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A study that looked at beverage consumption used sample sizes that were much smaller than previous national surveys_ One part of this study compared 20 children who...

Question

A study that looked at beverage consumption used sample sizes that were much smaller than previous national surveys_ One part of this study compared 20 children who were to 10 years old with 5 who were 11 to 13 The younger children consumed an average of 8.5 Oz of sweetened drinks per day while the older ones averaged 14.2 The standard deviations were 10.8 0z and 8.5 0z respectively Use younger children as population 1. (a) Do you think that it is reasonable to assume that these data are Normall

A study that looked at beverage consumption used sample sizes that were much smaller than previous national surveys_ One part of this study compared 20 children who were to 10 years old with 5 who were 11 to 13 The younger children consumed an average of 8.5 Oz of sweetened drinks per day while the older ones averaged 14.2 The standard deviations were 10.8 0z and 8.5 0z respectively Use younger children as population 1. (a) Do you think that it is reasonable to assume that these data are Normally distributed? Explain why or why not: (Hint: Think about the 95-99 rule, The distributions appear to be Normal according to the 68-95 99.- rule We can see that the groups have means and standard deviations_ two characteristics of Normal distributions_ The 68-95 99. rule suggests that the distributions are not Normal. If they were_ we would expect some of the children to drink negative amounts of sweetened drinks_ which does not make sense (b) Using the methods in this section, test the null hypothesis that the two groups of children consume equal amounts of sweetened drinks versus the two-sided alternative. Report all details of the significance- testing procedure with your conclusion: (Round your answers to three decimal places. SED = 1-260 P-value (c) Give 95% confidence interval for the difference in means_ (Round your answers to four decimal places: 215.019 4619



Answers

For data sets with approximately bellshaped distributions, we can improve on the estimates given by Chebychev's rule by using the empirical rule, which is as follows. For any data set having roughly a bell-shaped distribution: Approximately $68 \%$ of the observations lie within one standard deviation to either side of the mean. Approximately $95 \%$ of the observations lie within two standard deviations to either side of the mean. Approximately $99.7 \%$ of the observations lie within three standard deviations to either side of the mean. Concern the empirical rule. Malnutrition and Poverty. R. Reifen et al. studied various nutritional measures of Ethiopian school children and published their findings in the paper "Ethiopian-Born and Native Israeli School Children Have Different Growth Patterns" (Nutrition, Vol. $19,$ pp. $427-431$ ). The study, conducted in Azezo, North West Ethiopia, found that malnutrition is prevalent in primary and secondary school children because of economic poverty. The weights, in kilograms (kg), of 60 randomly selected male Ethiopian-born school children ages $12-15$ years old are presented in increasing order in the following table. $$\begin{array}{lllllllll} \hline 36.3 & 37.7 & 38.0 & 38.8 & 38.9 & 39.0 & 39.3 & 40.9 & 41.1 \\ 41.3 & 41.5 & 41.8 & 42.0 & 42.0 & 42.1 & 42.5 & 42.5 & 42.8 \\ 42.9 & 43.3 & 43.4 & 43.5 & 44.0 & 44.4 & 44.7 & 44.8 & 45.2 \\ 45.2 & 45.2 & 45.4 & 45.5 & 45.7 & 45.9 & 45.9 & 46.2 & 46.3 \\ 46.5 & 46.6 & 46.8 & 47.2 & 47.4 & 47.5 & 47.8 & 47.9 & 48.1 \\ 48.2 & 48.3 & 48.4 & 48.5 & 48.6 & 48.9 & 49.1 & 49.2 & 49.5 \\ 50.9 & 51.4 & 51.8 & 52.8 & 53.8 & 56.6 & & & \\ \hline \end{array}$$ Note: The sample mean and sample standard deviation of these weights are, respectively, $45.30 \mathrm{kg}$ and $4.16 \mathrm{kg}$ a. Use the empirical rule to estimate the percentages of the observations that lie within one, two, and three standard deviations to either side of the mean. b. Use the data to obtain the exact percentages of the observations that lie within one, two, and three standard deviations to either side of the mean. c. Compare your answers in parts (a) and (b). d. A histogram for these weights is shown in Exercise 2.100 on page $76 .$ Based on that histogram, comment on your comparisons in part (c). e. Is it appropriate to use the empirical rule for these data? Explain your answer.

