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Question 21random sample of 10 college sludents was drawn from large uiver sity standard deviation 0f 3.2 Their :1jes ale 22, 17, 27, 20,23,19 24, 18, 19,and 24 yea...

Question

Question 21random sample of 10 college sludents was drawn from large uiver sity standard deviation 0f 3.2 Their :1jes ale 22, 17, 27, 20,23,19 24, 18, 19,and 24 years, with sempleSuppose thal you want to lest whether (he population mean age diltets Iromn 20 Carlucl Ihis hypolhesis (esl al IhG 1% significance lvel To recoiva (ull credit You must sro" the following:null and allernative nypoinesesdescription the test statistic forula (Can reterenco localion on Iha tormula shool)ocationeclian t

Question 21 random sample of 10 college sludents was drawn from large uiver sity standard deviation 0f 3.2 Their :1jes ale 22, 17, 27, 20,23,19 24, 18, 19,and 24 years, with semple Suppose thal you want to lest whether (he population mean age diltets Iromn 20 Carlucl Ihis hypolhesis (esl al IhG 1% significance lvel To recoiva (ull credit You must sro" the following: null and allernative nypoineses description the test statistic forula (Can reterenco localion on Iha tormula shool) ocation eclian tegion (#ere 0n Ihe diagram) disirbulion (7 vorSuS wih dlogrees Ireudom Crilical value values calculalion 0f Ine tcst staiisiic (must sho; soM4 calcwlalon wolt showing nmheis plugged into Iormula) full conclusian as shoxn ie examp @ ViCeos including the p-value



Answers

A random survey of enrollment at 35 community colleges across the United States yielded the following figures: 6,414; 1,550; 2,109; 9,350; 21,828; 4,300; 5,944; 5,722; 2,825; 2,044; 5,481; 5,200; 5,853; 2,750; 10,012; 6,357; 27,000; 9,414; 7,681; 3,200; 17,500; 9,200; 7,380; 18,314; 6,557; 13,713; 17,768; 7,493; 2,771; 2,861; 1,263; 7,285; 28,165; 5,080; 11,622. Assume the underlying population is normal.
a.
i. $\overline{x}=$ _____
ii. $s_{x}=$ _____
iii. $n=$ _____
iv. $n-1=$ _____
b. Define the random variables $X$ and $\overline{X}$ in words.
c. Which distribution should you use for this problem? Explain your choice.
d. Construct a 95$\%$ confidence interval for the population mean enrollment at community colleges in the United States.
i. State the confidence interval.
ii. Sketch the graph.
iii. Calculate the error bound.
e. What will happen to the error bound and confidence interval if 500 community colleges? Why?

For today's question, we'll be using formulas five and six from this section of the textbook. And since we are using these two formulas as stated in the question, will be using this chart to sulfur our answer to find the mid points. What we do is we take the lower bound of a class and we add the upper bound of the class and we divide by two. So for the first class of this data are midpoint is 5.5 doing the same thing for the rest of the classes, we get 15.5 25.5 and 35.5, noting that the 35.5 comes directly from the question. And now the frequency is just the number of data in each class until we had 34 18, 17 and 11. Again coming straight from the question. Now, for this column, we just multiply the numbers from this column with this column getting us 187 to 79 4, 33.5 and 3 90.5. Now for this column we need to find the sample mean for the frequency uh deviation. So what we do is we take this some of should be a some we take the sum of all of the data in this column here, and then we divide that by the number of data, and the total number of data will be the sum of all of the different frequencies. So the sum of all of these numbers is 12 90 1290. And some of these numbers is 80 so we do 1290 divided by 80 and we get 16.125 So this is our sample mean? Now using this number, we can fill in this section of the chart and it may be easier if you go this way, filling out for the rest of the chart instead of this way. So we'll find our uh midpoint minus our sample mean for this one. And we get negative 10.625 Don't let the negative trip you up because we're going to be squaring it. So we get 1 12.89 and then we'll get 3838.26 And so please note that all we do is square this value for this column and then we multiply this value by the frequency for this column. Now filling it out for the rest of the data. We get negative 0.6 to 5.397 point 02 And then we get 9.375 87 89 1494.13 And you get three and yet 19.375 squared is 375.39 And multiplied by the frequency of 11, we get 4129.29 Yeah. Mhm. Yeah. So now solving for our standard deviation, because you mean something for a standard deviation, s views from up here, Formula six. So we take the square root of the some of this column here, because as you can see this correlates to this, which is 9468.7. And then we divide that by end minus one and then we found to be 80 so it's 79. And then when we divide those and take the square root, we get 10.9479 and then finding our sample variance. Mhm. We know that this is just our standard deviation squared, so it's 10.9479 squared, which gives us 119.8576 Thank you for listening, and that is going to conclude.

