...The probability distribution of the population data is called the population distribution. Tables 7.1 and 7.2 on page 309 of the text provide an example of such a distribution. The probability distribution of a sample statistic is called its sampling distribution. Tables 7.3 to 7.5 on page 311 of the text provide an example of the sampling distribution of the sample mean. 1. Sampling error is the difference between the value of the sample statistic and the value of the corresponding population parameter, assuming that the sample is random and no nonsampling error has been made. Example 7–1 on page 312 of the text exhibits sampling error. Sampling errors occur only in sample surveys. 2. Nonsampling errors are errors that may occur during collection, recording, and tabulation of data. The second part of Example 7–1 on pages 312 and 313 of the text exhibits nonsampling error. Nonsampling errors occur both in sample surveys and censuses. 3. a. [pic] b. [pic][pic] Sampling error = [pic] c. Liza’s incorrect [pic][pic] [pic] Sampling error (from part b) = –1.83 Nonsampling error [pic] d. | Sample | [pic] |[pic] | |15, 13, 8, 17 |13.25 |.92 | |15, 13, 8, 9 |11.25 |–1.08 | |15, 13, 17, 9 |13.50 |1.17 | |15, 8, 17, 9 |12...
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...Mercedes and BMW have been competing head-to-head for market share in the luxury-car market for more than four decades. Back in 1959, BMW (Bayerische Motoren Werke) almost went bankrupt and nearly sold out to Daimler-Benz, the maker of Mercedes-Benz cars. BMW was able to recover to the point that in 1992 it passed Mercedes in worldwide sales. Among the reasons for BMWs success was its ability to sell models that were more luxurious than previous models but still focused on consumer quality and environmental responsibility. In particular, BMW targeted its sales pitch to the younger market, whereas Mercedes retained a more mature customer base. In response to BMWs success, Mercedes has been trying to change their image by launching several products in an effort to attract younger buyers who are interested in sporty, performance-oriented cars. BMW, influenced by Mercedes, is pushing for more refinement and comfort. In fact, one automotive expert says that Mercedes wants to become BMW, and vice versa. However, according to one recent automotive expert, the focus is still on luxury and comfort for Mercedes while BMW focuses on performance and driving dynamics. Even though each company produces many different models, two relatively comparable coupe automobiles are the BMW 3 Series Coupe 335i and the Mercedes CLK350 Coupe. In a recent year, the national U.S. market price for the BMW 3 Series Coupe 335i was $39,368 and for the Mercedes CLK350 Coupe was $44,520. Gas mileage for both...
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...City Cents City Cents 1 Vancouver, BC 5.83 15 Edmonton, AB 10.12 2 Saskatoon, SK 6.77 16 Calgary, AB 15.32 3 Abbotsford, BC 6.32 17 Oshawa, ON 6.23 4 St. John's, NL 9.68 18 Halifax, NS 8.36 5 Winnipeg, MB 5.41 19 Moncton, NB 8.83 6 Sudbury, ON 5.83 20 Price George, AB 6.19 7 Ottawa, ON 8.08 21 Brantford, ON 6.65 8 Sherbrooke, QC 6.56 22 St-Jean, QC 5.25 9 Quebec, QC 5.74 23 Regina, SK 7.23 10 Whitehorse, YT 6.06 24 Montreal, QC 6.74 11 Toronto, ON 8.62 25 Yellowknife, NWT 6.43 12 Saguenay, QC 6.20 26 Kewlowna, BC 6.83 13 Hamilton, ON 6.91 27 Windsor, ON 6.69 14 Charlottetown, PE 10.60 28 Thunder Bay, ON 6.12 1. Determining the mean, variance and standard deviation of the provided sample: Mean: 7.3429 Variance: 4.4322 Standard Deviation: 2.1053 The mean, variance, and standard deviation for electricity prices of this sample is 7.3429, 4.4322 & 2.1053 cents respectively. 2. 95% confidence interval for the mean of the population utility rates: Confidence Interval: 0.77979 We're 95% confident that the true mean falls somewhere between 6.5631and 8.1227. 3. 90% confidence interval for the mean of the population utility rates, asssuming population standard deviation 2.09 Confidence Interval: 0.6497 ...
