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Point) Suppose a group of 700 smokers (who all wanted to give up smoking) were randomly assigned to receive an antidepressant drug or a placebo for six weeks. Of th...

Question

Point) Suppose a group of 700 smokers (who all wanted to give up smoking) were randomly assigned to receive an antidepressant drug or a placebo for six weeks. Of the 555 patients who received the antidepressant drug; 36 were not smoking one year later: Of the 145 patients who received the placebo, 47 were not smoking one year later: Test the null hypothesis Ho (p1 pz ) = 0 against the alternative hypothesis Ha (P1 p2) # 0. Use & = 0.01 The rejection region is Izl (b) The test statistic is z

point) Suppose a group of 700 smokers (who all wanted to give up smoking) were randomly assigned to receive an antidepressant drug or a placebo for six weeks. Of the 555 patients who received the antidepressant drug; 36 were not smoking one year later: Of the 145 patients who received the placebo, 47 were not smoking one year later: Test the null hypothesis Ho (p1 pz ) = 0 against the alternative hypothesis Ha (P1 p2) # 0. Use & = 0.01 The rejection region is Izl (b) The test statistic is z = The final conclustion is A. We can reject the null hypothesis that (p1 pz) = 0 and accept that (p1 ~ p2) # 0. B. There is not sufficient evidence to reject the null hypothesis that (p1 pz) = 0.



Answers

Test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim.
Among 198 smokers who underwent a "sustained care" program, 51 were no longer smoking after six months. Among 199 smokers who underwent a "standard care" program, 30 were no longer smoking after six months (based on data from "Sustained Care Intervention and Post discharge Smoking Cessation Among Hospitalized Adults," by Rigotti et al., Journal of the American Medical Association, Vol. 312, No. 7). We want to use a 0.01 significance level to test the claim that the rate of success for smoking cessation is greater with the sustained care program. a. Test the claim using a hypothesis test. b. Test the claim by constructing an appropriate confidence interval. c. Does the difference between the two programs have practical significance?

No. Okay, so we know that the number of patients given sustained care is 198. Among that, 82 0.8%. They're no longer smoking after one month. So 198 times 82.8 percent. Well, give us 163.94 Okay, so we can round that 264. Yeah. Andi making, then Right down. Arnel Hypothesis, which is given that P is 1 80 80% since the test claims that 80% of patients stopped smoking after giving state care and our culture bosses will therefore be P is not equal to your right June at a given level of significance. Which is your prince? Your one. Okay. Me? Yeah. So we will get a Z statistic value off 0.99 on the critical value for the normal area. Table movies equals plus or minus +257 eight A p value. 20 above output is just your 200.32 Sure. Yes. On. Since this is greater than our little of significance, we can include that there's not sufficient evidence to support the rejection of the claim that 80% off patients stop smoking when sustained care. What about

Okay. So the objective in this question is to test the claim that among smokers try to quit nicotine patch therapy, the majority are smoking year after the treatment. The null hypothesis h not, is that the probability is equal to 2.5 on the alternative hypothesis. H A is that the probability created in 0.5? Okay, you're also given that X is you got 39 and it's equal to 39 plus 32. She'll give us a sample size of 72. On that off level of significance is your 720.5 This just information. We're given the question self. Yeah, now can calculate Z statistic the statistic which is the proportion minus p over the square root off peak you of N v. Q Is one manus Okay, P is our no, not so playing all these values in We'll get value from the calculator, which is 0.83 on the P value. This test from the normal table is 0.203 on since the P value is greater than the level significance Alfa, which is your digital five. The null hypothesis is not rejected now for there is evidence to show that smoke among smokers try to quit with nicotine patches therapy. The majority are still smoking after a year of treatments, majority still smoking. That is what we can conclude.

In this question. I have been given a table. All right, Andi, I have to calculate. I think the mean is to be calculated by using that table. So if I use the table to find the mean the mean off those 32 these are 32 numbers, right? Yeah. These are ready to farmers focus. So their main turns out to be 14.83 4th expert is 14 0.834 My end is equal to 32. Have I been given the standard Division? Yes. Population standard Division 6.2 Sigma is equal to 6.2. And the claim? What is the claim? It takes smokers too. Quit permanently. Is there enough enough evidence to reject the claim that the meantime, it takes smokers to quit smoking permanently as 15 years? Okay, the mean is 15 years. So immune is equal to 15. This is going to be my null. And my alternative hypothesis will be that Mu is not equal to 15. All right, I need I need to know what I'm going to use. I'm going to use the sad statistic and I have all the information that I'm going to require in order to calculate this, so I'll simply put in the formula 14.83 minus 15 upon 6.2 by route off 32. Now I have to use the calculator to calculate this Just a moment. So this is going to be 14.83 minus 15 divided by 6.2, multiplied by route off 32. This is minus zero point. This is minus 0.155 This is my zed statistic. Okay, Now, if I calculate my people first of all, this is going to be a two tailed test because there's a not equal to sign. And what is my Alfa level? My al 50.5 So if this is a two tailed test night in these two tails, my total areas 0.5 which means in every single tail it will be 0.25 So let me do one thing. Let me find the critical value for 0.25 and this value turns out to be 1.96 which means that since it is symmetrical, this is minus 1.96 and this is plus 1.96 And since my Z value is between this interval, my Z value falls somewhere around here minus 0.155 I will fail to reject minor hypothesis. And if I want to find the p value for this, my P value for this is going to be P Valley for minus for the zero point. Uh, sorry for minus 0.1 point 55 The P value for this is just a moment in a zero point 88 My P value is turning out to be 0.8 ID, 0.88 right. This is my p value. I have used the people you calculate over you, and I can see that my p values greater than Alfa. So from here also, I can say that I will fail to the Jackman and hypothesis. So this was my null hypothesis. I will say that I do not have enough evidence to see or I do not have enough evidence to reject the claim that the meantime it takes the smokers to quit smoking permanently is 15 years. This is going to be my answer

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.


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