Statistics- Last name: Enter your LETTER answers
Question # 00117856
Posted By:
Updated on: 10/14/2015 12:30 PM Due on: 11/13/2015
| Last name: | First name: | ||||||||||
| Enter your LETTER answers HERE | |||||||||||
| ↓ | |||||||||||
| 1 | Note: | ||||||||||
| 2 | When done, using your last name and firstname, save THIS FILE as, | ||||||||||
| 3 | e270Lastname Firstname HW4 | ||||||||||
| 4 | (No space between E270 and Last name) | ||||||||||
| 5 | E-mail to | stowfig@iupui.edu | as an attachment. | ||||||||
| 6 | |||||||||||
| 7 | Example: | ||||||||||
| 8 | e270Smith Adam HW4 | YES | |||||||||
| 9 | e270 Smith Adam HW4 |  |
NO | ||||||||
| 10 | |||||||||||
| 11 | |||||||||||
| 12 | |||||||||||
| 13 | |||||||||||
| 14 | |||||||||||
| 15 | PAY ATTENTION | ||||||||||
| 16 | |||||||||||
| 17 | |||||||||||
| 18 | |||||||||||
| 19 | |||||||||||
| 20 | |||||||||||
| Questions 1-3 are based on the following "POPULATION" data (N = 63): | |||||||||||
| 19 | 89 | 91 | 20 | 95 | 36 | 63 | 12 | 79 | |||
| 67 | 54 | 61 | 16 | 39 | 39 | 63 | 61 | 42 | |||
| 31 | 84 | 62 | 76 | 40 | 23 | 10 | 16 | 84 | |||
| 76 | 76 | 63 | 17 | 78 | 14 | 50 | 86 | 65 | |||
| 79 | 97 | 39 | 85 | 96 | 21 | 36 | 70 | 69 | |||
| 84 | 67 | 21 | 66 | 79 | 87 | 31 | 58 | 77 | |||
| 66 | 68 | 59 | 84 | 39 | 61 | 76 | 66 | 47 | |||
| 1 | If sampling is done without replacement, the number of possible samples of size n = 5 from the above population is __________ | ||||||||||
| a | 7,380,289 | ||||||||||
| b | 7,028,847 | ||||||||||
| c | 6,677,404 | ||||||||||
| d | 6,343,534 | ||||||||||
| 2 | A random sample of size n = 5 is selected. The data in the highlighted cells above are shown below. | ||||||||||
| x | |||||||||||
| 95 | |||||||||||
| 12 | |||||||||||
| 84 | |||||||||||
| 66 | |||||||||||
| 31 | |||||||||||
| The mean and variance of this sample are: | |||||||||||
| a | x̅ = | 57.6 | s² = | 885.56 | |||||||
| b | x̅ = | 57.6 | s² = | 990.64 | |||||||
| c | x̅ = | 57.6 | s² = | 1238.30 | |||||||
| d | x̅ = | 58.2 | s² = | 1146.82 | |||||||
| 3 | For the sample means computed from the number of random samples determined in question 1, the expected value or mean of sample means is ______. | ||||||||||
| Use Excel rather than your calculator to do the calculations. | |||||||||||
| a | 60.26 | ||||||||||
| b | 59.85 | ||||||||||
| c | 58.25 | ||||||||||
| d | 57.54 | ||||||||||
| Questions 4-8 are based on the following: | |||||||||||
| The mean cost of getting a four-year college degree in a certain region of the country is $52,900 with a standard deviation of $8,500. Assume costs are normally distributed. | |||||||||||
| 4 | The fraction of costs in this region that fall within ±$4,500 of the mean cost is? | ||||||||||
| a | 0.4038 | ||||||||||
| b | 0.3955 | ||||||||||
| c | 0.3674 | ||||||||||
| d | 0.3293 | ||||||||||
| 5 | What fraction of sample means from samples of size n = 16 graduates fall within ±$3,000 from the population mean? | ||||||||||
| a | 0.8812 | ||||||||||
| b | 0.8414 | ||||||||||
| c | 0.8098 | ||||||||||
| d | 0.7722 | ||||||||||
| 6 | In repeated sampling of n = 25 graduates, what fraction of sample means would fall within±$4,500 from the population mean? | ||||||||||
| a | 0.9566 | ||||||||||
| b | 0.9606 | ||||||||||
| c | 0.9812 | ||||||||||
| d | 0.9920 | ||||||||||
| 7 | In repeated sampling of n = 25 graduates, the interval which contains the middle 95% of sample mean costs is: x̅₁ = ______, x̅₂ = ______ | ||||||||||
| a | $49,568 | $56,232 | |||||||||
| b | $50,266 | $55,534 | |||||||||
| c | $51,082 | $54,718 | |||||||||
| d | $52,009 | $53,791 | |||||||||
| 8 | In another region 10% of the x̅ values from samples of sizen = 25 areunder $49,500 and10% areover $55,500. From this sampling distribution information we can conclude that the population mean cost of a four-year college degree is μ = _______ and the population standard deviation is σ = ______. | ||||||||||
| a | $52,500 | $10,880 | |||||||||
| b | $52,500 | $11,719 | |||||||||
| c | $51,500 | $9,642 | |||||||||
| d | $51,500 | $9,146 | |||||||||
| Questions 9-11 are based on the following: | |||||||||||
| The mean annual Medicare spending per enrollee is $11,200 with a standard deviation of$3,100. Answer questions 9-12 based on the sampling distribution of x̅ for randome samples of sizen = 85 enrollees. | |||||||||||
| 9 | The fraction of sample means falling within ±$500 from the population mean is ______. | ||||||||||
| a | 0.8638 | ||||||||||
| b | 0.8288 | ||||||||||
| c | 0.7972 | ||||||||||
| d | 0.7656 | ||||||||||
| 10 | 95% of all x̅ values from samples of sizen = 85 deviate from the population mean of$11,200 by no more than ±$______. | ||||||||||
| a | $370 | ||||||||||
| b | $430 | ||||||||||
| c | $555 | ||||||||||
| d | $659 | ||||||||||
| 11 | In repeated sampling of n = 85 enrollees, the middle interval which contains themiddle 95% of sample mean spending is: x̅₁ = ______, x̅₂ = ______ | ||||||||||
