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For Questions 1-10, please identify the test statistic thatshould be conducted, I am double checking my work, please have thecorrect test statistic listed. 1. A res...

Question

For Questions 1-10, please identify the test statistic thatshould be conducted, I am double checking my work, please have thecorrect test statistic listed. 1. A researcher wishes to see if the average weights of newbornmale infants are different than the average weights of newbornfemale infants. The researcher recruits 10 male infants and 10female infants (via parental consent) to conduct her study.2. Two types of antibiotics (amoxicillin and penicillin) weretested to see how long (in minutes) e

For Questions 1-10, please identify the test statistic that should be conducted, I am double checking my work, please have the correct test statistic listed. 1. A researcher wishes to see if the average weights of newborn male infants are different than the average weights of newborn female infants. The researcher recruits 10 male infants and 10 female infants (via parental consent) to conduct her study. 2. Two types of antibiotics (amoxicillin and penicillin) were tested to see how long (in minutes) each remained in the bloodstream of patients with an infectious disease. The researcher also wanted to know if gender had an effect on how long the antibiotics remain in the bloodstream. 3. The per-patient costs (in thousands of dollars) for health care were collected in three different areas of Des Moines. Is there a difference in the means of the costs between the three areas? 4. A researcher wishes to see if there is a difference in the cholesterol levels of United States adults over 40 years of age and adults under 40 years of age. The researchers collects a sample of adults over 40 years of age and a sample of adults under 40 years of age to conduct his study. 5. A researcher conducted a study of three different diets and two different exercise programs. Seven randomly selected subjects were assigned to each condition for one month. The researcher wants to know if diet and exercise program have an effect on the amount of weight lost and if the interaction between diet and exercise program has an effect on weight lost. 6. The Medical Rehabilitation Education Foundation reports that the average cost of rehabilitation for stroke victims is $24,672. A researcher recruits a random sample of 35 stroke victims in Iowa who report an average cost of $35, 234 and a standard deviation of $6,349. Can it be concluded that the average cost of stroke rehabilitation in Iowa is different than that of the population? 7. The average 1-year old (both genders) is 29 inches tall with a standard deviation of 3 inches. A random sample of thirty 1-year olds in a large day care franchise resulted in a mean height of 27 inches. Can it be concluded that the average height of 1-years olds in the given sample differs from that of the population? 8. A random sample of 26 speech-language pathologists from Iowa reported an average salary of $45,350 with a standard deviation of $1,659, while a random sample of 26 occupational therapists from Iowa has an average salary of $50,215 with a standard deviation of $4,116. Is there a significant difference between the two groups’ salaries? 9. The number of protein grams per serving of four different kinds of meat (i.e., chicken, fish, beef, and pork) are collected. The researcher wants to know if there are differences between the kinds of meat for grams of protein per serving. 10. A researcher wishes to determine if an educational intervention will reduce the number of fat grams consumed among overweight children. To answer her question, the researcher recruits 10 participants and utilizes pre-post measurements surrounding her intervention phase.



Answers

This is a technique to break down the variation of a random variable into useful components (called stratum) in order to decrease experimental variation and increase accuracy of results. It has been found that a more accurate estimate of population mean $\mu$ can often be obtained by taking measurements from naturally occurring subpopulations and combining the results using weighted averages. For example, suppose an accurate estimate of the mean weight of sixth grade students is desired for a large school system. Suppose (for cost reasons) we can only take a random sample of $m=100$ students, Instead of taking a simple random sample of 100 students from the entire population of all sixth grade students, we use stratified sampling as follows. The school system under study consists three large schools. School A has $N_{1}=310$ sixth-grade students, School B has $N_{2}=420$ sixth-grade students, and School $C$ has $N_{3}=516$ sixth-grade students. This is a total population of 1246 sixth-grade students in our study and we have strata consisting of the 3 schools. A preliminary study in each school with relatively small sample size has given estimates for the sample standard deviation s of sixth-grade student weights in each school. These are shown in the following table: School A $N_{1}=310$ $s_{1}=3 \mathrm{lb}$ School B $N_{2}=420$ $s_{2}=12 \mathrm{lb}$ School C $N_{3}=516$ $8_{3}=6 \mathrm{lb}$ How many students should we randomly choose from each school for a best estimate $\mu$ for the population mean weight? A lot of mathematics goes into the answer. Fortunately. Bill Williams of Bell Laboratories wrote a book called $A$ Sampler on Sampling (John Wiley and Sons, publisher). which provides an answer. Let $n_{1}$ be the number of students randomly chosen from School A. $n_{2}$ be the number chosen from School $\mathrm{B}$, and $n_{3}$ be the number chosen from School $\mathrm{C}$. This means our total sample size will be $m=n_{1}+n_{2}+n_{3} .$ What is the formula for $n_{i} ?$ A popular and widely used technique is the following. $$n_{i}=\left[\frac{N_{i} s_{i}}{N_{1} s_{1}+N_{2} s_{2}+N_{3} s_{3}}\right] m$$ The $n_{i}$ are usually not whole numbers, so we need to round to the nearest whole number. This formula allocates more students to schools that have a larger population of sixth graders and/or have larger sample standard deviations. Remember, this is a popular and widely used technique for stratified sampling. It is not an absolute rule. There are other methods of stratified sampling also in use. In general practice, according to Bill Williams, the use of naturally occurring strata seems to reduce overall variability in measurements by about $20 \%$ compared to simple random samples taken from the entire (unstratified) population. Now suppose you have taken a random sample size $n_{i}$ from each appropriate school and you got a sample mean weight $\bar{x}_{i}$ from each school. How do you get the best estimate for population mean weight $\mu$ of the all 1246 students? The answer is that we use a weighted average. \begin{equation}\mu \approx \frac{n_{1}}{m} \overline{x_{1}}+\frac{n_{2}}{m} \overline{x_{2}}+\frac{n_{3}}{m} \overline{x_{3}}\end{equation}

