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The percentage of high school graduates applying to multiple colleges has been increasing over the past several years According to an association in a certain regio...

Question

The percentage of high school graduates applying to multiple colleges has been increasing over the past several years According to an association in a certain region, 31% of high school students applied t0 seven or more colleges in 2015 random sample of 331 high school graduates in 2018 was selected and 131 of them applied to seven or more colleges. Complete parts throughConstruct 90% confidence interval to estimate the actual proportion of high school graduates that applied to seven or more col

The percentage of high school graduates applying to multiple colleges has been increasing over the past several years According to an association in a certain region, 31% of high school students applied t0 seven or more colleges in 2015 random sample of 331 high school graduates in 2018 was selected and 131 of them applied to seven or more colleges. Complete parts through Construct 90% confidence interval to estimate the actual proportion of high school graduates that applied to seven or more colleges in 2018. The confidence interval has lower limit of 352 and an upper limit of 440 (Round to three decimal places as needed:_ What is the margin of error for this sample? The margin of error is (Round to three decimal places as needed:)



Answers

College costs rise" (October 29,2008 ), an article on the CNN Money website, gave the latest figures from the College Board on annual tuition, fees, and room and board. The average total figures are $\$ 34,132$ for private colleges and $\$ 14,333$ for public colleges. In an effort to compare those same costs in New York State, a sample of 32 junior students was randomly selected statewide from the private colleges and 32 more from the public colleges. The private college sample resulted in a mean of $\$ 34,020,$ and the public college sample mean was $\$ 14,045$ a. Assuming the annual college fees for private colleges have a mounded distribution and the standard deviation is $\$ 2200,$ find the $95 \%$ confidence interval for the mean college costs. b. Assuming the annual college fees for public colleges have a mounded distribution and the standard deviation is $\$ 1500,$ find the $95 \%$ confidence interval for the mean college costs. c. How do the New York State college costs compare to the College Board's values? Explain. d. Compare the confidence intervals found in parts a and b and describe the effect the two different sample means had on the resulting answers. e. Compare the confidence intervals found in parts a and b and describe the effect the two different sample standard deviations had on the resulting answers.

Now, this is the question that we have in this case. What is our end in his 20? So degree of freedom is going to become 19 inches 20 So D efs 19. This is the formula that I use for my standard deviation. Well, for by two. Since I'm constructing a 95% confidence interval, this is going to be 0.25 and I can again put in the values over here. I can find the value of the alphabet to using a table, substitute the values and I can get my confidence in trouble. Or I can simply use a software as I have done. Nobody care. Absolutely sorry. I have just simply put in the values. And this calculator has calculated all of the world is for me. You can see the sample mean the sample standard deviation on the confidence level of selected as 95. They can select whatever we want. A 2nd 95 and the conference of Devil s 4.96 to 8.44 point 964.96 to 8.4 This is my confidence interval now. Does it appear. The college students typically earned a bachelor degrees in four years. In four years. No, you can see that this confidence interval does not contain four years. So we can say that on an average means it is at 95% confidence. We can say that it takes more than 4.96 years to get a bachelor's degree. Is there anything about the data that would suggest that the confidence interval might not be a good result? Well, first of all, this is a very small sample size. This is a very small sample size, and the second thing is, uh, college Soon bachelor's degree based on data from Yeah, I think over here, the only problem is that the sample size is very small, so we cannot generalize it for the entire population.

