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A research study was conducted t0 examine the differences between older and younger adults on perceived life satisfaction pilot study was conducted to examine this ...

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A research study was conducted t0 examine the differences between older and younger adults on perceived life satisfaction pilot study was conducted to examine this hypothesis. Ten older adults (over the age of 70) and ten vounger adults (between 20 and 30) were give life satisfaction test (known to have high reliability and validity): Scores on the measure range from 0 to 60 with high scores indicative of high ' life satisfaction; low scores indicative of low life satisfaction. The data are

A research study was conducted t0 examine the differences between older and younger adults on perceived life satisfaction pilot study was conducted to examine this hypothesis. Ten older adults (over the age of 70) and ten vounger adults (between 20 and 30) were give life satisfaction test (known to have high reliability and validity): Scores on the measure range from 0 to 60 with high scores indicative of high ' life satisfaction; low scores indicative of low life satisfaction. The data are presented below: OldecAdulti Yquger Adults 24 22 35 24 What would be the hypotheses in this problem? Run an appropriate t-test and draw the conclusions Calculate the confidence interval for the difference in life satisfaction between older and younger adults



Answers

Taking It Easy In a Gallup poll conducted December $11-14,2003,455$ of 1011 randomly selected adults aged 18 and older said they had too little time for relaxing or doing nothing. (a) Verify that the requirements for constructing a confidence interval about $\hat{p}$ are satisfied. (b) Construct a $92 \%$ confidence interval for the proportion of adults aged 18 and older who say they have too little time for relaxing or doing nothing. Interpret this interval. (c) Construct a $96 \%$ confidence interval for the proportion of adults aged 18 and older who say they have too little time for relaxing or doing nothing. Interpret this interval. (d) What is the effect of increasing the level of confidence on the width of the interval?

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.

Once again, welcome to a new problem. This time we're dealing with proportions. And when you think about proportions, you could have a population and the population would have a population proportion population proportion and the population proportion would be something like the proportion of females in college campus, the proportion of females in a college campus and the population proportion is based on capital P. Um as opposed to the sample proportion, which is jihad, so p heart is X over and this is the sample proportion where you take the uh number of subjects of interest divided by the sample size. And for the most part, there's always a likelihood that you want to do an estimation of the population proportion based on the simple proportion and the estimation can have a margin of error and a confidence level. So for example, if you're 95% confidence Uh then it means your error is 5%. Or sometimes we call that alpha is 5%. And from the distribution point of view, what you're saying is that this middle part is .95 improbability. And then you split the tails, you get .025 on the left tail and then .025 on the right tail. And so the formula for confidence interval uh is the same as the hot plus minus margin of error. Where the margin of error is the same as the alpha over to Radical P Hut one -P had all of the end Members the alpha of the two For 95% confidence. From the tables would be 1.96. So that's the Z score. These twosies course, this would be the negative Z score and then this would be the positive Z score. That's why we have plus and minus over there. Uh So coming back to the problem, coming back to the problem, we have a situation where we're thinking about the report. So given, given the 2003 statistical abstract of the U. S. We see that uh the percentage of mm hmm smokers above 18 and what we're doing and this statistical abstract is too Uh huh. A design study I designed study collecting data own smokers and non smokers on smokers and nonsmokers. Mhm. And there's an estimate, there's an estimate of the proportion who smoke uh early there's an estimate of studies. So in part a. Mhm. Are determined the sample size becoming the sample size to estimate the proportion of smokers and the population uh with 95% confidence level And uh .02 error margin of error. So that's what a And then B mm hmm. Using using sample size requirements, whole suggestions if you want to call it that from part a mm. The study finds 5 20 smokers. Mhm. What is the points estimate of proportion of smokers? Mhm. In the population and the population. And then but see uh determined The 95 confidence interval for smokers in the population, 95% confidence interval of smokers uh in the population. So I want to jump right into it. We have the numbers to work with this problem. So we'll start by saying, Well this was actually 30%. So just remember that Start by saying the P hard in part AP heart is 30%, which is the same as .30. The margin of error is point to remember if you have the hard plus minus margin of error at the same 0.0.30 plus minus point or two. Uh and then your confidence level is 95%. And that means that if you're looking at a distribution, this middle part is .95 And then we split the two sides. This other side is .25 and this other side is also .25. And so he is the alpha over two is the same as 1.96. So we go back and say the margin of error is the alpha over too radical P hut one minus p hurt Oliver in, I want to solve for end. So we divide both sides by Uh zero off over to these two cancel out and we can switch it. So we have radical P heart one minus P hard all over in this is E of zero off over to the square both sides of the equation. And we end up having p hard one minus P hut Oliver in um This is going to give us E squared of a Z off over two squad. And then we multiply both sides by N. And also multiply both sides by the reciprocal of E squared ZR for over two on both sides. That way we can cancel out the uh we can cancel out the sample size is on this side. Then you can cancel out these two on this side. So now we have an appropriate formula for the sample size. So we have ZF over to uh, squirt and then we have a jihad one minus P hurt all of E squared. So if you go back to part A, you're going to see uh want to determine the sample size to estimate the proportion of the population, given that confidence level. So given that confidence level. So, uh, assuming, assuming p heart is 0.45. So we're doing the estimates and assuming Prd's uh huh 0.45. What's going to happen is that I'm going to get And being the same as 0/2 of the two squared and this part Becomes .025. And the reason why it's .25 is because if this is .5 And this fight is also .5, You get the product and you get .025. And then you want to divide that by obviously E squared. The next step would be to plug in the specific numbers 1.96 sq Times .025. And then we want to divide that by e squared, which is a point or two squared. And it's going to give us a sample size of approximately uh 2000 and 17. So that's the first part of the problem. In part a uh in the second part will give an X to be 520 and remember the sample size is to 17. So if you go back to the second question is using that sample size that we just got approximately 2017. The study finds 520 smokers. So what's the point estimate of the proportion? So what's the pr come back here? We know that he had is the same as uh X over. N Soapy hot is going to be the same as X being 5 20 And being 2017. So we get the p had to be equivalent to mm hmm, Approximately 0.25 78. Then the last part of the problem, this is Patsy. Uh we want to find the confidence interval and the confidence interval is p hard plus minus margin of error. already have the P had .2578 and then plus minus the margin of error which was mhm zero for two. So the margin of error. Was he off over to radical uh points. We actually want to displace it here. Zf over to body called p hut one minus p hut all over in. So we're gonna place it here Z alpha is 1.96 and then Radical P Hot is the same as 0.2578 And then 1 -1 0.25 78 11 is still going to 578. And then we want to divide that by 2017 and you can see that the confidence Interval and this is at 95%. If you simplify that On the lower side, it's going to give you 0.2387 and then on the Apple side is going to give you 0.2769. Hope you enjoyed the problem. Feel free to send any questions or comments and have a wonderful thing.

