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Comparative study of two new drugsand B 350 patients were treated with drugand 300 patients were treated with drug (The two treatment groups were randomly and indep...

Question

Comparative study of two new drugsand B 350 patients were treated with drugand 300 patients were treated with drug (The two treatment groups were randomly and independently chosen:) It was found that 241 patients were cured using drug and 222 patients were cured using drug Let p the proportion of the population of all patients who are cured using drug and Pz be the proportion of the population of all patients who are cured using drug Find 90% confidence interval for P1 Pz: Then complete the tabl

comparative study of two new drugs and B 350 patients were treated with drug and 300 patients were treated with drug (The two treatment groups were randomly and independently chosen:) It was found that 241 patients were cured using drug and 222 patients were cured using drug Let p the proportion of the population of all patients who are cured using drug and Pz be the proportion of the population of all patients who are cured using drug Find 90% confidence interval for P1 Pz: Then complete the table below: Carry your intermediate computations to at east three decimal places Round your responses ta at least three decimal places (If necessary, consult list_of jrmulas ) What the ower limit of the 9090 confidence interval? What the upper limit of the 909 confidence interval?



Answers

Use the data and confidence level to construct a confidence interval estimate of $p,$ then address the given question. In a program designed to help patients stop smoking, 198 patients were given sustained care, and $82.8 \%$ of them were no longer smoking after one month. Among 199 patients given standand care, $62.8 \%$ were no longer smoking after one month (based on data from "Sustained Care Intervention and Postdischarge Smoking Cessation Among Hospitalized Adults," by Rigotti et al., Journal of the American Medical Association, Vol. $312,$ No. 7). Construct the two $95 \%$ confidence interval estimates of the percentages of success. Compare the results. What do you conclude?

Now, in this case, it is given that out off. 333,000 and five adults, 81.7%. 81.7% off. People used at least one prescription medication. So what is this number? 81.7% of 3005 off 3005. This is going to be my party dancer to my park. If I use a calculator, this is going to be 81.7. Divided by 100 multiplied by 3005 Or this is approximately 2455 people. This is approximately 2455 people. Yes, this is my answer to Park eight. Now, what is part B five visas? I want to construct a 90% confidence in trouble. Estimate 90% conference and double estimate. 90%. Which means my Alfa by two is 0.5 0.5 This is my Alfa. By do so what is my Z Alfa biting Z Alfa by two is going to become 1.6449 1.6449 We can use a calculator to find this. Okay, Now we have 0.817 This is R P cap 0.817 So what will be a queue? Camp Que cap will be 18.3 or 0.183 Right. This will be 0.183 And we already know the formula for E is the margin of error. He is equal to Z Alfa by two, multiplied by route over. Speak up que capped by N n is 3005, right? And is 3005 adults. Okay, so if I use a calculator to calculate this, this turns out to be rude. Over 0.817 multiplied by 0.183 divided by 3005 And this is multiplied by 1.6449 multiplied by 1.6449 This is zero point 0116 This turns out to be 0.116 Okay, this is my margin off error. Now, if I want to construct a confidence interval what is this going to be first? If this is b the lower limit, what is a lower limit? The lower limit will be 0.817 minus 0.116 This is 0.80540 point 8054 or 80.54% age and 0.817 plus 0.116 182860.8286 This is my confidence in W now. Part C is asking us what the results tell us about the proportion off the college students who use at least one prescription medication. Just a moment. I think over here it is given to us that 3005 adults aged 57 through 85 you know, in a survey off adults, it was found that these many off them used one prescription medication and so on and so forth. So these 3005 subjects are actually adults who are from 57 to 85 years. So what does this actually tell us about the college students? Absolutely nothing, Right? So see is a great question. Question number sees a great question. It tells you absolutely nothing. And this is our answer. This tells us nothing about the college students. Why because these proportions are for adults off 57 to 85 years of age, so this is not answer.