In this problem, what we wanna do is sample 100 students from several schools in a school system to estimate the mean weight of sixth graders in that school system. So one way that we might do this is by dividing the school system up into strata which are not overlapping groups. So a reasonable way to do this might be dividing them up by school. So what we're given is some information about how many individuals, or how many sixth grade students are at each school as well as a sample standard deviation from some previous data. So the question might be, if we are limited to only choosing 100 students, how do we go about sampling or how many do we know to sample? How do we know how many to sample from each of these three schools to have everyone proportionally represented? So, if we have information about the sample standard deviations, one way that we might do that to calculate the sample size for each school is by multiplying that school or that stratagems number of individuals times the standard deviation of observations for that school or stratum, and then dividing by the sum of each, uh stratum size times the standard deviation. So in the denominator here we would have a number of people in the first school times, the sample standard deviation from that school, plus number of people in the second school times, that standard deviation plus a number of people in the third school times the third standard deviation. And again, this will only work if we have that sample data indicating what the standard deviation would be, and then what we do is we take this proportion and we multiply that by m which is the total that we want in the sample. So this will give us a proportion of individuals that we should choose, and then we scale that up by how many people we would like to have. So let's go ahead and calculate the sample size of students that we need from school a based on these data. So for school a Yeah, what I'm gonna do is take a number of people at school eh times standard deviation from school a divide by the sum of those quantities mentioned before. Then we multiply that by the sample size that we want, which here is 100. Okay, let's fill in some of this information. So the in the numerator, the number of people at school eh is 310 or number of sixth grade students. So we have 310 times the standard deviation which is three. This 100 can be multiplied in the, in the numerator, Excuse me. And then in the denominator, what we're going to have 3 10 times three in two times ehstuhs that's the number of people at school, two times that standard deviation. So we'll have times 420 times 12 plus 516 times six than simplifying that our numerator will become 93,000, and our denominator will become 9000 and 66. And then if you divide those you should get is 10.2581 Now remember that this is representing a sample size, that's representing a number of students and we can't choose fractional numbers or decimal numbers of students. So what we need to do is round this to the nearest whole number, which would give us 10 students, we would round this down. Mhm Okay, lets follow a similar process for school B trying to calculate the number of students from school B. It's going to be in too times S two and notice that our denominator is not going to change for in this formula. So we could just take the quantity we already computed for it. Not reinvent the wheel here. Save our selves a little bit of work. Okay, So end to number of students at school, to our school B is 420 the sample standard deviation from that school is 12. So we'll have 420 times 12 times 100 over 9000 and 66. And if you calculate that value, it's going to give you 55 0.5923 If we stop at four decimal places again, we need to round this to the nearest whole number. So because uh this is the decimal part is greater than five. Well, round this up to 56 students using traditional rounding rules. All right, let's do this one more. Time for school, be if you think you've got the hang of it, this might be a good time to pause the video and see if you can calculate it yourself. But what you'll do is multiply a number of students from school C times the standard deviation divide by 9066. Multiply by M which is 100 are total desired sample size using the quantities from the table on the last column. The way that I've had it that I haven't written out and our numerator will have 516 times six times 100 all over 9000 and 66. This should give you 34.1496 which will round down 2 34 students. So, it's always a good idea to check when you do these calculations and make sure that your number of students based on how you rounded adds up to the total sample size. So what we want to do is we want to check does 100? Sorry, just 10 from school A plus 56 from school B plus 34 from school. See do those add up to 100? And if you do that edition you'll see that. Yes they do. So we do have is our value M we did do our calculations correctly. It looks like, so that means that we are going to choose 100 or 10 students from school A. 56 students from school B and 34 students from school. See now that is part A of this problem. So I should probably label this up here. So this is part A. Which asked us about the sample sizes and we computed these three sample sizes. But now we're given some additional information and asked also calculate an estimate for the population mean. So the way that we're going to do this is using the formula given an estimate of the population meme is a each sample size from from each stratum over the total sample size times the sample mean gathered from your data. And we do that for each of the stratum, your strata. So the formula that we're going to be using his as follows. And of course we do need to be given some information. So we have all of our sample sizes. We have the total sample size that we want but we need to be given these sample means and we are. So X bar one is given as £82 expert. Two is £115 and X bar three is £90. So let's go ahead and fill in our formula. So the number of students that we chose from the first school was 10 and that's out of a total of 100 students times X one bar which is 82. Your ex bar one. Number of students we chose from school B is 56 out of 100 total, multiplied by the sample mean of 1 15. And then lastly a number of students in school three is 34 out of 100 X bar three is 9 90. Just 90 not 93. So then if you do this computation what you should get is 103.2 and because this is a an estimate of the mean it should have units with it. So our estimate is in pounds and that is how we can estimate the population mean weight of sixth graders in this school system.