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

The following is a solution in over 14 goodness of fit test where we look at the distribution of customer ages in a snoop report and see if it gives it agrees with the report from the sample. So the observed data values here uh were given to you in the book. So 88 1 35 50 to 40 76 1 28. I went ahead and just copy those down. And then for the expected, you gotta take the percent Times the sample size to get the actual expected value. So 12% times 519, for example, is 62.28. So I did that for all six categories there for the different age ranges. So now we're ready to go ahead and To answer these questions. So what's the significance level that they give? They give you that in the problem? It's a 1% significance level. So alpha's .01. And then we also need to write down what are null and alternative hypotheses are now typically for goodness of fit, test the Noles that these two distributions are the same. And then the alternative is that they're not the same. So in this particular case, bringing it back to this example, we're going to say the distribution of customer ages, uh and the snoop report, I'm not sure what a stupid port is, agrees with the sample report. Right? And then the alternative is that they don't agree. So, the distribution of customer ages and the stoop report does not agree with those of the sample report. So now we need to find the chi square value, but before we do that, we need to make sure that the expected values are greater than five and they are. So the expected values already found them, and you just take the percent times the sample size and they're all greater than five. So that's one of those conditions for inference they need to be at least five and they are so we can use the chi square distribution because of that with five degrees of freedom. Now why is it that five degrees of freedom? Well because there are six categories 1234566 categories minus one is five. That's always gonna give you the degrees of freedom. So now we can find basically everything else with software. So I'm gonna use the T. IT. for like using the T. 84 for smaller data sets. Um But you can use Excel or are or SAs or mini tab whichever you prefer. Or you can certainly use the formula. It might take you a little longer but you can use the formula. So I went ahead and typed these in already but I put an L1 I put the observed data values so you can see him there and then an L. Two. I put the expected data values you can see him there. So what I like to do sometimes is look at the difference here and these look to be kind of far apart. At least some of them do. So I don't know. I think it's been close but I don't think this is going to be a very good fit. Well let's see. So if we're gonna stat and then tests and it's one of the last ones I went up first. It's the D option if you're using A T. I. T. For. And it's the chi square geo F. Test that stands for goodness of fit test. L one is your observed, assuming you put it in. L one. L two is the expected degrees of freedom. Was five. Remember? And then we calculate and that's gonna give us the two values that we need to. The chi square value. Is this first number right here? 15.651 15 651 And then the P. Value, I don't know if you saw it there but we'll go in round. It's about .008 0.8 So just what I thought it's less than alpha. So anytime the P values less than alpha we this is the beauty of the P value. Matthew explicitly compared the P value with the alpha value. And if it's less than alpha then you always reject the null hypothesis. So we're going to reject h. Not so we're rejecting the claim that the distribution of customer ages and the snoop report agrees with the sample report. So that's how we conclude that will go and say there is. Oops. Well, I didn't mean to do that. Yeah, I can just type it in I guess. Mm. Sorry. Okay. I'll just I'll just actually write this out so there is okay, sufficient evidence to suggest the distribution of customer ages does not agree with that of the sample report. Okay. So these two distributions are not the same.


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