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...4 problem set 1: Normal Probability Distributions Page.285 Ex 6,8,10,12 6. x = 80, z=80-10015 = -1.33 z= 0.0918 1-0.0918 = 0.9082 8. x = 110, z=110-10015 = 0.67 z= 0.7486 z= 75-10015 = -1.67 z= 0.0475 0.7486-0.0475= 0.7011 (shaded area) 10. z= 0.84 (shaded) z= -0.84 x= 100+(-0.84∙15) = 87 (rounded) 12. . z= 2.33 x= 100+(2.33∙15) = 135 (rounded) Page 288 Ex 34 34.Appendix B Data Set: Duration of Shuttle Flights a. Find the mean and standard deviation, and verify that the data have a distribution that is roughly normal. Mean= 25317115 = 220.15 Standard Deviation=115253172-(25317)2115(115-1) = 86 (rounded) The normal distribution is 115 b. Treat the statistics from part (a) as if they are population parameters and assume a normal distribution to find the values of the quartiles 1,2 and 3. Mean= 220.15 Standard Deviation= 86 Q1 = 220.5 + (-0.67 ∙ 86)= 162.53 Q2= 220.5 + (0.00 ∙ 86) = 220.5 Q3=220.5 + (0.67 ∙ 86) = 277.77 Page.300 Ex 20 Quality Control: Sampling Distribution of Proportion after constructing a new manufacturing machine. 5 prototype integrated circuit chips are produced and it is found that 2 are defective (D) and 3 are acceptable (A). Assume that two if the chips are randomly selected with replacement from this population a. After identifying the 25 different possible samples, find the proportion of defects in each of them, then use a table to describe the sampling distribution of the proportions...
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...standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution...
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...Lecture 7. Sampling Distributions. Statistical Inference: Using statistics calculated from samples to estimate the values of population parameters. Select Random Sample Sample for (statistic) Calculate to estimate Becomes Population Parameter. BASIC Example: Soft Drink Bottler μ=600, σ=10. Normal Distribution. What is P(X>598)? p(x<598) . Sampling Dist.of the Mean – Distribution of all Possible Sample Means if you select a sample of a certain size. μX= μ. μ = i=1NXiN (formula for mean) . σ = i=1N(Xi-μ)2N Although you do not know how close the sample mean of any particular sample selected comes to the pop mean, you know that the mean of all possible sample means that could have been selected = the pop mean. Standard error is calculating the probability of a certain amount of error. EXAMPLE: Standard error is . As n increases decreases. CENTRAL LIMIT THEOREM: Regardless of shape of individual values in distribution; as long as sample size is large enough the sampling distribution of the mean will be approximately normally distributed with μX= μ and σX= σ . For most population distributions n ≥ 30 will be large enough. For symmetric population distributions, n ≥ 5 is sufficient. For normal population distributions, the sampling distribution of the mean is always normally distributed EXAMPLE: SAMPLING DISTRIBUTION OF THE PROPORTION. π is the proportion of items in the population with a characteristic of interest. p is the sample proportion...
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...easier and more efficient. If you had a good introductory quantitative methods course, and like the text that was used, you may want to refer to it whenever you feel in need of a refresher. If you feel uncomfortable with standard quantitative texts, this reference is for you. Our aim is to present the essential quantitative concepts and methods in a self-contained, nontechnical, and intuitive way. Our approach is structured in line with requirements for the CFA program. The material included is relevant to investment management by the ICFA, the Institute of Chartered Financial Analysts. We hope you find this appendix helpful. Use it to make your venture into investments more enjoyable. 1006 Appendix A 1007 A.1 PROBABILITY DISTRIBUTIONS Statisticians talk about “experiments,” or “trials,” and refer to possible outcomes as “events.” In a roll of a die, for example, the “elementary events” are the numbers 1 through 6. Turning up one side represents the most disaggregate mutually exclusive outcome. Other events are compound, that is, they consist of more than one elementary event, such as the result “odd number” or “less than 4.” In this case “odd” and “less than 4” are not mutually exclusive. Compound events can be mutually exclusive outcomes, however, such as “less than 4” and “equal to or greater than 4.”...
