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