Your SlideShare is downloading. ×
MTH120_Chapter8
MTH120_Chapter8
MTH120_Chapter8
MTH120_Chapter8
MTH120_Chapter8
MTH120_Chapter8
MTH120_Chapter8
MTH120_Chapter8
MTH120_Chapter8
MTH120_Chapter8
MTH120_Chapter8
MTH120_Chapter8
MTH120_Chapter8
MTH120_Chapter8
MTH120_Chapter8
MTH120_Chapter8
MTH120_Chapter8
MTH120_Chapter8
MTH120_Chapter8
MTH120_Chapter8
MTH120_Chapter8
MTH120_Chapter8
MTH120_Chapter8
Upcoming SlideShare
Loading in...5
×

Thanks for flagging this SlideShare!

Oops! An error has occurred.

×
Saving this for later? Get the SlideShare app to save on your phone or tablet. Read anywhere, anytime – even offline.
Text the download link to your phone
Standard text messaging rates apply

MTH120_Chapter8

269

Published on

Published in: Technology, Education
0 Comments
0 Likes
Statistics
Notes
  • Be the first to comment

  • Be the first to like this

No Downloads
Views
Total Views
269
On Slideshare
0
From Embeds
0
Number of Embeds
0
Actions
Shares
0
Downloads
8
Comments
0
Likes
0
Embeds 0
No embeds

Report content
Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
No notes for slide

Transcript

  • 1. 8- 1 Chapter Eight Sampling Methods and the Central Limit TheoremGOALSWhen you have completed this chapter, you will be able to:ONEExplain why a sample is the only feasible way to learn about apopulation.TWODescribe methods to select a sample.THREEDefine and construct a sampling distribution of the samplemean.FOURExplain the central limit theorem.McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 2. 8- 2 Chapter Eight continued Sampling Methods and the Central Limit TheoremGOALSWhen you have completed this chapter, you will be able to:FIVEUse the Central Limit Theorem to find probabilities of selectingpossible sample means from a specified population.McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 3. 8- 3 Why Sample the Population? The physical impossibility of checking all items in the population. The cost of studying all the items in a population. The sample results are usually adequate.McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 4. 8- 4 Why Sample the Population Contacting the whole population would often be time-consuming. The destructive nature of certain tests.McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 5. 8- 5 Probability Sampling A probability sample is a sample selected such that each item or person in the population being studied has a known likelihood of being included in the sample.McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 6. 8- 6 Methods of Probability Sampling  Simple Random Sample: A sample formulated so that each item or person in the population has the same chance of being included. Systematic Random Sampling: The items or individuals of the population are arranged in some order. A random starting point is selected and then every kth member of the population is selected for the sample.McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 7. 8- 7 Methods of Probability Sampling  Stratified Random Sampling: A population is first divided into subgroups, called strata, and a sample is selected from each stratum. Cluster Sampling: A population is first divided into primary units then samples are selected from the primary units.McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 8. 8- 8 Methods of Probability Sampling In nonprobability sample inclusion in the sample is based on the judgment of the person selecting the sample. The sampling error is the difference between a sample statistic and its corresponding population parameter.McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 9. 8- 9 Sampling Distribution of the Sample Means The sampling distribution of the sample mean is a probability distribution consisting of all possible sample means of a given sample size selected from a population.McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 10. 8- 10 EXAMPLE 1  The law firm of Hoya and Associates has five partners. At their weekly partners meeting each reported the number of hours they billed clients for their services last week. Partner Hours 1. Dunn 22 2. Hardy 26 3. Kiers 30 4. Malinowski 26 5. Tillman 22McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 11. 8- 11 Example 1  If two partners are selected randomly, how many different samples are possible? This is the combination of 5 objects taken 2 at a time. That is: 5! 5 C2 = = 10 2! (5 − 2)! There are a total of 10 different samples.McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 12. 8- 12 Example 1 continued Partners Total Mean 1,2 48 24 1,3 52 26 1,4 48 24 1,5 44 22 2,3 56 28 2,4 52 26 2,5 48 24 3,4 56 28 3,5 52 26 4,5 48 24McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 13. 8- 13 EXAMPLE 1 continued  Organize the sample means into a sampling distribution. Sample Frequency Relative Mean Frequency probability 22 1 1/10 24 4 4/10 26 3 3/10 28 2 2/10McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 14. 8- 14 EXAMPLE 1 continued Compute the mean of the sample means. Compare it with the population mean.  The mean of the sample means is 25.2 hours. 22(1) + 24(2) + 26(3) + 28(2) µX = = 25.2 10McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 15. 8- 15 Example 1 continued  The population mean is also 25.2 hours. 22 + 26 + 30 + 26 + 22 µ= = 25.2 5 Notice that the mean of the sample means is exactly equal to the population mean.McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 16. 8- 16 Central Limit Theorem For a population with a mean F and a variance 2 the sampling distribution of the means of all possible samples of size n generated from the population will be approximately normally distributed. The mean of the sampling distribution equal to T and the variance equal to 2/n.McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 17. 8- 17 Point Estimates  Examples of point estimates are the sample mean, the sample standard deviation, the sample variance, the sample proportion. A point estimate is one value ( a single point) that is used to estimate a population parameter.McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 18. 8- 18 Point Estimates If a population follows the normal distribution, the sampling distribution of the sample mean will also follow the normal distribution. To determine the probability a sample mean falls within a particular region, use: X −µ z= σ nMcGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 19. 8- 19 Point Estimates Ifthe population does not follow the normal distribution, but the sample is of at least 30 observations, the sample means will follow the normal distribution. To determine the probability a sample mean falls within a particular region, use: X −µ z= s nMcGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 20. 8- 20 Example 2 Suppose the mean selling price of a gallon of gasoline in the United States is $1.30. Further, assume the distribution is positively skewed, with a standard deviation of $0.28. What is the probability of selecting a sample of 35 gasoline stations and finding the sample mean within $.08?McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 21. 8- 21 Example 2 continued The first step is to find the z-values corresponding to $1.24 and $1.36. These are the two points within $0.08 of the population mean. X −µ $1.38 − $1.30 z= = = 1.69 s n $0.28 35McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 22. 8- 22 Example 2 continued X −µ $1.22 − $1.30z= = = −1.69 s n $0.28 35McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 23. 8- 23 Example 2 continued Next we determine the probability of a z-value between -1.69 and 1.69. It is: P (−1.69 ≤ z ≤ 1.69) = 2(.4545) = .9090 We would expect about 91 percent of the sample means to be within $0.08 of the population mean.McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.

×