| a | $10,896 | $11,504 | |||||||||
| b | $10,541 | $11,859 | |||||||||
| c | $10,239 | $12,161 | |||||||||
| d | $9,841 | $12,559 | |||||||||
| 12 | In the previous question, to reduce the margin of error such that the middle 95% of all sample means deviate from the population mean by no more than ±$250, the minimum sample size is ______. | ||||||||||
| a | 708 | ||||||||||
| b | 660 | ||||||||||
| c | 624 | ||||||||||
| d | 591 | ||||||||||
| The following binary data represent the students taking E270, where "1" is for students who are business majors and "0" for other majors. | |||||||||||
| 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | |||
| 1 | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | |||
| 1 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | |||
| 1 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | |||
| 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | |||
| 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | |||
| 1 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 0 | |||
| 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | |||
| 1 | 0 | 0 | 0 | 1 | 0 | 1 | 1 | 0 | |||
| 0 | 1 | 1 | 1 | 0 | 0 | 1 | 0 | 1 | |||
| 1 | 1 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | |||
| 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | |||
| 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | |||
| 0 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | |||
| 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | |||
| 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | |||
| 1 | 1 | 0 | 1 | 1 | 1 | 0 | 1 | 0 | |||
| 1 | 0 | 1 | 0 | 1 | 1 | 0 | 1 | 1 | |||
| 1 | 1 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | |||
| 1 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 0 | |||
| 1 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | |||
| 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | |||
| 13 | If we take repeated samples of size n = 40 from this population of E270 students, the expected value of sample proportions would be ______. | ||||||||||
| Use Excel!! | |||||||||||
| a | 0.812 | ||||||||||
| b | 0.737 | ||||||||||
| c | 0.681 | ||||||||||
| d | 0.641 | ||||||||||
| Questions 14-17 are based on the following information | |||||||||||
| Among all adult Indiana residents 83% are high school graduates. Answer questions 13-17 based on the sampling distribution of p̅ for random samples ofn = 400 Indiana residents. | |||||||||||
| 14 | The fraction of sample proportions obtained from samples of size n =400 that fall within±0.04 (4 percentage points) from the population proportion π is _________. | ||||||||||
| a | 0.9266 | ||||||||||
| b | 0.9342 | ||||||||||
| c | 0.9668 | ||||||||||
| d | 0.9826 | ||||||||||
| 15 | The fraction of sample proportions obtained from samples of size n = 800 that fall within±0.03 (3 percentage points) from π is _________. | ||||||||||
| a | 0.9762 | ||||||||||
| b | 0.9652 | ||||||||||
| c | 0.9500 | ||||||||||
| d | 0.9266 | ||||||||||
| 16 | The lower and upper ends of the interval which contains the middle 95% of all sample proportion obtained from samples of sizen = 500 are: p̅₁ = _______, p̅₂ = _______ | ||||||||||
| a | 0.780 | 0.880 | |||||||||
| b | 0.787 | 0.873 | |||||||||
| c | 0.791 | 0.869 | |||||||||
| d | 0.797 | 0.863 | |||||||||
| 17 | In the previous question, in order the obtain a margin of error of ±0.02 (MOE = 0.02) for the middle interval that contains the middle95% of all sample proportions, the minimum sample is: n = ______. | ||||||||||
| a | 1422 | ||||||||||
| b | 1356 | ||||||||||
| c | 1128 | ||||||||||
| d | 1064 | ||||||||||
| Questions 18-20 are based on the following information | |||||||||||
| Just before a mayoral election a local newspaper polls 450 voters in an attempt to predict the winner. Suppose that the candidate Johnny Comlately has52% of the votes in a two-way race. | |||||||||||
| 18 | What is the probability that the newspaper’s sample will predict defeat of Johnny Comlately? | ||||||||||
| a | 0.2485 | ||||||||||
| b | 0.2119 | ||||||||||
| c | 0.1977 | ||||||||||
| d | 0.1751 | ||||||||||
| 19 | In repeated polling of n = 450 voters, 95% of sample proportions would deviate fromπ = 0.52, in either direction, by no more than ______ (or _____ percentage points). | ||||||||||
| a | 0.046 | (4.6 percentage points) | |||||||||
| b | 0.041 | (4.1 percentage points) | |||||||||
| c | 0.037 | (3.7 percentage points) | |||||||||
| d | 0.035 | (3.5 percentage points) | |||||||||
| 20 | In order to make the probability of wrongly predicting defeat at most 5%, the minimum number of voters to be included in the sample should be n = ______? | ||||||||||
| a | 944 | ||||||||||
| b | 1068 | ||||||||||
| c | 1482 | ||||||||||
| d | 1679 | ||||||||||
-
Rating:
/5
Solution: Statistics-Last name: Enter your LETTER answers