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First-semester GPAs for random selection of freshmen at a large university are shown: Estimate the true mean GPA of the freshman class with 90% confidence: Assumeo ...

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First-semester GPAs for random selection of freshmen at a large university are shown: Estimate the true mean GPA of the freshman class with 90% confidence: Assumeo 0.62 Round intermediate and final answers to two decimal places Assume the population is normally distributed.2.7 2.9 2.0 3.2 1.9 2.8 2.2 4.0 1.9 2.8 2.0 2.7 3.9 3.8 3.5 2.7 3.1 3.8 3.0 3.2 2.8 2.7 2.5 1.9Download data

First-semester GPAs for random selection of freshmen at a large university are shown: Estimate the true mean GPA of the freshman class with 90% confidence: Assumeo 0.62 Round intermediate and final answers to two decimal places Assume the population is normally distributed. 2.7 2.9 2.0 3.2 1.9 2.8 2.2 4.0 1.9 2.8 2.0 2.7 3.9 3.8 3.5 2.7 3.1 3.8 3.0 3.2 2.8 2.7 2.5 1.9 Download data



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Freshmen GPAs First-semester GPAs for a random selection of freshmen at a large university are shown. Estimate the true mean GPA of the freshman class with $99 \%$ confidence. Assume $\sigma=0.62$ $$ \begin{array}{llllll}{1.9} & {3.2} & {2.0} & {2.9} & {2.7} & {3.3} \\ {2.8} & {3.0} & {3.8} & {2.7} & {2.0} & {1.9} \\ {2.5} & {2.7} & {2.8} & {3.2} & {3.0} & {3.8} \\ {3.1} & {2.7} & {3.5} & {3.8} & {3.9} & {2.7} \\ {2.0} & {2.8} & {1.9} & {4.0} & {2.2} & {2.8} \\ {2.1} & {2.4} & {3.0} & {3.4} & {2.9} & {2.1}\end{array} $$

Okay, so we know that the mean and standard deviation Incoming freshmen's high school GPS are you is 3.4 some deviation is your 35 on that? They're 25 students. The first step is to just check for the conditions read demonisation condition on 10% tradition. Okay, So the right for the radicalization condition, we can assume that the sample size is independent and randomly selected population on for a 10% condition. We can assume that 25 students is less than 10% of all students. So now we know that it is okay to use the a normal model to describe the sample distribution model because according to essential limit here, t the theoretical mean and standard deviation for the sample size. And are you on signal over and respectively, where you is the truth Mean and Sigma's truce, that education So the theoretical mean GPS for one of these seminar groups and a standard deviation for the sample size. And he is equal to 25 Can you tell us is equal to you four standard deviation. Why is equal to signal over which is your 0.35 five? Does your yourself. Okay, Now we just have to use to 68 95 1917 68 five percent route. It just says that distribution is union model symmetric union model. No trick. Then approximately 68% of observations will be within one standard deviation of the mean 95% 2 standard deviations. Nine Inference. Three standard deviations from the meat. So this means that approximately 68% of the observations will mean freshman high school GPS between 333 3.47 Because that's just two standard deviations from the mean on approximately 99.7% of the GPS observed nine and 3.61 that's just three. Standard deviations from the mean on 68% will be between do you point to six on 3.54

Probably 26 wants us to find how large of a sample is needed in order to estimate the true mean of undergraduate GPS. So we're needing to find a sample size and and we need to do so within its our error 0.25 with a 99% confidence level and were also given a standard deviation of 1.2. Now the Z value associated with a 99% confidence level is 2.58 And with the following formula, we would determine n or a sample size of we just plug her values in 2.58 multiplied by 1.2, all divided by our error, which is your 0.25 and we will square this quantity. Simplifying our N is equal to 12.38 quantity squared. And when we square 12.38 we have 153.26 and again, since we're dealing with sample size, we want to round or sample size up to the very next hole number, which is 1.54 undergraduates, especially because we're dealing with students or people. We have to round up to the next hole number, and that's my sample size

In this exercise, we're going to be regarding the data given here on college G. P. A. As arising from a census of all freshmen at a small college at the end of their first academic year of college study. And using this we're going to compute the population mean. Um uh um um So what we want to do is to go to the data tab and click on data analysis and select the descriptive statistics tool and then we click ok next for the input range. We want to click on the entire fast column and we also need to ensure that our data is grouped by columns and that there's a label in the fast room which is College G. P. A. For the output range. I want to click on C. three where all the results will come then actually cocaine. Yeah. So we have the results from the descriptive statistics too. And what we're looking for is the mean for the population, Which in this case is 2.47? Yeah, In part B of the question, we're only going to regard the fast 36 students as a random sample. So I'm going to select the 1st 36 students. uh so that's fast 36 students, a random sample of 36 students And use that to construct a 95% confidence interval for the mean of all their 1000 gps and check whether it actually captures the population mean. Now in a similar way, um going to click on the data tab and click on data analysis and select the restrictive statistics and click OK next, for the input range, I'm only going to pick The Fast 36. So after selecting the 1st 36, I want to click on F three where the output would go and now We have the output for 36 students, the fast 36 students. And Also we have the mean for the 36 students and the sample standard deviation. Next We're going to calculate the 95% confidence interval. Um using that sample Now the 95% uh confidence interval he's obtained by the following formula X bar plus or minus. There's a it's co corresponding to the given level of significance, multiplied by s of the scourge of n. In this case we assume that the value of the population um standard deviation is not known. Now we want to substitute the values accordingly and for expert we have 2.45 or four a six. When you don't talk to two decimal places plus or minus. The critical value observed that corresponds to the given level of significance is 1.96. Then we multiply that By S which is 0.76 and divided by the square root of 36. So when we simplify that, it's going to be 2.46 plus or minus 0.25. And this here is the 95% confidence interval for the population mean based on the sample that has been given here With the fast 36 students. Next, we want to check whether this um confidence interval captures the population. Mean that we had obtained in step in part A no, We need to add and subtract 0.25 from both sides so that we can get the confidence interval. So this will become 2.21 And 2.71. So this is the confidence interval range And you see as you can see the value 2.4 seven is within that interval. And so we can conclude yes, the confidence interval covers the population mean given

We have the data set given below here and using it, we want to find the sample mean X. Bar, the sample standard deviation S. And constructed 99% confidence interval for the population we knew, assuming the population is normally distributed. So since we have this data we can find expire and s using the appropriate definitions expire, is that some of the data divided by n 2.353 and s. Is the square root of the sum of deviations of the mean square divided by n minus one. In this case 1.3 Now with X. Bar and sample standard deviation S, we can use a student's t distribution to construct this confidence interval. So we can identify the appropriate T score. So for the degree of freedom and minus one equals 14 and critical values um providing probability 140.99 for the confidence interval, we can use the students to table either on google or in a textbook to identify T C equals 2.977 Now, with this T. C, we can identify the margin of error E, which is given by this formula plugging in our T C. S and N. We obtain E equals 0.572 And now plugging into our confidence interval formula, which is that new falls between sample mean minus C, and sampling plus E gives us our confidence interval. That new is between 1.781 and 2.9 to 5 with 99% confidence.


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