Good morning. Uh, this question that we have in front of us right now offers us the enrollment for 35 community colleges across the United States and asks us to do several things. The first thing that it asks us to do is to divide the data up into five even intervals. Um, call that classes in statistics. And so, in order to do that, the first thing we need to do is identify the maximum value and the minimum value. The maximum value for this data set is 28 1001 165 and the minimum value is 1263. By subtracting the mom from the maximum, we find our range, the range of our data. How Why does it spread? And the spread his 26,902 to divide that into five. Even inter rules we divide by five, giving us 5380.4 Now. Um, what we're doing when we divide by five is we're trying to determine our class with health. How big are our classes? And so we always want to take this number even if it was 5380 even. And round up to the next number, we round up 5000 381. That gives us our class That told us what we're gonna add to our first class limit to get to our next one. So for five intervals, we start with our minimum data value. Okay? And we take that 5381 to determine what the beginning of my next class is next, even if get 6644. So I add 5381 again for 12,000 in 25 and added again for 17,000 406. And my last class is going to start with 22,000 787. Um, those air the lower limits for each one of my classes came, um, In order to determine the upper limits for each one of my classes, I start one below the 2nd 1 So you're toxicity for just a minute? Um, one below that is gonna put me at 6643. Okay, One below that. 12,000 and 24 17,405 and 22,000 786 Now for my upper class limit. I just take this class with here, um, and I add it to this one right here and get my last one for 28,000 167. Oh, now, this is where my five even hold 12,060. 343,644 to 12,000 and 24 so on now, too organized. The data into those five categories are even intervals. We need to basically tally up how many pieces of data fall into each category. Um, and so the first piece of data that we see it says 6400 in 14. Well, that would fall here in this first class, and then the next piece of data is 50 1550. That would also fall in that first class. And so we go through and basically tally up. Um, how many pieces of data falling to each class We do that correctly. We should figure out that 19 pieces of data pull into that first class nine going to the second one in the third. We're in the fourth, two in the fifth, and we can double check our work by adding up all of these all of these pieces of data and well, if we get 35 then we did something correctly. Hopefully, right. Um, because there were 35 pieces of data given to us initially. So now part B of this question asked us to construct a history, Graham and, um, to construct a history. Graham, With this data, there's couple different ways that people do it. But one thing that's common is the frequencies always go on vertical act frequency. So in this case, that would be how many colleges? Ah, um, intervals are enrollments within each one of those intervals. And so here I go, to see the highest number of frequencies that I have is 19. So I think I'm going to go by fives. And when you set up this vertical axis, it's really important that won her lowest numbers at the bottoms into so an equal increments. So if I'm going to go by fives, I need to go five and 15. I want you like that. I can't. Um change the intervals between tally marks as we go up. And now along the bottom of my history. Graham, some textbooks will call for cross foundries. I'ma call for midpoint. I'm in this case. I think we're gonna do class boundaries. Um, so class boundaries basically require us to meet halfway between classes so that they can