the money compared to the proportion off the money the proportion of women in China given the same amount in smaller bits. If you've given smaller bits, you tend to spend more. Or in other words, when you given ah single large bill, you tend to spend less. So let's see ah, how we're going to do that. And we're going to do it in two ways. Using the hypothesis test and using them confidence interval method. So when we consider the data we have, um, toe took samples. The first sample was given a single of bill large pill Uh huh. And the second sample was given the same amount of money, but in smaller bills. So out off the 75 women that were given the large bill. 60 of them spent some or all of the money. And out of the 75 women that were given smaller bills, 68 spend some or all of the money. So at the 0.10 point 05 level of significance, we're going to test the clean that a smaller portion, a smaller proportion of women who was given, um, the large bill spend some or all off the money. So then I'll hypotheses will be p one is equal to p two. The alternative hypothesis is P one iss less than p two. So when we walk out, this will be a 21 tales test on the critical widely, that will be negative 1.64 So we can now walk out the value of the test statistic by such student values accordingly, so that we get the correct value of set. In this case, it is negative 1.85 zero so we can compare the test statistic and the critical value. In this case, we can share the critical region, which is there, left hand side off negative Z 1.64 And when we look at the critical there. Test statistic. We see that it is within the critical region and therefore we reject the now hypothesis and by rejecting the knowledge what this is, we conclude that there is sufficient, um, evidence to support the claim that when given a single large bill, a smaller proportion of women in China spend some or all of the money compared to the proportion of women in China given the same amount in smaller bills. Now let's do the same comparison, but this time using the confidence interval method. So when we subjected the values into the formula, imagine of error. E is given by 0.0 936 and now we can compare, uh, then I can substitute their other values into their confidence, our interval limits. And we see that the law in limit is negative zero point 2006 on the upper limit. His negative 0.134 And when we look at this confidence interval, we notice that the confidence in terms for limits do not include zero, and the confidence interval limits on Lee have negative values. And so it appears that the proportion off women who spent summer. All the money s smaller for the sample that was given a larger bill that the sample that was given small bits. So in the last part of the question, we're supposed to change the significance level two zero point 01 and see whether the conclusion will change. So when we set the confidence, the level of significance 001 in the critical body upset will be negative. One negative too. 0.33 on when it is negative. 2.33 Maybe at this point, so when you get you 2.33 we notice that the critical region does not contain the test statistic off negative 1.850 and therefore changes the conclusion. So we would have to fail to reject. The non hypotheses are in that case. And hence we would not have sufficient evidence to support the claim that for the proportion for Freeman, who spent some or all of the money, was smaller when the large bills were given, as compared to when the smaller buildings were given. So yes, there is are changed in the conclusion. Once we change the level of significance to 001

Problem 21 which is a it snowed if that new one is more than or equal to Muto, and each one or the authority hypothesis is that anyone is larger than YouTube. So the degree of freedom, which is equal to anyone, plus 20 to minus two she's 19 plus 25 minus soldiers equal 42. So the critical value in the rules degree of freedom equal to 42 offer is equal toe open one one tape. So using Table five, the critical values equal to 1.3 or three eso the full statistic, which is equal toe the square dudes and one minus one times this one squared course any to minus one point is to square over in one plus any to minus two, which approximately equal toe eight going toe one mine. So so the test statistic is equal to T is equal toe x one x one Bar minus X To bar over on the gold standard deviation time square root off one over and one plus one over and to which approximately equal to 42.295 eso as this value is bigger than 1.3 or three. So we reject that complaint processes

In this problem we're going to be looking at to smoking cessation programs. One of them is the sustained cab, and the other one is the standard camp in the sustained care. We had 198 smokers going through the program on out of those 198 smokers, 51 no longer smoke after six months and for this standard care program among 189 smokers, fatty one no longer smoking after six months. So we're going to use the 0.1 significance level to test the claim that the rate of success for smoking cessation is greater with the sustained care program are compared to them standard care program. So the first part of the question we're going to test the claim using a hypothesis test. After that, we're going to test it using the confidence interval. And then we're going to see whether there's a difference between the two programs, uh, in terms off the proportions. Okay, so let's begin and state the hypothesis on. In this case, the non hypothesis is P one equals P two, which implies that the two proportions have are equal and the alternative hypotheses is P one is greater than P two. So this means that according to the hypothesis that we have a higher proportion off people who no longer smokey the sustained care program compared to the standard care program. So this being, uh, one tales test, the critical value will be 2.33 So we need to substitute the values into the test statistic, and we have the following proportion. For those in the sustained care, we have 51 longest, smoking out off 188. And for the standard care, we have 30 the longest, walking out off the 199. And to get the calculated value of that, we need to substitute these values into the formula. And here p one hot in decimal form is 0.258 and we need to subtract p too hot, which is 0.151 Okay, then the difference. We subtract zero from the difference, which is assumed to be, uh, zero, according to the Nile hypothesis. So then, from there we get the square it off. P one p bar Cuba, developed by N one plus p. Baquba Weber. And to So PBA is given by the sum of all the, uh all the X X one and next to divided by anyone and into So what we need to have here is the sum of 51 and that he divided by the summer of 188 and 199. And when you walk that out together, P bob is 0.204 divided by N one, which is 188 on, we have to multiply the PBA, Cuba and Cuba is one minute 0.204 which is going to be 0.796 So the new Morita is repeated in the next fraction. So it's 0.204 time 0.796 divided by N to an end to is 199. And when we simplify the value off the test statistic that he's 2.6 five now, we can compare the calculated value of that and the critical value off that and you can see. But the critical value is 233 so we can share the right side and the calculated value of that is within the critical region, which is 2.65 And since that is the case, we have to make the conclusion to reject the null hypothesis. No, this means that there is sufficient evidence to support the clean, that the rate of success is for the smoking cessation is greater with the sustained care program compared to them the standard care program. Next, we're going to test the same claim by constructing, uh, confidence interval. And in this case, we're going to construct 98% confidence interval. And for that we need to get the margin of error. E is in the formula given, and when you walk that out you will get e equals 0.9 0.9 35 And when you substitute into this formula for P one heart minus B p one hut minus be too hot minus e, you will obtain the following confidence interval. So to be 0.1 35 it's less than P one main US P two, which is less than zero point 2005 now. In this case, we noticed that the confidence interval limits do not contain zero. And that means that there is significant, a significant difference between the two proportions, as we have had really seen in the test off hypothesis, meaning we could reject the null hypothesis off having the proportions being the same. So because the interval consists off positive numbers only, it appears that the success rate for the sustained care program is greater than the success rate for the standard care program. Lastly, in Passy, we're going to see whether the difference between the two programs have practical significance. So if you check the the percentages, for example, for P one it is 25.8%. And for P two hot, this is 15 0.1%. And based on this sample, the success rates off the programs our 5th 25 points, 8% and 15.1%. And that difference does appear to be substantial. There's a there's a difference. So the difference between the programs does appear to have practical significance, because it's about a difference off 10% between p of P one heart and P to heart


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