1 / 11

Sample Variability

Sample Variability. Consider the small population of integers {0, 2, 4, 6, 8} It is clear that the mean, μ = 4. Suppose we did not know the population mean and wanted to estimate it with a sample mean with sample size 2. (We will use sampling with replacement)

dugan
Télécharger la présentation

Sample Variability

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Sample Variability Consider the small population of integers {0, 2, 4, 6, 8}It is clear that the mean, μ = 4. Suppose we did not know the population mean and wanted to estimate it with a sample mean with sample size 2. (We will use sampling with replacement) We take one sample and get sample mean, ū1 = (0+2)/2 = 1 and take another sample and get a sample mean ū2 = (4+6)/2 = 5. Why are these sample means different? Are they good estimates of the true mean of the population? What is the probability that we take a random sample and get a sample mean that would exactly equal the true mean of the population? Section 7.1, Page 137

  2. Sampling Distribution Each of these samples has a sample mean, ū. These sample means respectively are as follows: P(ū = 1) = 2/25 = .08P(ū = 4) = 5/25 = .20 Section 7.1, Page 138

  3. Sampling Distribution Shape is normal Mean of the sampling distribution = 4, the mean of the population Section 7.1, Page 138

  4. Sampling Distributions and Central Limit Theorem Alternate notation: Sample sizes ≥ 30 will assure a normal distribution. Section 7.2, Page 141

  5. Central Limit Theorem Section 7.2, Page 144

  6. Central Limit Theorem Section 7.2, Page 145

  7. Calculating Probabilities for the Mean Kindergarten children have heights that are approximately normally distributed about a mean of 39 inches and a standard deviation of 2 inches. A random sample of 25 is taken. What is the probability that the sample mean is between 38.5 and 40 inches? P(38.5 < sample mean <40) = NORMDIST 1 LOWER BOUND = 38.5 UPPER BOUND = 40 MEAN =39 ANSWER: 0.8881 Section 7.3, Page 147

  8. Calculating Middle 90% Kindergarten children have heights that are approximately normally distributed about a mean of 39 inches and a standard deviation of 2 inches. A random sample of 25 is taken. Find the interval that includes the middle 90% of all sample means for the sample of kindergarteners. Sampling Distribution NORMDIST 2 AREA FROM LEFT = 0.05 MEAN = 39 ANSWER: 38.3421 NORMDIST 2 AREA FROM LEFT = .95 MEAN = 39 ANSWER: 39.6579 The interval (38.3 inches, 39.7 inches) contains the middle 90% of all sample means. If we choose a random sample, there is a 90% probability that it will be in the interval. Section 7.3, Page 147

  9. Problems Problems, Page 149

  10. Problems Problems, Page 150

  11. Problems Problems, Page 151

More Related