The following is a solution in number 18 which is the goodness of fit test for Poisson distribution. Uh This involves like bacteria in the mouth or something. And the lambda the mean is 2.8 or 2.80. And we're asked to find the probability that zero occur whatever it is, I think it's like bacteria colonies or something. Zero probability that one occurs 234 and then greater than or equal to five. So you can use the formula but I like to use the software so I'm gonna use the C. I A. T. Four. And if you go to second bars which is the distributions we can go to this possum pdf a sense for the probability density function. And then we're just gonna type in the mean which is mu in this case at lambda, they call it lambda for the present. And then the x. value is just what I'm trying to find. So this is the probability of zero given that the mean is 2.8 and that gives us .0608. So .0608. Okay I'll do that one more time and then the other ones I'll just copy down. So then the possum pdf 2.8 and then this time I'm going to make it one And then we press enter and then it's .17 03 Let's say. So .1703. Okay so that's what you're gonna do and then you're going to change it to 234 So I'll just go and copy these down whenever you do that. For the two should get 20.2384 for the three, you should get 30.22 to 5. And then for the probability of four you should get 40.1557 Now for greater than equal to five it's one minus the things that you just found. So what you can do is you can just add 0123 and four together. These these probabilities and you take one minus that and that will give you the probability that it's great and equal to five because we don't have enough time to do probability of five plus probability 6789 all the way up to infinity. You know, we don't have enough time. So Instead it's quicker if we just take the total which is one minus the probability is that we don't want which is zero through four. Now fortunately um there is a function of the calculator that adds it up for you. If you want to use it you don't have to but it's called the person C D f. C D F stands for the cumulative density function. So it's one minus the person C. D. F. So if we go second distribution and you may have seen it already it's the person CDF actually you know what when the second quit because I forgot to do one minus so one minus posen CDF. And remember the mean was 2.8. Now instead of the X. Value being I'm not gonna say 01234 I'm just gonna say four. So the calculator automatically knows that it's zero plus one plus two plus three plus four. The probabilities associated with that And we paste it in there and that gives us .15-3. Okay so let's write that down some point. So equals .15-3. So those are all the probabilities. The second part of the support be. It is finally expected value if the sample size N was 100. Now that's kind of nice because the math is super easy. We're just gonna take and I'll just show it for the first one we take the sample size which is 100 times each associated probability. And that's going to give you the expected value. So really all you gotta do is just move the decimal place over twice to the right and that's gonna give you 6.08 and then 17.03 when we multiply 23.84 22.25 15 57 And then greater than equal to five is 15.23. So those are all the expected values And that's going to lead us to our goodness of fit test. So they observed that was the chart that was given so zero that we observed 12 to have zero. We observed 15 to have one, we observed 29 to have two and so on, and so forth. So I wrote those down now. I got to write down the expectancy. Remember the expected For each of these. 6.08 17.03 23.84 22-5. So these are all the expected values from the previous part, Part B that I can plug in now. Okay, so now I need to find the chi square test statistic and I can go and do this now. The degrees of freedom is always in minus one or the number of categories. So k minus one. So 123456 That means there are five degrees of freedom, So 6 -1. Okay, so let's go back to the calculator Or you can use um you know, again, any software you want or you can certainly use the formula that they give you. But this is That's not right, this is a lot easier. So if you get a stat and edit you see already put those in. So L1 I put the observed And then L. two I put the expected. Now if you have something different or if you plug them in differently that's fine. But I just do L one observed L two is expected to go back to stat and then tests and it's the very close to the very bottom, that's the chi square G. O. F. Test which stands for goodness of fit test. And as you can see there I do L one for observed L. Two for expected. So if you have something different there, if you put observed in L. Three let's say you just change that to L three degrees of freedom. Number was five already said that. And then calculate and that gives us the chi square value of 13 point 138, let's say 13 .138. And that's that. Okay. The last part is we actually finish our test. So let's just look at this p value, that's gonna be the easiest route. So the P values .022. So let's go and write that down. P values 0- two. And we're going to explicitly compare that to the alpha value. In this case it's less than alpha, it's less than 0.5 point two, is less than 20.5 anytime the alpha value or the p values less than alpha. You reject. H not well, we didn't formally write down what the, you know, h not was but it's understood in a goodness of fit test, the H not is that these two distributions are the same, the observed and the expected are about the same. And we're rejecting that. So in this case, because we're saying that it follows a person, a person distribution, we can say this data, I think it's bacteria does not follow a person distribution because we're rejecting that meaning, we're accepting the the alternative hypothesis and the alternative hypothesis would be that this does not follow a Poisson distribution.