But this problem refers to a survey of 1000 adults, and it was concluded with 95% confidence that from 49% to 55% of Americans believe that big time college sports programs corrupt the process of your higher education institutions. So and was 1000 and we were given a confidence interval at the 95% confidence level of 0.49 up 2.55 So if we think of that in terms of a number line, the low end is 49% the high end is 55% and part A is asking you to calculate the point estimate and the point estimate is identified with the variable p prime. And every time you do a confidence interval, the point estimate is found right in the middle of your confidence interval. So in order to find P prime, we could average the two endpoints of our confidence interval. So we'll take 0.49 plus 0.55 and we'll divide by two. And in doing so, you're going to get a point estimate off 52% or 520.52 The other part of this problem or part A asked you to find the error bound, and the error bound of the proportion is going to be that wiggle room. So it's the distance from the center to each end point of the confidence interval so we can get that by taking the high end of the confidence interval minus the point estimate which gets you three. Or you can go from the point estimate 0.0.52 and subtract the low end. And either way, we get an error bound of 0.3 In part B of this problem, it's asking, can we, with this 95% confidence, conclude that more than half of all American adults believe this? And the answer is no. We cannot conclude that more than half, and the reason being would be because our confidence interval spans from 49% up to 55%. So that means your true proportion can be anywhere in there in that interval. And since the interval goes as low as 0.49 we might be here, which could be less than half so. Therefore, no. We cannot conclude that more than half of all American adults believe this based on this confidence interval as we go into part, C. Parsi is asking you to construct a 75% confidence interval. So we're going to use the P prime that we found in part A and that was 0.52 And we're going to use the fact that we surveyed 1000 people and we need to find the confidence interval well. In order to find the confidence interval, we will need to find the error bound of that proportion, using the Formula Z of Alfa over to multiplied by the square root of P prime times Q prime over N. And in order to calculate the Z score associated with this elf over to, we will need to draw our bell shaped curve, which then puts 75% confidence into center. And our Alfa is the part of the curve that is not accounted for in that confidence interval. So there's 25% of the curve still unaccounted for, and because the bell shaped curve is symmetric, each tale will have half of that or 0.1 to 5. So in each tale we can put a 0.1 to 5 or 12.5%. And then the Z score associated with this left boundary can be found by doing. You're in verse norm on your graphing calculator. And when you use inverse norm, it asks you for three parameters. It asks you for the area in the left tail, which is 1.1 to 5. It asks you for the mean of the standard normal curve. And the standard normal curve has the mean of zero and the standard deviation of the standard normal curve, which is one. So I'm going to bring in my graphing calculator and to access inverse norm, you're gonna hit the second button, the variables button and number three, we're gonna type in the area that's in the left tail, followed by the mean, followed by the standard deviation of the standard normal curve. And we're getting a Z score of approximately negative 1.15 So that negative 1.15 is the Z score associated with the left boundary of the confidence interval. And because of the symmetric nature of the bell, the right boundary is going to be positive 1.15 So, in order to find the error bound of that proportion, we're just going to use 1.15 as RZ and the P prime was 0.52 We're going to multiply that by Q Prime and P Prime and Q Prime must add up to one so Q Prime would be 10.48 and R N was 1000. So our 75% confidence interval error bound is going to be 750.182 So we're not finished yet. We still have to generate our confidence interval. So to generate our confidence interval at the 75% confidence level, we're going to take the point estimate and we're going to subtract the error and we're going to take the point estimate and we're going to add the error. So our estimate was five to so we will subtract 0182 and then we'll take 0.52 and we'll add 0182 So for part C, our confidence interval will be 0.5018 up 2.5382 So the final part of this question is part D. Can we, with 75% confidence, conclude that all American adults believe this, and the answer to that part is yes. We can conclude with 75% confidence that at least half of all Americans believe that big time sports programs corrupt the process off the higher education system. And the reason being would be. Here's our point estimate of 0.5 two, but this time our interval on Lee goes down as low as point 5018 And if the true proportion is found in here, it's always above one half or 10.5. So, yes, we can conclude with 75% confidence that all American adults believe that.

So the first thing you should do is put all those numbers in the calculator. Um, for a you're gonna indeed, some of that information. So put it all on your list. Goto won their stats. Uh, a someone asked for the mean Mean is 800 8000 629 people, so he rounded it up. Um, the sample standard deviation is 6944. And how many colleges we look that was 35 and end by. This one would then be 34 for being, we have to define X and X bar. So for X, it's the number of students enrolled in college. That's the number of students. Basically, what we collected an ex bar would be the sample mean of the 35 colleges, So just be the sample C s from distribution. Since we have the sample standard deviation, we have a tea distribution with 34 degrees of freedom. And for our parameters, we have 8629 and 6944 over the square root of 35. He wants the confident terrible, the air bound, and then a sketch of the graph, so 8629 closer minus. So there's a lot of ways to look up this critical value. I did it with the P I calculator, so I have 2.32 time 6944 over the square root of 35. So 86 to 9 plus or minus 2300 85.6 So this is your air bound and doing the addition. Subtraction gives us 6243 0.94 for the lower bounds and the upper mound, the 11,000 14 0.6 grass. We have the normal curve number 95% sure that the true number of enrollees and community colleges between these two, so approximately 6 to 44 2 11,014 e What if we use 500 colleges, so if you have more samples, you'll have a smaller interval, a more narrow interval