Now in part A. Okay, if the therapists are just guessing randomly. Okay. Given that Emily used a coin toss to select either her right hand or her left hand, what exactly would be the proportion of correct responses if the test therapists made random guesses? What is the probability of getting it right? It is 0.5, right. The probability is 50%. It's a 50 50 chance. So it's 0.5. Now, let us move on toe Part B in part B. We want to say what is the best point estimate of the therapists success rate? It has given that into 80 trials 1, 23 times they were correct. So it is going to be 1. 23 by 2. 81 23 by 2. 80. Right, 1 23 by 2. 18. So what is this? 123 divided by 280 This is 43% around. This is around 0.43924392 This is around 44%. Okay, this is the best point point estimate now moving on to part. See what is Park Si se now? Party wants us to construct a 99% confidence interval estimate off proportion off correct responses made by the touch therapists. So in this case, this is my peak cap. This is my peak cap. So what is going to be my cue cap? Que ca will be one minus speaker. Or I can write this as 0.56070 point 5607 Okay, now the formula for he is Z Alfa by route over P cap que capped by N N s to 80 anarchists. And what is myself? How I do it is going to be 2.57582 point 5758 Why? Because I want to construct a 99% confidence in my Alfa by two. In this case, alphabet to in this case is 0.5 So if I just do this calculation, this is going to be rude. Over 0.4392 multiplied by zero point 5607 divided by to 80 and after multiplied is about 2.575 point 5758 This is 0.763 Okay, let me illustrate the direct answer. 0.7 16. 0.763 Okay. And what is my pick up? 0.4392 Now, if I want to calculate the interval, my first lower limit will be pick up minus E, which is going to be 0.439 to minus 0.763 This is 0.3 69 0.36 to 9 less than p. And this is less than now. The addition of these two. What is the addition of these two? Plus 0.7 60. This is 0.51550 point +5155 Okay, Now, using the Emily sample results, we just had to construct an idea and person confidence interval to assume the proportion of correct response is made by the therapists. So this is going to be our answer. This is the confidence interval. We are 99% confident. Okay, We are 99% sure that the two proportion is going to lie between these two values. And this is our answer

Are We want to find the sample mean extra the sample standard deviation s and construct the 90 confidence interval about the population of meaning you or data below. Reassuming population is normally distributed. And once we compute these particular values, we want to interpret our findings to start off with, let's find, expiring s. Remember that to do so we use appropriate definitions for X bar that is the some of the data X I divided by the end. The number of data points. This computer expert equal 62.3 Ash is defined as the square root of the sum of deviations about the moon square, divided by n minus one. Which in this point computes to 8.0. Next let's find the TC value or rather the critical T value from the student's T distribution needed to compute this confidence interval to do so, we have to use a tea table, A tea table takes as input to important things, the degree of freedom in this case nine and the confidence level in this case .9 to output the correct value, we find we use a tea table either from google or on a textbook and doing so, we find T C. Equals 1.833 next we want to compute the margin of Arafat's interval, given by the formula on the left, plugging in T C. S and end for this problem. We obtain E equals 4.6 and we can compute the confidence interval. Now, given the following formula experiments, E is less than you as less than expert plus E plugging in our experts and our E gives confidence interval 57.7 is less than you is less than 66.9, which we can interpret to mean. We are 90% confident that the new the population mean is between 57.7 and 66.9.

So we we would be assuming that the mean of the diet coke is equal to the mean weight of the regular coke. And alternately we think that the diet coke is actually less than the regular coke. And we have our data that's given to us in the example of two sample sizes of 36. So really we're assuming that the difference between the two is zero, so that the mean of diet coke minus the mean of regular coke is zero. And then we're going to get a difference that's actually negative. That's what this is our actual difference. And we want to find what this likelihood is and this will be our p value. So let's find our test statistic, our test statistic which we're going to use the conservative estimate and find that 35 degrees of freedom. So we need to know what this test statistic is for for this specific value. So we're going to take our mean of the two, which is that 20.78479 minus the mean of the other, which is that 0.81682 And then we're going to divide that by the square root of And we're gonna take our standard deviation 439 squared divided by the sample size. And again this standard deviation 0.751 square divided by the sample size. And this gives us a test statistic that is negative 22.9 And this is very, very uh Mhm incorrect. As far as where I place that difference. That difference is like way down here in this distribution. So the likelihood of getting this type of test statistic for that difference, Mhm is approximately zero. So we have strong evidence to reject the null. We have strong evidence to reject the null and claimed that the diet coke does have a lesser weight, lesser weight than regular coke. So we have strong evidence of that. That doesn't matter what significance level you use. But we were supposed to use 5% and then it asks us in part B will give us the appropriate confidence interval. If we're going to use this idea. And if we were doing a confidence interval, we would need to use 5% of the lower tail, 5% of the upper tail. We'd actually need a 90% confidence interval to look at that. Not a 95% because we would want 5% of the air at the left tail, 5% at the other tail. And so on my table, I looked at my table and mind skips from 32 to 40. I don't have 35 degrees of freedom. And so I looked for the confidence interval number. So I would have 5% in the upper tail or you would have down at the bottom, it would say a 90% confidence interval and I have this value comes out to be 1.697 Now your book might be more accurate than mine. So if you can get more accurate you would and so we would take the difference that we got and so we would have the point 78479 minus the 0.81682 plus or minus. And then we would use our T star value for that. Now, if we use software will get a different value and then we still use that same standard deviation of the 0.439 squared over the sample size plus 0.751 squared over the sample size. And I'm going to leave that for you to do the calculation of finding what these two numbers are. But we'll find that it basically does not include zero. So these numbers are going to be different now. Part C. Asked why why would this happen? Well, the diet coke uses artificial sweetener versus the regular cope. He uses sugar and this must be more dense. So it's not that you're getting a lesser volume, you're just getting a lesser mass because of the sugar being dancer. Mhm. Okay.


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