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Chapter 8

i. n. g. S. m. l. a. p. Methods. &. Central Limit Theorem. Chapter 8. 1. 2. 3. 4. 5. Chapter Goals. When you have completed this chapter, you will be able to:. Explain under what conditions sampling is the proper way to learn something about a population.

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Chapter 8

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  1. i n g S m l a p Methods & Central Limit Theorem Chapter 8

  2. 1. 2. 3. 4. 5. Chapter Goals When you have completed this chapter, you will be able to: Explain under what conditions sampling is the proper way to learn something about a population. Describemethods for selecting a sample. Defineandconstructa sampling distribution of the sample mean. Explainthe central limit theorem. Use the central limit theorem to find probabilities of selecting possible sample means from a specified population.

  3. Sampling We use sampleinformationto make decisions or inferences about the population. Two KEY steps: Choice of a proper method for selecting sample data & 2. Proper analysis of the sample data (more later) KEY 1.

  4. Sampling

  5. Sampling KEY 1. If the proper method for selecting the sample is NOT MADE … the SAMPLE will not be truly representative of the TOTAL Population! … andwrong conclusions can be drawn!

  6. …of the physical impossibilityof checking all items in the population, and, also, itwould be too time-consuming …the studying of all the items in a population would NOT becost effective …the sample results are usually adequate …the destructive nature of certain tests Why Sample the Population? Because…

  7. Sampling Techniques withReplacement withoutReplacement Each data unit in the population is allowed to appear in the sample nomore than once Each data unit in the population is allowed to appear in the sample more than once Non-Probability Sampling Probability Sampling Each data unit in the population has a known likelihood of being included in the sample Does not involve random selection; inclusion of an item is based onconvenience

  8. Sampling Methods ...each item(person) in the population has an equal chance of being included …items(people) of the population are arranged in some order. A random starting point is selected, and then every kthmember of the population is selected for the sample …a population is first divided into subgroups, called strata, and a sample is selected from each strata …a population is first divided into primary units, and samples are selected from each unit

  9. Sampling Terminology “Sampling error” … is the difference between a sample statistic and its corresponding population parameter “Sampling distribution of the sample mean” … is a probability distribution consisting of all possible sample meansof agiven sample size selected from a population Example

  10. Sampling The law firm of Hoya and Associates has five partners. At their weekly partners meeting each reported the number of hours they billed their clients last week: Example If two partners are selected randomly… how many different samples are possible?

  11. Sampling 5 If two partners are selected randomly… how many different samples are possible? Objects …taken 2 at a time …for a Total of 10 Samples! Using 5C2…

  12. Sampling 5 If two partners are selected randomly… how many different samples are possible? Objects 5C2 = 5! = 2! (5 – 2!) = 10 Samples

  13. Sampling 22 (22+22)/2 = 28 (26+30)/2 = (26+26)/2 = 26 (26+22)/2 = 24 (30+26)/2 = 28 (30+22)/2 = 26 (26+22)/2 = 24 24 (22+26)/2 = 26 (22+30)/2 = 24 (22+26)/2 =

  14. Sampling Example …continued Organize the sample means into a Sampling Distribution 10 Samples 22 1 1/10 24 4 4/10 3 26 3/10 28 2 2/10

  15. Sampling m 22 1 X 24 4 3 26 28 2 Example …continued Compute the mean of the sample means. Compare it with the population mean = 22(1)+ 24(4)+ 26(3) + 28(2) 10 = 25.2

  16. Sampling Note + + + + 22 26 30 26 22 m 5 = Example …continued The population mean is also the same as the sample means…25.2 hours! = 25.2

  17. Sampling Mean (µx ) Standard Deviation (standard error of the mean) / n X CentralLimitTheorem The sampling distributionof the means of all possible samples of sizen generated from the population will beapproximately normally distributed! Sampling Distributions: µ Variance 2/n

  18. samplemean samplestandard deviation samplevariance sampleproportion Sampling PointEstimates A point estimate is one value ( a single point) that is used toestimate a population parameter Examples More

  19. Sampling - m X s n - m X Z = Z = s n PointEstimates Population follows… the normal distribution Population does NOTfollow… the normal distribution The sampling distribution of the sample meansalso follows the normal distribution If the sample is of at least 30 observations, the sample WILL follow the normal distribution Probability of a sample mean falling withina particular region, use: Probability of a sample mean falling withina particular region, use:

  20. Sampling CentralLimitTheorem Chart 8 – 6 Results for Several Populations

  21. What is the standard error of the mean? /n Formula  40 = 80 / Using the Sampling Distribution of the Sample Mean Data… Suppose it takes an average of 330 minutes for taxpayers to prepare, copy, and mail an income tax return form. A consumer watchdog agency selects a random sampleof 40 taxpayers and finds the standard deviation of the time needed is 80 minutes = 12.6

  22. Data… Suppose it takes an average of 330 minutes for taxpayers to prepare, copy, and mail an income tax return form. A consumer watchdog agency selects a random sampleof 40 taxpayers and finds the standard deviation of the time needed is 80 minutes. A What is the likelihood the sample mean is greater than 320 minutes? nswer… Using the Sampling Distribution of the Sample Mean

  23. Data… * average of 330 minutes *random sampleof 40 * standard deviation is 80 minutes 1 a1 - - m 320 330 X = = z 80 40 s n What is the likelihood the sample mean is greater than 320 minutes? Using the Sampling Distribution of the Sample Mean Formula = 0.79 330 320

  24. Data… * average of 330 minutes *random sampleof 40 * standard deviation is 80 minutes 2 a1 What is the likelihood the sample mean is greater than 320 minutes? Look up 0.79 in Table Using the Sampling Distribution of the Sample Mean Required Area = 0.2852 + .5 = 0.7852 a1 =0.2852 330 320

  25. Sampling Distribution of Proportion The normaldistribution (a continuous distribution) yields a good approximation of thebinomialdistribution (adiscrete distribution) forlarge values of n. Use when np and n(1- p ) are both greater than 5!

  26. m Mean and Variance of a Binomial Probability Distribution np = s 2 m Formula s 2 p - np Formula ( 1 ) =

  27. A Based on this information, what can we say about the company’s claim? Sampling Distribution of Proportion A multinational company claims that 55%of its employees are bilingual. To verify this claim, a statistician selected a sample of 60 employees of the company using simple random sampling and found 48% to be bilingual. The sample size is big enough to use the normal approximation with a mean of .55 and a standard deviation of (.55)(.45)/60 = 0.064 np = 60(.55) = 33 n(1- p ) = 60(.45) = 27

  28. …continued - m X = z s 1 2 A a1 Look up 1.09 in Table Sampling Distribution of Proportion Formula Z= (0.48 -0.55) / 0.064 Z= -1.09 .55 .48 a1 =0.3621 Required Area = .5 – 0.3621 = 0.1379 or 14%

  29. …continued - m X = z s 1 2 A Look up 1.09 in Table Sampling Distribution of Proportion Formula Conclusion There is approximately a 14% chance that the company’s claim is true, based on this sample. Z= (0.48 -0.55) / 0.064 Z= -1.09 a =0.3621 Required Area = .5 – 0.3621 = 0.1379 or 14%

  30. Sampling Distribution of Mean Suppose the mean selling price of a litre ofgasoline in Canada is $.659. Further, assume the distribution is positively skewed, with a standarddeviationof $0.08. What is the probability of selecting a sample of 35 gasoline stations and finding the sample mean within $.03 of the population mean?

  31. - - $. $. 689 629 $. $. 659 659 = Data… mean selling price is $.659 SDof $0.08 Sample of 35 gasoline stations Probability ofsample meanwithin$.03? $ 0 . 08 35 $ 0 . 08 35 - m - m X X z z = = 2 1 Find the z-scores for .659 +/- .03 s n s n 1 Step = 2.22 = -2 . 22 = Sampling Distribution of Mean i.e. 0.629 and .689 .629 .689

  32. We would expect about 97% of the sample means to be within $0.03 of the population mean. Data… mean selling price is $.659 SDof $0.08 Sample of 35 gasoline stations Probability ofsample meanwithin$.03? 2 Find areas from table… z Step = -2.22 1 Required A= .9736 z = 2.22 2 Sampling Distribution of Mean a1 = .4868 a2 = .4868

  33. www.mcgrawhill.ca/college/lind for quizzes extra content data sets searchable glossary access to Statistics Canada’s E-Stat data …and much more! Test your learning… Click on… Online Learning Centre

  34. This completes Chapter 8

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