And this particular question we have given us that there are 3000 institutions and we have taken a sample of $400 per foot. So are the leader has been given to us. So I want to write it down here. The sample size and is 400 schools and, uh, the average in Norman him. Take expert is giving us 3700 then Understanding division for the sample is also given me just six powers of 500. And, uh, the commission has put a give or take off standard error, so that standard at all is given as 3 25 on the estimate. So I'm going to start with the first part here. The question say's that for the first part, they have asked that an approximate 68 core person of confidence in double for the average enrollment off all 3000 institutions from 3375 to 4 to five. So meaning to check the statement is true or false. So if I start with the explanation now, I want to see that, uh, the what have you done is that the 60 person off the glass and debris for the average in government. What? Steven, it wasn't a question for the average enrollment off all the 3000 institutions. Eyes running from him saying 3375 tau four or five. So therefore, I can see that the interval that is given to us, which is ah, 3375 to 4025 is one standard editor away from the average in government. Yeah, So I can therefore say that one standard error. Oh, God. Response to 68% off confidence and double in the normal level. So we can therefore see that the statement is don't now moving on to the next part for but me. The question states that if the status institution takes a simple random sample of 400 institutions out of 2000 and goes on a standard error either way from the average enrollment off, wondered sample schools still be about 60% chance off the double, which will cover the everyday enrollment off all 3000 schools. So he's supposed to check if this is a true or false statement, so I can see that it is already Norm you found in party that 60% off samples cover the population average off endorsement from 3375 to 4 or five. So therefore, control that the statement is true. Someone who already learned that from party it is known waas that 68% off all samples cover the population average off a Norman from 3375 to 4025. So therefore, I will say that is given statement for the parties. Also true moving onto the bar. See, the question states that about 60% off the schools in the sample had enrollment in this change off 2700 plus or minus 6500. So you need to check if this is a true or false statement. So if I start with the park, See, uh, we can say again that you know, it's already been proved here and Barbie again that 68% off schools in the sample had enrollments in the range. You can see it's a strong 23375240 Goofier. So that means the statement of what they've given his force. So we therefore say that, uh, the Steve Moon that a boat. 68% with schools in the sample had enrollments in the range 3700 plus or minus 6500. So we say that this statement is false because we've already proved earlier that it is from 2375 to 4025. Now, will you go on to the next part the butt d for the Barbie, The question states that it is estimated that 68% off the 3000 institutions off higher learning in the U. S. S and all between 50,007 dead minds Treat my friends, which is 3375 and it's 2700 plus t 25. So which is for water Firestones? So, uh, you want to prove this part in but a and that is a double order given to us. So we need to again check on dhe side effects are true or false statement. So we will repeat the same thing that it can be said that 60% off the class in trouble for the Aborigine woman. All 2000 scores. Isn't that angel off 3375. This is the range toe 4000 and 25. This is given to us already. So therefore, women say that it may be correct that 60% off. So I'm concluding now that 60% off the 3000 institutions also off higher learning in the U. S. On gold between, say, 3700 plus or minus 3 25 students. So that means the therefore say that this statement is also true. Then we wanted the next part. That's the last part for this question. But e we're supposed to check the normal cough can be used to figure confidence levels. Here are the door because the data don't for the normal cope. So Teoh, right? My answer on this part I'm saying that the statement that the normal cough can be used to figure out conference levels as the later for a normal coach here and well, you see that the statement is falls. So the reason you what is that? Uh, when the sample size increases. So in this case, the detail approximates toe normal distribution. Norman disc um, Yushin. So since I would see the sample spice eyes over here is 400 so which is, uh, large enough to follow normal distribution. That is where we control that. The statement. What they've given in the question is for us.


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