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.

The following is solution video to number 26 which is two sample T tests comparing meantime loss in the workforce with stressors and intimidators. And you just got to be really careful here. The only kind of weird thing is They kind of switch up the order on you so X one represents the intimidators. In the next two represents the stressors. And we want to see if the stressors, so I'm actually gonna write that down. So this is intimidators and this is stressors. Okay, So it says is there enough evidence to suggest that the stressor is greater than the intimidator? So keeping that in mind, we're going to write less than some you won Is less than you two, Even though it says greater than but the orders just switched, if that makes sense? So the greater than in this case means less than for our alternative, since the order is opposite. Okay, so the first part of this is just verifying that these means. And standard deviations are four and 2.38 for intimidators, and then 5.5 and 2.784 stressors. So I went ahead and it says to use a calculator, so I'm using a T 84 if you go to stat and then edit here are the list, so L one is the intimidator and then L two is the stressor. So if you go back to stat over to couch and then it's one of our stats, we can change that to L one and this will be the four and the 238 that was over here. So four and 238 And then go back to stat couch. One bar stats, change that to L two calculate, and that's where we get to 55 and the 27855 to 78 So we verified that that is the way to do it. So then the second part of it is to kind of sort of not formally conduct a hypothesis test but essentially come up with the conclusion here and it is gonna be a two sample T test. We're gonna use the calculator since we already have the data input anyway. So it's two sample T test for the alpha is point oh five. And the alternative we already talked about, the alternative is less than not greater than but less than so. Um if you go back to stat and then we're gonna air over two tests and it's the fourth one down is the T. Test. Two sample T test because we don't know the standard deviation for the population, We don't know sigma, we only know as um so sincerity of the data, I'm just going to keep it as data instead of doing the summary stats. L one is the first list. L two is the second list and you can keep those frequencies as one and then we're gonna change this alternative to less than you two. And then pulled is usually no unless they tell you otherwise and then we're gonna go and calculate and you can have that test statistic there if you want but you really don't need to because the p value is all you need. So it's 0.14. So let's write that down. So the p values 0.14 and what we do is we explicitly compare the P value with the alpha. In this case 0.14 is greater than point oh five. So that means whenever the p values greater than alpha, we fail to reject H not so we failed to reject h. Not whenever the p values greater than alpha at the p values less than alpha, then we will reject the null hypothesis. So keep in mind that the null hypothesis is generally always saying that these two means are equal. We can we're failing to reject that, so we're accepting that to be the truth. So let's go ahead and complete this hypothesis test. And we're gonna say there is not sufficient evidence to suggest that the mean time lost due to stressors is greater than the meantime lost due to intimidators. That's spelled right? I have no idea. Looks like it's not so intimidators. Okay, that's why math teachers aren't spelling enthusiast. So intimidators um now you could also say there is not sufficient evidence to suggest that the meantime lost due to intimidators is less than the meantime loss due to stressors. If you want to keep to that less than thing. I just wrote it like this because that's the way the book had it. So there is not enough evidence to say that these two means are different or that one means more than the other. So we're accepting the truth that these two means are probably about the same.

We want to conduct the pair differences. Test that alpha equals 5% confidence. Testing the claim that population mean expert A. Is greater than population index. Barbie. Were given pair data samples A and B. Here and we assume amount, shapes and distribution on the right. Have already concluded or rather computed the differences mundi bar sample size and and the standard deviation differences SD as 4.5, 10 and 4.12 respectively. With this info we can proceed to do five texas below to compute this test first. That the value of the requirements using student's T. Distribution as well as hypotheses. So the requirements on that because the distribution specified degree freedom and mine as well as nine are null and alternative hypotheses are nude equals zero moody greater than zero and we have alpha equals 00.5 is our confidence. Next. It's gonna be the test at the P value. The test that is T equals D. Bar over sdo route and equals 3.45 And the tea table. This gives P between 0.5 point 0005 Thus we can include since P is equal to alpha, that we reject the null hypothesis H not which means that we have evidence that moody is greater than zero.


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