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FIB The enrollments of the 13 public universities in the state of Ohio are Iisted below:EnrollmentCollege University of Akron Bowling Green State University Central...

Question

FIB The enrollments of the 13 public universities in the state of Ohio are Iisted below:EnrollmentCollege University of Akron Bowling Green State University Central State University Unlversity of Cincinnati Cleveland State University Kent State University Miami Universlty Ohio State University Ohlo University Shawnee '- State Unlversity University of Toledo Wright State University Youngstown State Unlversity26,106 18,864 1,718 44,354 17,194 41,444 23,902 62,278 36,493 4,230 20,595 17,460 12

FIB The enrollments of the 13 public universities in the state of Ohio are Iisted below: Enrollment College University of Akron Bowling Green State University Central State University Unlversity of Cincinnati Cleveland State University Kent State University Miami Universlty Ohio State University Ohlo University Shawnee '- State Unlversity University of Toledo Wright State University Youngstown State Unlversity 26,106 18,864 1,718 44,354 17,194 41,444 23,902 62,278 36,493 4,230 20,595 17,460 12,512 Is thls sample or population? Whal Is the mean enrollment? What the medlan enrollment? What Is the range of the enre Iments? Compute the standard devlation;



Answers

A survey of enrollment at 35 community colleges across the United States yielded the following figures:
6414; 1550; 2109; 9350; 21828; 4300; 5944; 5722; 2825; 2044; 5481; 5200; 5853; 2750; 10012; 6357; 27000; 9414; 7681; 3200; 17500; 9200; 7380; 18314; 6557; 13713; 17768; 7493; 2771; 2861; 1263; 7285; 28165; 5080; 11622
a. Organize the data into a chart with five intervals of equal width. Label the two columns "Enrollment" and "Frequency."
b. Construct a histogram of the data.
c. If you were to build a new community college, which piece of information would be more valuable: the mode or the mean?
d. Calculate the sample mean.
e. Calculate the sample standard deviation.
f. A school with an enrollment of 8000 would be how many standard deviations away from the mean?

Let's look at this question. Listed below are student evaluation ratings off courses where reading is fight for excellent. Alright, you want 90% confidence interval 90% confidence interval means war Let's go here to a template again, first of all and in this case is 15 and will be 15. So my degree of freedom will be in minus one which will become 14 My sample standard deviation I simply substitute the values here Use this formula and I get my sample standard deviation and my Alfa by two I'm calculating 90% confidence. So my Alfa by two will be 0.0 fight 0.0 fight I will calculate the value off p Alfa by two by using my Alfa by two and degree of freedom on substitute the values here and get the value for my confidence in double. I can do all of this or I can just use a software like over here. I'm using this website. I've just put in the values This is calculated everything for me. I can see the sample mean I can see the standard deviation. My confidence level is 90. So my confidence level is 90 let me select 90 and my confidence intervals. 3.6. 94.153 point 69 to 4.15 3.69 to 4.15 What can I say? What is the question? What does the conference will tell us about the population? College students? The population of college students in access were with well, with 90% confidence. I can say that the true population mean lies between these two values. Okay, the true population means lies between these two Valuev's.