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.

So we're looking at people wearing seatbelts and not wearing seatbelts Children and finding out how many days they stay in the I. C. U. And so we're going to assume that those students with a seat belt have equal stay in the hospital as those who do not wear a seatbelt. And alternately that the seat belted Children have a mean stay less than those who were not seat belted. And so we're assuming that the difference between these two seat belt minus not seatbelt is zero and we're actually getting something that's negative and this will be our p. Value. So let's look at the data and let's get our test statistic and we have sample sizes to sample sizes. One sample size was 1 23 and the other was to 90. So that was the for the seat belt and this is for the non seatbelt. And so we're going to use degrees of freedom of 122 to be conservative. And they're relatively large sample sizes anyway. And so let's find our test statistic and we have our 0.83 minus are 1.39 And then we're going to divide that by the standard deviation of the seat belted group squared, divided by the sample size, and then the standard deviation of the non seat belted Children divided by the sample size of it. And when we do that calculation, we get the test statistic comes out to be negative 2.330 And so now we want to find, and that's so that's what this value is. We want to find the likelihood, if the difference is actually zero or the means are equal, how likely is this number or more extreme to come up? And so I'm going to use my uh my T c D E f to find this mighty CDF. And I'm going to use negative uh minority has negative one times 10 to the 99th hour in it. And then our upper value is going to be that negative 2.330 My degrees of freedom is again the 1 22. And I will paste and let that do the calculation. And I have a value of a p value of about 1%. And that is less than my 5% significance level. So I definitely have sufficient evidence to reject the null meaning that we believe that the Children that had seat belts on have a smaller uh stay in the uh in ICU than uh those that are non seatbelt. And again you write my sons for that. So now we want to find the confidence interval and the appropriate confidence intervals. Since this was a one tailed test and we had 5% in this tale, the confidence interval has 5% down here, 5% up there. We would need to look at a 90% confidence interval. And so my table does not have a T. V. Star value for 122 degrees of freedom. So I'm going to use my inverse T. Value my inverse teeth. And on the inverse T. I want the area below it to the point oh five. So I'm going to put in the 0.5. It will be finding basically this low limit which will be symmetrical. What the upper one and then our degrees of freedom are that 1 22? And we get that value to be Yeah. Uh And this was 1 22 comes out to be negative 1.657 And so now let's get that confidence center, let's look at the difference first. So we need this difference the 0.83 minus 1.39 And that difference comes out to be negative 0.56 plus or minus. And we have that T star value 1.657 because we have one down here and went up here and then we have this same uh same standard deviation we had here. So let's see if I get 1.77 squared and that is a 3.6 squared mm And our sample sizes are 1 23 and 2 90. Okay, And let's get that margin of air, Yeah. Mhm And 1.657 times square root of 1.77 squared divided by 1 23 plus 3.6 square divided by 2 90. And that margin of air comes out to be 0.398 and two. And it keeps going on but I'm just going to store that is X. And so let's get those two values. We have the negative 20.56 minus the X. Value. And that comes out to be negative 0.958 And then we can change that into an addition sign. And we find out that that is still negative and notice that that does not include zero. So we would again from this interval see that the difference does not appear to be zero as he appears to be negative. So it does appear as though um that the seat belts make a difference seatbelts beautiful. They're good that they appear to have a lesser chance of a lesser stay in the hospital in the ICU for Children.

