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Consider drug testing company that provides test formarijuana usage. Among 280 tested subjects, results from 30 subjects were wrong either false positive or false n...

Question

Consider drug testing company that provides test formarijuana usage. Among 280 tested subjects, results from 30 subjects were wrong either false positive or false negative) Use 0.10 significance evel to test the claim that less than 10 percent of the test results are wrong:Identify the test statistic for this hypothesis testThe test statistic for this hypothesis test is (Round to two decima places a3 needed:)Identify the P-value for this hypothesis testThe P-value for this hypothesis test is (R

Consider drug testing company that provides test formarijuana usage. Among 280 tested subjects, results from 30 subjects were wrong either false positive or false negative) Use 0.10 significance evel to test the claim that less than 10 percent of the test results are wrong: Identify the test statistic for this hypothesis test The test statistic for this hypothesis test is (Round to two decima places a3 needed:) Identify the P-value for this hypothesis test The P-value for this hypothesis test is (Round to three decima places a3 needed ) Identify the conclusion for this hypothesis test OA. Fail to reject Ho There is sufficient evidence to warrant support of the claim that less than 10 percent of the test results are wrong: Reject Ho There is not sufficient evidence to warrant support of the claim that less than 10 percent of the test results are wrong: Fail to reject Ho There is not sufficient evidence to warrant support of the claim that less than 10 percent of the test results are wrong:



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. Use the P-value method unless your instructor specifies otherwise. Use the normal distribution as an approximation to the binomial distribution, as described in Part 1 of this section. The company Drug Test Success provides a " 1-Panel-THC" test for marijuana usage. Among 300 tested subjects, results from 27 subjects were wrong (either a false positive or a false negative). Use a 0.05 significance level to test the claim that less than $10 \%$ of the test results are wrong. Does the test appear to be good for most purposes?

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

In this problem, we're going to be considering the outcomes off a clinical trial where subjects were given two different types of treatment. The first sample was given OxyContin on from the first sample. We have 52 subjects developing museum out off the 227 subjects. For those who received the placebo, we have five who developed nausea out off 45. So in this test, we're going to use toe approaches. They have policies, method on the confidence interval method. For the hypothesis method, we are going to be considering the critic the critical value for the 0.5 level of significance. And since we're only testing for a difference between the rates of nausea, then we will have a two tails test where the conflict where they now hypotheses this p one is not equal to be to, and the alternative before this is his P one is not equal to P two. So the critical value corresponds to this. Um, test is plus or minus 1.96 So now we can work out the test statistic using the formula given for that. And when we walk it out, completed one use. That is 1.7 seven or five. Now we can compare the step test statistic and the critical region. So here we have the critical region. She did negative 1.96 and positive one 0.96 So we notice that the test statistic 1.1775 It's less than the critical value positive 1.96 that leads us to making the decision to fail. To reject been now hypothesis by failing to reject another. Thank goodness is we conclude that there is not sufficient evidence to support the clean, that there is a difference between the rates of nausea for those treated with oxytocin on those given a placebo. Now we proceed to the confidence interval method and we need to work out the margin of error e by substituting the values in the formula and when we do that correctly, the E should be zero point 1068 and the critical interval. The confidence interval limits are zero point zero 112 and 0.2 248 So when you look at these numbers, you notice that confidence intervals do not include zero confidence interval in it do not contain zero, so it appears that the two proportions are not equal simple. According today, our confidence interval method, the to of the two proportions are not equal, which is different from what we have obtained in the previous approach using the hypothesis test. So now we can move on to the next month of the question where we're supposed to tell whether nausea appears to be an adverse reaction resulting from OxyContin. And when we compare the proportions off, those who developed knows it. We have 22 point 9% from those who received OxyContin, and for those who did not receive OxyContin, we have 11 0.1%. So based on the hypothesis test, and it appears like there is no difference between the two proportions, so there's no significant difference between the two proportions.

Yeah. Has a percentage of 18-25 year olds using marijuana changed since the publication stating it was 13.6%. Well, our first thing to do is look at the randomly selected people done recently, of the 1283 randomly selected 18-25 year olds, 205 of them said that they currently use marijuana, which is about 16%. Now, in terms of our criteria to see if we can use the one property test are we have to have a simple random sample which we have. And the sample size is so big. That's gonna when we multiply the sample size by our hypothesized population proportion, it's definitely going to be bigger than five. So we are going to go on. Those Are null hypothesis is that our population proportion is 13.6%. And then our alternative hypothesis is going to be the population proportion is not 13.6. See if there has been a change from that last publication. So for our Z score, we're going to have our sample proportion minus the hypothesized population proportion. And I'm plugging in my values here for P zero and for N. And when you run this calculation, you get a Z score of 2.51. And when we looked at him up in the table in the back of the book is the score of 2.51 corresponds to .994. But We're going to subtract that from one. This will give us our high end probability, which is About .006 And then we want not equal to so we actually have to double this because this is a two tailed test, which is going to be .012. Now we are looking at a 5% significance level. This .012 is definitely smaller than .05. So we are going to reject H. O. Which tells us essentially, we have enough evidence to claim. The population proportion of 18-25 year olds using marijuana has changed From the 13.6%. It is not that 13.6% anymore.

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.


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