share a bar because one of the our share a limit here a boundary? Um, because one of the requirements of the hist a gram is that when we construct our bar graph, which is what instagram is, the bars touch. And so I would have been a do is with my lowest class limit, which was 1263. I am going to subtract five, 1000 262.5. And now, with the upper class limit, I'm going to add point by 66 43. Wait, Fry. And I'm gonna do that with all of my upper class limits on down the line here. So I'll have 12,000 24 white farm, and then I will have 17,000 of 18,000 400 and 5.5 22,000 786.5. And last but not least, I have my for most class limit. 28,000 167. Okay. And so now the bar that we construct right here is gonna represent the dead between those two enrollments. And we had 19 colleges, it had, um, enrollments between 1263 and 6000 643. So we create a bar that goes not quite up to 45. The little cricket do better on paper, I'm sure with a ruler if you need it. But, um, my next class has nine pieces of data in it, and so I'm gonna go just below 10. Men draw bar, that's a little better. And then, um, I have just one enrollment between that 12,000 and 24 12,000 and 25 then the 17,405. I have Forbes mongo little bit higher, but not quite to five that are. And then too for my last are your Houston. Graham should look similar to that. Um, the next thing it asks us is if we were to build a community college, which would be more valuable. The motor that mean you go through this data, there's not one piece of data that appears more often to the other. But when we find the mean, I think we'll find that it's not necessarily representative, uh, most colleges either. So when we look at the history, you can tell this class is the one that has the most occurrence. And, um, that would be more of most community colleges. In order to find the mean, we want to use either a graphing calculator or one that allows you to enter all of our enrollments into it into a table and then find our one variable. So once you do that, um, you can look at your one variable statistics and you'll notice our ex bar or as we know in statistics, are our represents our sampling that is 8628.7. Other piece of information. Look for here. Let her e. S the strip you is our sample standard deviation. So when we're looking through our one variable statistics, we want to look for the letter s, um because that represents a sample standard deviation bond. That would be in this 6000 943 believed eight s. So we now have our sample mean in our sample standard deviation on the next part that it asks us for is how many standard deviations with the data value or the enrollment of 8000 be away from the means. How far away is 8000 from the mean in terms of standard deviations? And so, with statistics, we call back a Z score. Okay, Ours thescore tells of how many standard deviations away from the mean of specific data value is, and there's a formula for that. Um, it looks like this where we take our data value X. In this case, that's 8000. We subtract the mean, which in this case is 8628.7. And then we divide by our standard deviation 6943. I want A That gives us a Z score of negative 0.9 What that means is that, um, first off, a negative Z score tells us that the data value we're evaluating is below demean. And since it's negative 0.9 It tells us it's not very far below the mean. It's not even one standard deviation below the mean. It's just a little bit below me. And you can see that if you look at those numbers right, we're evaluating 8000 and, um, our means 1000 628. Our standard deviation, Uh, 6900 right? That's 600 difference. There between, um, the 8000 and 8600 is not much compared to that 6900. Almost seven standard deviation. That should complete this question.