We want to compare a CT scores in USA and in Ohio basically want to know if the average a city score is the same in the US as it is in the state of Ohio. So we're gonna use a procedure presented at page for 91 of your book. Is it started Step ones just for this problem with any use? Wanted to note that you will say and to for Ohio their average simple. The average a CT scoring the sample gathered for him. The USA is 21.4, whereas in Ohio the sample average was 20 point a. In the US, this simple was based off. 1000 students are clearly greater than 30 whereas in Ohio, 500 students by hundreds course where take a 500 is also greater did 30. And the standard deviation both in the U. S and in oil is the same. And it's three. What's the iPod disease? What are the I parties? Well, they're easy h zero. We think that the average score is the same in the US and in Ohio and our alternative hypotheses we think that they're different in Ohio and compared to the rest of the United States. So once again, we have it too. Tail test. Now we need to find our critical values so we can figure out what's our rejection area. That's figured it's out right now In step two, I'm going to use that Step two eso Alfa is 0.5 Therefore, Alfa over 20 for 0 25 Which means that the critical value bad you our plus minus 1.96 So we have minus. If we go back to our of the graph, we have minus 1.61 point six. So we have older death unnecessary to make a decision. Except we don't have Z. We're gonna go computed right now. Z is use the formula again. Your books are excellent bar minus X to bar the difference Naming of population difference of simple difference. The friends sample average or mean average. So this is a difference. You mean of the sample difference of the mean of the population of that divided Bible? Wrong order the variance in the US over the simple side of the U. S. With the variance in Ohio Square where a simple size taking Ohio. Remember, this is aged zero. So we know that this is zero. Now we just plug and play the value. Just plug the value state that we gather. That's the born. So we have 21.4 minus 20 point a over three where divided by 1000 plus three squared, divided by 100 who put everything in the square room. I'm gonna that you plug all those number in your calculator or your computer and Indian. You should have three points sitting fight his logic all seems it is very small compared to 1000. And still there's small comfort 500 before everything in the square. We is very small. That's what we get Z, that's quite big. So we have Z of 3.75 which is created in 1.96 We go back toward ugly graph of the beginnings of 3.7 by were clearly in their rejection area. So we reject h zero, which means that they mean a C T score in US order, you would say is different then. That mean a C T score in o quick recap. We gathered all of our data for this problem and we figure out our fantasies. We figured out our critical values, which to make our decision We compute Z and we compared see to the critical value. And we figured out that were rejecting the hypothesis. Which means average the mean a CT score in the U. S. Is different than the moon. A city's Court, Ohio.

