1 / 28

Sampling Distribution of a Sample Mean

Sampling Distribution of a Sample Mean. Lecture 28 Section 8.4 Fri, Oct 28, 2005. Sampling Distribution of the Sample Mean. Sampling Distribution of the Sample Mean – The distribution of sample means over all possible samples of the size n from that population. With or Without Replacement?.

cate
Télécharger la présentation

Sampling Distribution of a Sample Mean

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. Sampling Distribution of a Sample Mean Lecture 28 Section 8.4 Fri, Oct 28, 2005

  2. Sampling Distribution of the Sample Mean • Sampling Distribution of the Sample Mean– The distribution of sample means over all possible samples of the size n from that population.

  3. With or Without Replacement? • If the sample size is small in relation to the population size (< 5%), then it does not matter whether we sample with or without replacement. • The calculations are simpler if we sample with replacement. • In any case, we are not going to worry about it.

  4. Example • Suppose a population consists of the numbers {1, 2, 3}. • If we select samples of size n = 3, find the sampling distribution ofx. • Draw a tree diagram showing all possibilities.

  5. The Tree Diagram 1 1 1 2 4/3 3 5/3 1 4/3 2 1 2 5/3 3 2 1 5/3 2 2 3 7/3 3 1 4/3 1 2 5/3 3 2 1 5/3 2 2 2 2 3 7/3 1 2 3 2 7/3 3 8/3 1 5/3 1 2 2 3 7/3 3 1 2 2 2 7/3 3 8/3 1 7/3 3 2 8/3 3 3

  6. The Sampling Distribution • Of the 27 possible samples (equally likely), we find that • 1 sample has an average of 1. • 3 samples have an average of 4/3 = 1.333. • 6 samples have an average of 5/3 = 1.667. • 7 samples have an average of 2. • 6 samples have an average of 7/3 = 2.333. • 3 samples have an average of 8/3 = 2.667. • 1 sample has an average of 3.

  7. The Sampling Distribution • The sampling distribution ofx is

  8. Probability Histogram 7/27 6/27 5/27 4/27 3/27 2/27 1/27 1 2 3

  9. Probability Histogram 7/27 6/27 5/27 4/27 3/27 2/27 1/27 1 2 3

  10. Experiment • We may simulate this on the TI-83 using the function randInt. • The expression randInt(1, 3, 3) will select 3 random numbers from the set {1, 2, 3}, with replacement, and put them in a list.

  11. Experiment • For example, randInt(1, 3, 3) gives {3, 3, 2} which has a sample mean of 2.667. • Generate a list of 100 such sample means, each from a sample of size 3.

  12. Experiment 50 40 30 20 10 1 2 3

  13. Fit a Normal Curve 50 40 30 20 10 1 2 3

  14. Observations • The distribution appears to be centered at . • The distribution appears to be approximately normal (maybe).

  15. Experiment • Now generate a list of 100 sample means, each from a sample of size 30. • Draw the distribution.

  16. Experiment 50 40 30 20 10 1 2 3

  17. Fit a Normal Curve 50 40 30 20 10 1 2 3

  18. The Central Limit Theorem • Begin with a population that has mean  and standard deviation . • For sample size n, the sampling distribution of the sample mean is approximately normal with

  19. The Central Limit Theorem • The approximation gets better and better as the sample size gets larger and larger. • For many populations, the distribution is almost exactly normal when n  10. • For almost all populations, if n 30, then the distribution is almost exactly normal.

  20. The Central Limit Theorem • If the original population is exactly normal, then the sampling distribution of the sample mean is exactly normal for any sample size. • This is all summarized on pages 536 – 537.

  21. Example • Suppose the heights of all adult males has • Mean 68 inches. • Standard deviation 3.6 inches.

  22. Example • The set of sample means from all possible samples of size n = 4 will have • Mean 68 inches. • Standard deviation 3.6/4 = 0.18 inches. • But the distribution will (probably) not be normal.

  23. Example • The set of sample means from all possible samples of size n = 100 will have • Mean 68 inches. • Standard deviation 3.6/100 = 0. 36 inches. • And the distribution will be normal.

  24. Example • If we collect a sample of 100 adult males from this population, what is the probability that their average height will be at least 67 inches?

  25. Example • The sampling distribution ofx • Is normal (because n 30), • Has mean 68, • Has standard deviation 0. 36. • Therefore, P(x > 67) is given by normalcdf(67, E99, 68, 0.36), which is 0.9973.

  26. Let’s Do It! • Let’s Do It! 8.9, p. 541 – Probability of Accepting the Shipment. • Let’s Do It! 8.10, p. 543 – Mean Grocery Expenditure.

  27. Estimating the Population Mean • Example 8.12, p. 543 – Estimating the Population Mean Grocery Expenditure. • The sampling distribution ofx is approximately normal with • x = $60. • x = $35/100 = $3.50. • Based on the Empirical Rule, 95% of all samples have a mean within $7.00 of $60, that is, between $53 and $67.

  28. Estimating the Population Mean • Suppose that we do not know the true population mean for grocery expenditures, but our sample mean is $58. • What is our best estimate of the true mean? • Point estimate: $58. • Interval estimate: Allow 2 standard deviations either way: $58  $7.

More Related