Hi. This question We are trying tow us like that Instagram here. As you can see here, we can, um, approximate that as a normal distribution. Since we have this bill shape and the mean is 935 on the standard deviation. Here's one on the 66 and he is the probably blood which also confirms that we can and substance assume that there are follows the A normal distribution. So after that that hey is the 95% confidence interval and they use this website and we found that, um, the communists fl interval is from 833 to 10 4 e and the interpretation for confidence and servants meals, um, 95 ah, percent sure. That's my mean is somewhere in that internal.

All right, we're looking at the number of hours watched by some students, and that's what this data is, and I've sorted it uh into a column. And so I can easily do some calculations with it. And they were asked to figure out or determine if this normal probability plot which is created from the data, is Yeah, it tells us uh that could be normal, and the answer is yes, because the way these these probability plots work is that if you have your point, your data points within these bounds, you can assume it's normal. There's a really good chance is from a normal distribution. And so according to the normal probability plot, which I recreated here, Yes, we're told to find the sample mean and sample standard deviation. And so I found those here, using the spreadsheet function average and standard deviation. So that's what those are. And then using part B estimates, we are told to graph the role model for the distribution. So, what I did here, and I'll do that to and did that twice. Why do it once in a spreadsheet, when you can do it twice for the main standardization, I just want to draw a normal model for the distribution. So we'll go ahead and do that. And that is part C. So, here's my normal curve in the mean, is 20 point four, which is the mean. And then we're gonna add a standard deviation to get the next answer. Here's uh huh who is that? And that is going to be this is plus, I'm just going around this to 10 point five Most 10.5. That's gonna give us, what's that going to be 30 14 for the first standing ovation and then we'll had another one up there and that's gonna give us 40 yeah 41 point 31 point nine. Um And then you do it again That and that's another plus 10.5. And that's gonna give us, What's that going to be? 51.4 I believe when he's double check. 52 please. 52 52.4 Okay. And then we've got let me do that subtract. Okay, let me do keep someone consistent here -10. It's going this way and there's Is that? And that's gonna be 10 that's five. I'm just gonna keep that before I'm just going around at the 20.9 because I'm going to keep track of the thousands of hours. I watched tv Mhm. There's the second Last another 10 so that's going to be 10 points. Oh that is 10, whoops, I didn't it's color so it's gonna be negative. So this is a it's negative. Uh Mhm 0.1 Mhm. 12 Yeah 10 for you for this 10. Mhm Okay. You know what, I don't know why my brain sometimes our brains after a long day of a hard time calculating. So there we get into negative values. And uh I was talking to a friend of mine who just who had to statistics with back in college, That's a negative 10 point, right? And uh you know sometimes statistics just doesn't give the values that don't make sense. You have to be okay with it because you can't have negative time. I mean that doesn't work. But still that's our values. These are normal drawing. And then we want to figure out the party the probability that a student has watched between 20 and 35 hours. So for that I'm going to use the spreadsheets again. So this is D. D. And that's uh Z scores. And if you're called Z scores Z is X minus mu over sigma. And I'm gonna use a spreadsheet for this. So we need to find the Z score of 20 and the Z score of 35. And then use the normal probability distribution table. Or maybe even the a computer program to calculate it. So to do the area. So that's what we're going to do. So let's get a picture. I was like pictures, pictures are nice. So here's the here's the curve And we want to know between 20 which is a little less than the mean. So here's the mean. Didn't black, Here's the mean. And then we want to know between 20 Which is roughly here in 30, which is up here, we don't know this stuff. And so what we need to do is find the Z score and then the area under the curve which will give us this red area, and then Well, look up the Z score for this, this is the 3rd, 35, and then we'll get this green area. So what we do then, is to do the green like this, and then we're gonna subtract off the red, and that will give us, I was like equations with pictures, these are kind of fun, something like this. God. There you go. That's what we're looking for. This is the this purple pinkish era we want. So I did that for 20, and actually I just use this normal distribution function in the spreadsheets, which you put in the X. Bar or the the sample point, the mean, which we calculated the standard deviation, which we also calculated, and then it gives us the appropriate air into the curve, as opposed to doing the Z score. And then the look up. So let's do that for both of those were the difference, we subtract them. So the probability uh the probability is point um This .44, really? And then e the last part we want to use the model that we created earlier, Uh to see if the probability of picking a student who watched more than 40. Oh, Hey, that's a lot. So if we use this idea of the picture, it's gonna be hears this and 40s up here. So more than 40s over here. So how do we get that? Well, up to 40, is this green stuff. So to get the black, what we do is we take the whole curve which I'll cover in red with the whole curve. Mhm. Excuse my drawing The whole this is the whole curve and then we're gonna subtract off the green and then we'll be left with our area that we want. Which is this little black Nevin over here. So to do that uh we do this I do the same thing with the normal distribution function in the spreadsheet. Put in the appropriate mean and discrimination. And then That gives us this number .965 is the green area here. And then to calculate the area that we want is one minus that value which is 10.3. So the probability of picking a student Who watches more than 40 hours a week as a point of three. Okay. Probably that Bill t. There you go.


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