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Coal comdany wanis determine 9596 confidence interval estimate for the average daily tonnage ol coal Inal [nnks Assuming Ire cornpany reporls Inat Ihe staridard dev...

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Coal comdany wanis determine 9596 confidence interval estimate for the average daily tonnage ol coal Inal [nnks Assuming Ire cornpany reporls Inat Ihe staridard devialion al dally output 200 tons; hov many days should sampic S0 tnat thc margin of crror bo 39. tons less?Filty students AA nralla fconamics dasc Aller Ine (Irsi Pssa eramnalan rhndoa sampic five papcrs was sclcctcd. Tho gradcs wcrc 60, 75, 80, 70,and 90. If thorc wero 200 students the class, what iould be the 90%0 contidence interval

coal comdany wanis determine 9596 confidence interval estimate for the average daily tonnage ol coal Inal [nnks Assuming Ire cornpany reporls Inat Ihe staridard devialion al dally output 200 tons; hov many days should sampic S0 tnat thc margin of crror bo 39. tons less? Filty students AA nralla fconamics dasc Aller Ine (Irsi Pssa eramnalan rhndoa sampic five papcrs was sclcctcd. Tho gradcs wcrc 60, 75, 80, 70,and 90. If thorc wero 200 students the class, what iould be the 90%0 contidence interval for the mean grade al the students the class? 1057 6331 86.01 65.22 t0 34.55 783 85,217 234 64.34 t0 85.56 enouah intormannn ceiemine Question 1.25 pts Question 2 "25 pts pesparcner intetested determining the average numner of years employees company with the company: past inormnaticn Ghous siangard deviaton months; what size sample should bc sclected 50 that at 9580 confidence thc margin etor will be months lcss? The monthly starting salaries of students who receive an MBA degree have standard deviation of S110. What size sarple should selected I0 Oblain 95 probability estimating Ihe Iean monthly incomc within S20 Icss? 1058



Answers

Business Week surveyed MBA alumni 10 years after graduation (Business Week, September $22,$ 2003 . One finding was that alumni spend an average of $\$ 115.50$ per week eating out socially. You have been asked to conduct a follow-up study by taking a sample of 40 of these MBA alumni. Assume the population standard deviation is $\$ 35 .$
a. Show the sampling distribution of $\overline{x},$ the sample mean weekly expenditure for the 40 MBA alumni.
b. What is the probability that the sample mean will be within $\$ 10$ of the population mean?
c. Suppose you find a sample mean of $\$ 100 .$ What is the probability of finding a sample mean of $\$ 100$ or less? Would you consider this sample to be an unusually low spending group of alumni? Why or why not?

All right, were given some data about graduates and there salaries 10 years after graduation, and we've divided them into data for men and data from women. So for part A, given the statistics Ah, we're given a sample of 40 men, and we want to find the probability that our sample mean will be within 10,000 of our population. Means hold on. Gonna crack that notation. There we go on that might need his penmanship, but you get the general idea, right? So let's find our standard deviation of the sampling distribution. It's gonna be the standard deviation over sample size. So that's gonna be, uh, 40,000 divided by the square root of 40. That equals 6324 56 We're gonna find a Z lower and see upper like so. So for a C lower, uh, it's gonna be negative. 10,000 divided by our center deviation, uh, for the sample. Yeah, our sampling distribution, and then for upper, it's gonna be 10,000 positive. When you compute these out, you get negative 1.58 from 1.58 Comparing this to our normal probabilities table. This gives you probability. Lower 0.571 probability, upper 0.9429 Uh, this means our probability is gonna be probability upper minus probability, lower. Which is zero point 8858 Right party. Uh, this time we're looking at a sample of 40 women, and now we need to find the probability that our sample mean that we find is within 10,000 of the mean for the women. So once again, we're going to find our standard deviation sampling distribution. That's gonna be this time. We're looking at the statistics for the women. So this is going to be 25,000 over square root of 40. Calculate that out. That's 3952.47 are Sorry. 847 Anyway, let's find a Z lowers the upper. So the els you again. That's gonna be negative. 10,000 over our standard deviation of the sampling distribution for women. This one's gonna be 10,000 positive. So these are equal negative. 2.53 2.53 respectively. This is P. L. Looking at our table Once again, this lower probability is from zero point 0057 Upper probability 0.9943 Finding a probability, which is probability Oper minus probability, lower. You get, um, zero point 9886 part C s us two compared these to given explanation why one is higher than the other. And we see that part. He is greater. This is because standard deviation for men is greater then that for women I'm gonna have to move that up on. And because the standard deviation is greater, this means that this is a small This is smaller relative to our standard deviation, which means this let fewer standard deviations away from the mean, as opposed to this. All right, I'm gonna move to a different page party. Now we have sample of 100 men. We want to find the probability that air sample mean is within. Ah, I believe it is sorry. It's not within this time. We need to find a sample mean that is greater than our population mean minus 4000. All right, so let's find our sampling distribution. Ah, standard deviation. So that's referring back to hear. That's 40,000 square root of 100 square. 100 is tense. And this just 4000. All right, now we just see score for 4000 against 4000. So this could be because we are looking at 4000 less than the mean That's gonna be negative. 4000 top standard deviation is 4000. So this is negative 1.0 And if we look at our table, this corresponds to a probability is zero point 1587 and there you have it.

Problem 22 or with the data values from the smallest to the largest, which is hundreds, hundreds on one 1/6. Then 115 115 141 one or two and one for three and 14 green and one for five and one for seven and one for seven and 1 15 and 1 52 on 1. 53 and 1 55 and 1 57 and 1 61 63 and 1 64. So to remind the critical value using Table and J, who's offer is equal to opening one and then is equal to 20 so the critical value is equal to find, so add one toe critical. Various okay is equal to five plus one, which is six to the boundaries, off the confidence interval for the million or then six smallest and the sixth largest value insulting data list, which is 1 41 and 1 53

Yeah. In this question, we want to estimate the average difference in the percentage of foreign revenue for tech companies and basic consumer product companies. After you've verified the mean and standard deviations for the two types of companies, we can plug these numbers into our confidence interval formula. And notice here I'll answer question D I'm using T. Star, the T distribution because this is quantitative data, and we do not know the population standard deviations. All we know are the sample standard deviations. Now I've plugged in my means, my standard deviations and my sample sizes, but Mighty Star, Our smallest sample sizes 16, which means our degrees of freedom is going to be 15. And we want an 85% confidence interval for degrees of freedom 15 and R. T. Star for that is 1.517 Yeah. And now I'll use my calculator. Find the difference in the means which is 1806. And then when I multiply 1.517 by this big square root here, we're going to get 5.42 is our margin of error. And when we add and subtract 5.42 from 18.06, it's going to give us a confidence interval from 12.64 up to 23.48. And what this means is that we are 85% confident. The average difference in the percentage of foreign revenue for the two types of companies is between these two numbers. And notice these two numbers are both positive. Which will tell us that our yeah. Mhm. Tech companies are most likely to have a higher percentage of foreign revenue than our basic consumer product companies. Okay. But

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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