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A market researcher is doing a study about tar contents incigarettes. A random sample of 21 filtered king-size cigarettesresulted in a sample mean tar content of 13...

Question

A market researcher is doing a study about tar contents incigarettes. A random sample of 21 filtered king-size cigarettesresulted in a sample mean tar content of 13. 3 mg with acorresponding standard deviation of 3. 7 mg. Another independentrandom sample of 8 nonfiltered king-size cigarettes resulted in asample mean tar content of 24 mg with a corresponding standarddeviation of 1. 7 mg. Test whether the data provides enoughevidence to conclude that the true mean tar content of nonfilteredcigaret

A market researcher is doing a study about tar contents in cigarettes. A random sample of 21 filtered king-size cigarettes resulted in a sample mean tar content of 13. 3 mg with a corresponding standard deviation of 3. 7 mg. Another independent random sample of 8 nonfiltered king-size cigarettes resulted in a sample mean tar content of 24 mg with a corresponding standard deviation of 1. 7 mg. Test whether the data provides enough evidence to conclude that the true mean tar content of nonfiltered cigarettes is significantly higher than the mean tar content of filtered cigarettes. Set up the null and alternative hypothesis. Estimate the p-value of the test and give your conclusion using a significance level of 0. 01.



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. In one study of smokers who tried to quit smoking with nicotine patch therapy, 39 were smoking one year after the treatment and 32 were not smoking one year after the treatment (based on data from "High-Dose Nicotine Patch Therapy," by Dale et al., Journal of the American Medical Association, Vol. 274, No. 17). Use a 0.05 significance level to test the claim that among smokers who try to quit with nicotine patch therapy, the majority are smoking one year after the treatment. Do these results suggest that the nicotine patch therapy is not effective?

So we're given data on carbon monoxide diffusing capacity and we were provided data and there happened to be 20 pieces of data and I've taken those 20 pieces of data and I've placed them into my calculator. So the first thing we want to do is we want to calculate the average of those and the standard deviation. So I'm going to hit stat. I'm going to move over to the calculate menu one variable statistics of all of the data. We've placed enlist one and you will see that our average and we're gonna call our average y bar is equal to 89 85475 And our sample standard deviation is 14 9035 251 And what we want to do here is we want to check or test whether the data indicates that the mean dl reading for current smokers is lower than 100. So we're trying to test whether the mean is less than 100. So if we're going to test that, we have to write a null hypothesis and an alternative hypothesis and your alternative hypothesis is a statement of inequality. So therefore our mu is less than 100 would be the alternative hypothesis. Now the null hypothesis is always a statement of equality. So our statement of equality will be that mu is equivalent to 100. So based on the way we have written are null and alternative hypothesis, we can determine the nature of the test. And since our alternative hypothesis was a less than symbol, this is going to be a left tailed test. So we want to test this information using a significance level of 0.1 so alpha is 0.1 And what that is saying is we're putting a 0.1 in the left tail. So we are going to use a T. Test on this. So in order to use the T. Test on this, we need to find our T. Critical value that is going to separate the two parts of our curve. When we are on this side of the boundary, we are in our reject region and when we're on the other side we are in the fail to reject region. So we need to find that critical value. Now you do have a chart in the back of your textbook and that chart is going to talk in terms of the area in the right tail. So we could see a 0.1 associated with a T value. And it also refers to the degrees of freedom. And we find our degrees of freedom by taking the sample size and lowering it by one. So our degrees of freedom is going to be 19. So we're going to look in the row of that T distribution chart where were referring to degrees of freedom of 19. And we are going to look under the area in the right tail and we're going to get a 2.539 Now, since our T distribution is bell shaped and symmetric and zero would be right where the peak of the bell is, our critical value is going to be the negative of what we just found in the chart. So our T value are critical value is going to be a negative 2.539 So the next thing you want to do is you want to find your standardized test statistic and to find your standardized test statistic, we're going to use the formula T equals why bar minus mu divided by S over the square root event. And ry bar was 89.85475 Mu. Based on our null hypothesis was 100. S. Was 14.9035251 And our sample size was 20. And when you put that into your calculator you are going to get a T. Value of negative 3.44 So if you go back to the bell shaped curve and think of the base of it as being a number line and zero is in the center and are critical value. T. Was negative 2.539 Then the standardized test statistic we just found would be back here at negative 3.44 which means we are in the reject region. So therefore our decision is going to be to reject the no hypothesis. So when the question said, does this data indicate that the mean dl reading for current smokers is lower than 100. Since we are rejecting that null hypothesis, we could say yes, there is evidence to support the claim that the mean D. L. Reading for current smokers is lower than 100. So that was part A. We now want to look at Part B and part B. Is talking about the P value and we want to bound the P value. So part B. Let's talk about what the P value is. Your P value is the smallest value of alpha, which is that level of significance for which the null hypothesis can be rejected. And in our instance, R p value is going to be the probability that R. T. Score is less than a negative 3.0 44 So what you're going to need to do for this one is we are going to look at the chart again, the T distribution chart in the back of your textbook and we're not going to only focus on one number. We're going to focus on the entire row where the degrees of freedom is 19. And with that entire row we have a T. Score for an area of 0.100 We have a T. Score with associated with the area of 0.50 We have one associated with 10.25 We have one associated with 10.10 and we have one associated with 0.5 And the T values that go along with these would be negative 13 to 8, 1.7 to 9, 2.93 2539 and 2.861 Now, because we said that this was a left tailed test and our critical value was a negative value. What we're going to do here is we're going to turn all of these into negatives because the t distribution is symmetric. So if the area in the right tail is 0.100 and that's associated with the T. Value of 1.3 to 8. If that area is in the left tail it would be associated with a negative 1.3 to 8. And our T score that we came up with was a negative 3.44 which would end up falling right here. So what we're basically saying is that it's somewhere between a T. Value associated with 8.5 and a T. Value associated with a zero which would be a negative infinity. So we could then say that our T. Score which was equivalent to a negative 3.0 for four, was less than a negative 2.861 but yet it was greater than a negative infinity. So therefore our P value would be bound by an area of 0.5 and a zero. And then the final part for part C. We want to use the app to find the exact P value. So there isn't an app associated with your text. And when you look at that app you're going to see a bell shaped curve at the top of it, you're going to see a place for you to fill in the degrees of freedom. And at the bottom you're going to fill in a place for X. And a place for the probability. And we know that our degrees of freedom in this particular problem is 19 and we know that our T. Score is a negative 3.44 But what we're going to do is we're going to talk about that being on the right side, which would then give us an area which is your answer to this. The area would be a point 003 Now, another approach we could have done for this particular one is if you did not have the app, you could always use the T score, cumulative density function in a calculator, and that would require you to provide the lower boundary of your shaded area, the upper boundary of your shaded area and your degrees of freedom. So, since our particular problem was a left tailed test and our T value is negative 3.44 We are referring to the left tail here which would go back to a negative infinity. So when we use our calculator, since we can't put negative infinity and we'll just put a super small number in. So we're gonna say negative one times 10 to the 99th. The upper end of that shaded area is negative 3.44 And our degrees of freedom was 19. So let me show you where you can find that function as well. So I'm gonna bring in my calculator and you would hit the second button and the bears button to access the distributions menu and it's number six in this menu. So are lower boundary was negative one times 10 to the 99th Power. Our upper value was our tea score are standardized, test T score and our degrees of freedom was 19 and we get a similar um P value of 0.33 So whether you use the app or you use the graphing calculator, you are getting the same p value of 0.3