Hi. So we're gonna go over problems 79 of Chapter 10 of introductory statistics. So Problem 79 is talking about a student who claims that the bean enrollment at four year colleges is higher than the me enrollment at two year colleges in the United States. So we're going to just write that down. The, uh, mean enrollment on a four year colleges is higher than that at two year colleges. Now, in order to test this claim to surveys were conducted, one survey took place for the two year colleges, and the other survey took place four year colleges. Now for the two year case, 35 2 year colleges were surveyed, Serena say, and one is 35. That's our sample size. Of those 35 the average enrollment SE expert one was 5000 68 and thestreet andere deviation for that sample. Se s one was 4700 and 77 for the four years there were also 35 4 year colleges served age, so we'll say end two is equal to 35. Their average was 5400 and 66 and their standard deviation was 8191. That is all the information that the problem statement gives us. So with this, we're first going to define a null hypothesis and an alternative hypothesis that as a and B now quick refresh on Knoll and, uh, attorney of hypothesis, the, uh, no hypothesis will always be dealing. We'll always look like qualities. So we're gonna have equal or greater than equal or less than or equal to like that. And the alternative hypotheses will be strict inequalities. Well, I'm not equal or strictly Ah, strictly greater than or strictly less than so. Looking back at our claim here, the claim is that the mean enrollment at four year colleges is strictly greater than strictly greater than, ah, the mean aroma at two year colleges. The strictly obviously sounds like a alternative hypothesis. So we're gonna make our alternative hypotheses be this claim here. Sorry to say that X to bar, which is the mean enrollment at four year colleges is strictly greater than X one bar, which is the mean enrollment at two year colleges over. We are going to rewrite this as X one bar is strictly less. Inpex to bar is he has this exact same thing, just sort of flip. Now that we ever alternative hypothesis, what we can do is, uh, determine our no hypothesis by simply making it the opposite. So x one bar is gonna be greater than or equal to x two bar. Now that we have those, we want to look for the random variable that represents our problem here. It's right off the bat. You might, ah, notice that. It kind of seems like they're almost two random variables here right now. You're comparing the mean enrollment of one population to the mean enrollment of another. So these two things that were comparing are almost random variables in of themselves. So will say that the first random variable is like X capital x one bar because we're looking at the average and the other mean variable will say his capital X bar to Now, the way we're looking at these averages is stated right here. We're comparing the two to each other. We're actually going to rewrite this in terms of right random variables. So basically comparing are random variables like this and then we're gonna further rewrite that as capital X one bar minus Capital X to bar his lesson zero. And this right here is our random variable. You can see we have both random variables on one side of the inequality. And because of that, we can basically just group it as a single and a variable. This single random variable is a difference of tour in variables and his these two random variables represent means So it's difference. Uh, mean enrollments of mean enrollment, I should probably say, at two year colleges and four year colleges and the United States. So this right here is our random variable and rash again. And to say that is part C and move on to part D because now that we have around variable, you want to look at what sort of distribution it's gonna be used for this test. So looking at the structure of our problem here, see, we have two populations. We do not know standard deviation for your population. I should say that thes standard deviations here are of the samples of these 35 samples here. They're not the standard deviations off populations, our So we do not know the population, standard deviations and what we're looking for are these averages three population averages. So to us, that sounds like a scenario that we see in chapter 10 Section one and the type of distribution we use. There is what's called a students T distribution now reviewing over sectional. And in Chapter 10 you'll notice that these ah t distributions in order for in order to really find them, there's a couple steps need to do. We need to find the the standard aRer we need the T value. And we also need to know that degrees of freedom uh, each of these steps is fairly complicated. So in order to expedient that process, what we can do is actually use our scientific calculator here. So if we go to scientific calculator, go to stat good over two tests and head down to number four the to sample A T test thickener on that, go over to stats select, And from here, you can actually, uh, just input all the information that the problem statement already gave us. So for the two year R X bar, one was 5068 inter deviation are two years was 4777 and the sample size was 35. Likewise, for the four years we had 5466 and we had a standard deviation of 8191. So now here, Chris is asking for the knoll. High earners are the alternative hypothesis. Remember, we re wrote our alternative hypothesis as X bar one less than expert too. So that corresponds over to mu one being less than YouTube. For guys like that going to keep cooled off the variances air. Not gonna be pooled on this. And then we're gonna go ahead and draw instead of calculating, can see here that the normal curve is being drawn right here. And the shaded area is the P value. And where this shaded area stops, that's where the Z score is. We have an entire P value represented by the shaded area, and it stops at the sea score or I'm sorry. I should say t score. Since this is a student's T distribution, Uh, they stopped at the T score and that looks about right. We have a T score of native 0.2483 That looks like it's about a negative quarter over from this year accident. And this whole area under the curve is 0.40 to 4, which corresponds to a T value 40.24%. We can actually go ahead and write those. So e, you have a t value her T score of negative zero point 24 83 And for F, we have a P value of 40 0.2 4%. Obviously, this graph pretty much already answers part G. So we're gonna move on to Part H, uh, which is looking at, uh, actually, before we do anything on porridge, we need to define a significance level because the problem statement never gave us significance level. So we're gonna say they're Alfa is just a standard, which is 5% now. What we can do is compare Alfa two r p and we really easily see that Alfa is actually much less than P. 5% is much less than 40.24%. Because of that, what we can do is decide whether or not to reject or not reject our no hypothesis since r p is so much greater than our Alfa we are going to on. I'll phrase it like this decline to reject the no hypothesis that should make sense. Uh, because we have a P value so large, it seems pretty likely that, uh, what with p values telling us is that it's pretty likely that Thea, the fact that the expert to happen to be larger than expert one was likely just do to randomness. Now the P value were less An Alfa could say that that's actually very unlikely. And it's evidence towards rejecting age. Not, however, because of p values so large. Uh, we're pretty confident that, um, that was the fact that the average of the four years being greater than the average of the two years was likely just a fluke. Uh, dude randomness. So in conclusion, in conclusion, what we can say is that the I mean enrollment, the mean enrollments at two year colleges is likely. At least the same has four year colleges in the United States, and I say at least the same because remember, are no hypothesis was that X Bar one is greater than or equal to export to, so it could be EQ so they could have the same average enrollment. That's, um, fairly like it. And it's also fairly likely that the two years actually have a higher average enrollment than the four years so we can could basically conclude that the students claim is very likely to be wrong. It's not 100% certain, but it is pretty likely seeing as how we have a P value of 40.24% which is very high. So, uh, that is basically, ah, all to this problem, and I hope you were able to follow along easily enough and understand how we got these answers and why're they are correct. Thank you for listening, and I hope to see you against you.


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