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.............................................................. 1 1 Introduction ............................................................................................................................. 2 2 Methodology ........................................................................................................................... 4 2.1 ARCH and GARCH ......................................................................................................... 4 2.2 EGARCH ......................................................................................................................... 4 2.3 GJR-GARCH ................................................................................................................... 5 2.4 Distributions ..................................................................................................................... 5 2.5 Information criterions ....................................................................................................... 6 3 Data ......................................................................................................................................... 7 3.1 Descriptive statistics ......................................................................................................... 7 4 Results ..................................................................................................................................... 9 4.1 Evaluation of the models ....................................
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...Some basic relationships of Probability Read ASW Chapter 4 or LR Chapter 4 | 2 | Theories of Probability - Classical theory, Relative Frequency theory, Axioms, Addition rule, Multiplication rule, Rule of at least one, Concept of Expected number of Success – Numerical Problems & Applications Case Problem: Hamilton County JudgesSIP: Morton International - Chicago, Illinois Read ASW Chapter 4 or LR Chapter 4 | 3 | Bayes Theorem – Theory, Problems & Applications, Probability revision using tabular approach Read ASW Chapter 4 or LR Chapter 4 | 4 | Probability Distribution - Meaning of Probability Distribution, Type of Probability Distribution, Need (Application) for Probability DistributionRead ASW Chapter 5 or LR Chapter 5 | 5 | Discrete Probability Distribution - Binomial Distribution – Applications, Numerical Problems, Excel & SPSS functions, Poisson Distribution...
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...Z-Scores A z-score, which is symbolized as z, is the statistic that relates the distance a score is relative to its mean when measured in standard deviation (SD) units (Heiman, 2012). The importance of a z-score is that it enables one to analyze data relative to scores. Z-Scores allow us to determine whether a particular score is equal to the mean, below the mean or above the mean of a bunch of scores, and how far a particular score is away from the mean. Evaluating the scores of Eric’s to determine different z-scores, we use the following computations that he computed where it takes a mean of 17 minutes with a standard deviation of 3 minutes to drive from home, park the car, and walk to his job. , Next we determine the z-score relative to the mean and the SD, to analyze the difference in time to accomplish these steps on different days. One day it took Eric 21 minutes to get to work, and computing the means from the minutes by the SD results in a z of +1.33, which tells you that Eric's time to get to work is 1.33 standard deviations from the mean. The z is positive because it is above the mean, and demonstrates that it took longer for Eric to leave home, park his car and walk to his job when the raw value of time was 21 minutes. On another day in which it took Eric 12 minutes to get to work, the z value resulted -1.66, demonstrating that the time it took Eric to get to work was below the mean, or it took less time. Observing the time in comparison to the mean, it is obvious...
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...78583 * 2 × Error (b)n for Factor M 2 × 9.78586 * 2 × Error (a)n for Factor PP 2 × 4.96756 * 2 critical-t values → t at 2 and t at 4 df i.e. 4.303 and 2.776 * Could ask: do ANOVA and t-test, or ANOVA and interpret results from F; Standard error for the difference (a or b); Conclusion: levels differ/do not differ at 1% etc. NS 13 – Non-parametric tests * Parametric tests for data with normal distribution (t, F or X2 distribution) * Non-parametric tests for * Categorical data, * Quantitative data divided into class intervals, * Small data sets, * Data sets without repetition of the TMTs. * Non- parametric tests * Medians, not Means * Usually rank your data * Single sample: * Sign test (No assumptions about distribution) * Rank test (assumes data comes from symmetrical distribution) * Wilcoxon’s symmetry test * For 2 independent samples * Mann-Whitney U test (Assumes distributions have same shape and equal...