So we're going to assume that for these non smokers, that those people who are exposed to second hand smoke, I have going to have equal lovers of Kolkata Ning, uh uh as those who are not exposed to secondhand smoke, alternately, we think that that exposed group is going to have a higher level. And We know that we need to find our test statistic. And since our sample size is 40, for each, will use the conservative measure. and we have the difference between our two groups, which we look at these numbers and they look different. However, we look at the samples standard deviations, we find they are very large, so 138.08 sq divided by the sample size, And then the 62.53 squared Divided by the sample size. And when we do that calculation, we find that that test statistic comes out to be 1.8455. And so we're assuming that the difference between the two groups Is equal to zero. But we're getting a test statistic that's up here at like 1.8455. And again, this is the test statistic. The actual difference is like 40 something, but we want to find this which will be our p value. So what is the likelihood if these two are equal or their differences? 0? What's the likelihood of getting a test statistic that is greater than or equal to this value. And I used my T C D E. F. To find that and I got a p value of 0.36 and that is definitely less than 5%. So a five significance level we would have evidence to reject to reject the null and claim that the mean of the experimental does seem to be higher is higher than that uh group of non I said, experimental. I mean, the exposed group than the non experimental group that they don't end up having exposure. Now let's find a confidence interval, the appropriate confidence interval. Because we're using five significance for a one tailed test, five down here for a confidence interval. We would also want five in the higher level. So we really want to find a 90 confidence interval. And again, I'm going to use this degrees of freedom of 39. Now, my table does not have that. I was using, it has degrees of freedom for 30 and 40. And probably we could very safely use this one for 40 if the degrees of freedom is 39. But I'm actually going to look up and use my inverse nor inverse T Button. So if I go to inverse T and go to second and distribution and I go to inverse T And I'm going to plug in an area of 25. Let me type the Button in the area .5. And then I'm going to type in my degrees of freedom of 39 and then find out what that gives me. It tells me that that lower T value is For 39° of freedom is negative 1.68. And we would round that to five. And I believe if you look up this when it's 1.684, so notice that these are very very close. So when we find that uh that difference and let's just find what this difference is. This difference if we subtract eight and then five becomes a three A three and the five minus two, Three becomes a two. And then these two have a difference of 44. So that difference between those is 44.23 plus or minus. And then we put our test statistic, R r Rt star value here and then we'll use that times the square root of and we have that first standard deviation was again very large squared over 40 Plus the 62.53 squared over 40. And let's find this margin of air first. Okay. And so I have that 1.685 times the square root of now enter this in 1 38.8 squared divided by 40 Plus that 62.53 squared divided by that 40. And I get that margin of air to be 40 0.38 And so let's find the two values and I'm going to just store that value as X. And so I have 44.23- the x value. And that gives me 3.85. And then I can go back and second entry and just change that subtraction sign to an addition sign And I get 84.61. And so we're 90 confident that the actual confident that the actual difference between those two is somewhere in here and notice it does not it doesn't include zero. Which means we definitely do not think that they're equal, means we think they are different. So in part C. It says, can we conclude we definitely look at the group that had exposed proof that they definitely have seemed to have a higher level of that uh that narcotic or the drug the nicotine offset of them, the non exposed group.

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

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


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