Solution number 17. And this is a, an interesting problem with the Poisson distribution. And the Poisson distribution is the type of discrete distribution where the events are random and rare. And uh, this, in this case it's a traffic accidents, Daily traffic accidents, and there is an average, that's what we use this lambda for. So λ,, which is the average of the mean is 1.72 accidents per day. Were asked to find the probability that zero accidents occur and the probability that one occurs To occur three occurs and then greater than four. Curse. Now you can use the formula, but again, I like to use the uh, software. So what I'm gonna do is I'm gonna go to second distribution and then I'm gonna go down to the present pdf and then here it asks from you, or sometimes it lasts for lambda, and that's 1.72 And then the X value, I'm just going to find the probability that zero occurs, And that gives me .1791. I'm going to go in round here some point 1791. So you can do this with any type of software or you can just use the formula. Although the formula can take awhile .1791. And I'm gonna do it one more time. Just show you whether they're the second bars for distribution. And then I went to the Plaza pdf, right? That's the probability density function. So pdf. And the mu the mean is that's the land of 1.72. And this time we're gonna find the distribution or the probability that one occurs and it's about Point Let's say .308 point 308. And that's what you're gonna do for, you know, basically all the rest of them. Until you get to the greater than so zero point 2649, I'll go ahead and give you these answers here and then 0.15 one night. So then uh to get something that's greater than what you're gonna do is you're gonna take one go all the way up to infinity and save some time. We're just going to take one minus the four that we've already found. 0123 So there's another function in the calculator we can use is called the plaza CDF. The cumulative density function. And we're going to go up to three. So the CDF calculates the probability of zero plus the probability one plus probability two plus probability of three whenever you do CDF of three, so one minus that. Or you could just do one minus these four numbers here whichever you like. So one minus. And then 2nd distribution. And I'm going to go to the present CDF. And the main remember was 172 and then the X value. Now I'm not gonna put 0123 I'm just gonna put the three And it automatically calculates 0 1, 2 and three combined. And whenever you do that that should be your answer. So .0962 0962. You might get 61 if you just use these numbers here but you get the same thing if you do one minus and these all added together. Next up we find the expected value and any time you find the expected value just take the sample size times the probabilities. So for zero, remember there were 90 days that we looked at. So 90 times the probability of zero was remember that .1791? That should give you 16.119. And then all of these are gonna be the same 90 times something. And those some things are the probabilities. So .308 that's going to give you 27.72. So that means we can expect about 27.72 days where there are there's one wreck And then 90 times the point 2649. That should give us 23 841 And then 90 times .1519. He was this 13671. And the 90 times 0.0 962 Gives you 8658. Okay, so those are the probabilities. Those are the I'm sorry those are the expected values. And then the part c we find we're gonna find our use the good as fit test. So they observed that was the chart that was given and the expected we actually just found. So we're just gonna copy those expected values down 27.72 23-841 13.671 and then eight 658. So those are the expected values where we take the probabilities times the sample sizes. Okay? So now we can go back to our calculator and do the goodness of fit test. So if you go to stat and then edit you can see that here in L. One. I put the observed values and then L. Two I put the expected values. And then if you go to stat tests it's the chi square goodness of fit test. And the observed is L. One the expected sell to or you can use the formula or any other software. And degrees of freedom was four. So the degrees of freedom, that's actually another answer. The degrees of freedom of four. Since there are five categories there of 0123 and then greater than equal to four. So we can calculate that And that gives us a chi square value of about 12 509 So chi square equals 12.509 And then it also asked for degrees of freedom. The degrees of freedom, like I said, was equal to four because it's five minus one. Okay, so the p value let's look back at the P value, it's about .01. Let's go at some .014 and that is greater than the alpha, barely, but it is greater than the alpha. So whenever the p value is greater than the alpha than we fail to reject to reject H not. Which means in this case, actually, in all these goodness of fit test cases, the null hypothesis is that the distribution does in fact follow whatever we're talking about in this case, we're failing to reject that. So this follows a person distribution so the traffic accidents, and per day in this particular area, it does in fact follow a Poisson distribution.


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