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...CHAPTER 6: THE NORMAL DISTRIBUTION AND OTHER CONTINUOUS DISTRIBUTIONS 1. In its standardized form, the normal distribution a) has a mean of 0 and a standard deviation of 1. b) has a mean of 1 and a variance of 0. c) has an area equal to 0.5. d) cannot be used to approximate discrete probability distributions. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: standardized normal distribution, properties 2. Which of the following about the normal distribution is NOT true? a) Theoretically, the mean, median, and mode are the same. b) About 2/3 of the observations fall within 1 standard deviation from the mean. c) It is a discrete probability distribution. d) Its parameters are the mean, , and standard deviation, . ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: normal distribution, properties 3. If a particular batch of data is approximately normally distributed, we would find that approximately a) 2 of every 3 observations would fall between 1 standard deviation around the mean. b) 4 of every 5 observations would fall between 1.28 standard deviations around the mean. c) 19 of every 20 observations would fall between 2 standard deviations around the mean. d) all of the above ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: normal distribution, properties 4. For some positive value of Z, the probability that a standardized normal variable is between 0 and Z is 0.3770. The value of Z is a) 0.18. b) 0.81. c) 1.16. d) 1...
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...Case Study#2 The XYZ Company Katharine Rally is the vice president of operations for the XYZ Company. She oversees operations at a plant that manufactures components for hydraulic systems. Katharine is concerned about the plant’s present production capability. She has reduced the decision situation to three alternatives. The first alternative, which is fully automation, would result in significant changes in present operations. The second alternative, which is semi-automation, involves fewer changes in present operations. The third alternative is to make no changes (do nothing). As a manager of the plant management team, you have been assigned the task of analyzing the alternatives and recommending a course of action. The capital investment and annual revenue for the first two alternatives are shown in the following table: |Alternative |Capital Investment |Future Sales |Annual Revenue | |A |$300,000 |Good |$250,000 | | | |Average |$100,000 | | | |Poor |$50,000 | |B |$85,000 |Good |$100,000 | | | ...
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...MAT 540 Quiz Answers 1) Deterministic techniques assume that no uncertainty exists in model parameters. Answer: TRUE Diff: 1 Page Ref: 489 Main Heading: Types of Probability Key words: deterministic techniques 2) Probabilistic techniques assume that no uncertainty exists in model parameters. Answer: FALSE Diff: 1 Page Ref: 489 Main Heading: Types of Probability Key words: probabilistic techniques 3) Objective probabilities that can be stated prior to the occurrence of an event are classical or a priori. Answer: TRUE Diff: 2 Page Ref: 489 Main Heading: Types of Probability Key words: objective probabilities, classical probabilities 4) Objective probabilities that are stated after the outcomes of an event have been observed are relative frequencies. Answer: TRUE Diff: 2 Page Ref: 489 Main Heading: Types of Probability Key words: relative frequencies 5) Relative frequency is the more widely used definition of objective probability. Answer: TRUE Diff: 1 Page Ref: 490 Main Heading: Types of Probability Key words: relative frequencies 6) Subjective probability is an estimate based on personal belief, experience, or knowledge of a situation. Answer: TRUE Diff: 2 Page Ref: 490 Main Heading: Types of Probability Key words: subjective probability 7) An experiment is an activity that results in one of several possible outcomes. Answer: TRUE Diff: 1 Page Ref: 491 Main Heading: Fundamentals of Probability Key words: experiment ...
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...University of Phoenix Material Distribution, Hypothesis Testing, and Error Worksheet Answer the following questions. Questions that are answered without the work will not receive full credit. When a question says explain or describe, please DO NOT copy word for word from a reference. You need to explain the concept so I know you understand what it means. For questions requiring material from Statdisk, make sure to turn labels on, take a screen capture (CTRL-Print Screen on most Windows-based computers), and paste the image into the worksheet. Crop the image as appropriate. 1. Describe a normal distribution in no more than 100 words (5 point). Answer: A normal distribution is a continuous random variable distribution with a bell shape, and has only two parameters: the mean, and the variance. A normal distribution can be represented by the formula: y=e^(-1/2)(x-μ/σ)/(σ√2pi). The mean can be any positive number and variance can be any positive number, so there are an infinite number of normal distributions. The shape of the distribution when graphed is symmetrical and bell-shaped. Use this information to answer questions 2-4. Following a brushfire, a forester takes core samples from the ten surviving Bigcone Douglas-fir trees in a test plot within the burn area, and a dendrochronologist determines the age of the source trees to be as follows (in years): 15 38 48 67 81 83 94 102 135 167 2. Construct a normal quantile plot in Statdisk, show the regression...
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