1 / 63

Chapter Seven Introduction to Sampling Distributions

Understandable Statistics S eventh Edition By Brase and Brase Prepared by: Lynn Smith Gloucester County College. Chapter Seven Introduction to Sampling Distributions. Review of Statistical Terms. Population Sample Parameter Statistic. Population.

marydiaz
Download Presentation

Chapter Seven Introduction to Sampling Distributions

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. Understandable StatisticsSeventh EditionBy Brase and BrasePrepared by: Lynn SmithGloucester County College Chapter Seven Introduction to Sampling Distributions

  2. Review of Statistical Terms • Population • Sample • Parameter • Statistic

  3. Population the set of all measurements (either existing or conceptual) under consideration

  4. Sample a subset of measurements from a population

  5. Parameter a numerical descriptive measure of a population

  6. Statistic a numerical descriptive measure of a sample

  7. We use a statistic to make inferences about a population parameter.

  8. Principal types of inferences • Estimate the value of a population parameter • Formulate a decision about the value of a population parameter

  9. Sampling Distribution a probability distribution for the sample statistic we are using

  10. Example of a Sampling Distribution Select samples with two elements each (in sequence with replacement) from the set {1, 2, 3, 4, 5, 6}.

  11. Constructing a Sampling Distribution of the Mean for Samples of Size n = 2 List all samples and compute the mean of each sample. sample: mean: sample: mean {1,1} 1.0 {1,6} 3.5 {1,2} 1.5 {2,1} 1.5 {1,3} 2.0 {2,2} 4 {1,4} 2.5 … ... {1,5} 3.0 There are 36 different samples.

  12. Sampling Distribution of the Mean p 1.0 1/36 1.5 2/36 2.0 3/36 2.5 4/36 3.0 5/36 3.5 6/36 4.0 5/36 4.5 4/36 5.0 3/36 5.5 2/36 6.0 1/36

  13. Sampling Distribution Histogram | | | | | | | | | | | | 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6

  14. Let x be a random variable with a normal distribution with mean  and standard deviation . Let be the sample mean corresponding to random samples of size n taken from the distribution .

  15. Facts about sampling distribution of the mean:

  16. Facts about sampling distribution of the mean: • The distribution is a normal distribution.

  17. Facts about sampling distribution of the mean: • The distribution is a normal distribution. • The mean of the distribution is  (the same mean as the original distribution).

  18. Facts about sampling distribution of the mean: • The distribution is a normal distribution. • The mean of the distribution is  (the same mean as the original distribution). • The standard deviation of the distribution is  (the standard deviation of the original distribution, divided by the square root of the sample size).

  19. We can use this theorem to draw conclusions about means of samples taken from normal distributions. If the original distribution is normal, then the sampling distribution will be normal.

  20. The Mean of the Sampling Distribution

  21. The mean of the sampling distribution is equal to the mean of the original distribution.

  22. The Standard Deviation of the Sampling Distribution

  23. The standard deviation of the sampling distribution is equal to the standard deviation of the original distribution divided by the square root of the sample size.

  24. The time it takes to drive between cities A and B is normally distributed with a mean of 14 minutes and a standard deviation of 2.2 minutes. • Find the probability that a trip between the cities takes more than 15 minutes. • Find the probability that mean time of nine trips between the cities is more than 15 minutes.

  25. Find this area 14 15 Mean = 14 minutes, standard deviation = 2.2 minutes • Find the probability that a trip between the cities takes more than 15 minutes.

  26. Mean = 14 minutes, standard deviation = 2.2 minutes • Find the probability that mean time of nine trips between the cities is more than 15 minutes.

  27. Find this area 14 15 Mean = 14 minutes, standard deviation = 2.2 minutes • Find the probability that mean time of nine trips between the cities is more than 15 minutes.

  28. What if the Original Distribution Is Not Normal? Use the Central Limit Theorem.

  29. Central Limit Theorem If x has any distribution with mean  and standard deviation , then the sample mean based on a random sample of size n will have a distribution that approaches the normal distribution (with mean  and standard deviation  divided by the square root of n) as n increases without bound.

  30. How large should the sample size be to permit the application of the Central Limit Theorem? In most cases a sample size of n = 30 or more assures that the distribution will be approximately normal and the theorem will apply.

  31. Central Limit Theorem

  32. Central Limit Theorem • For most x distributions, if we use a sample size of 30 or larger, the distribution will be approximately normal.

  33. Central Limit Theorem • The mean of the sampling distribution is the same as the mean of the original distribution. • The standard deviation of the sampling distribution is equal to the standard deviation of the original distribution divided by the square root of the sample size.

  34. Central Limit Theorem Formula

  35. Central Limit Theorem Formula

  36. Central Limit Theorem Formula

  37. Application of the Central Limit Theorem Records indicate that the packages shipped by a certain trucking company have a mean weight of 510 pounds and a standard deviation of 90 pounds. One hundred packages are being shipped today. What is the probability that their mean weight will be: a. more than 530 pounds? b. less than 500 pounds? c. between 495 and 515 pounds?

  38. Are we authorized to use the Normal Distribution? Yes, we are attempting to draw conclusions about means of large samples.

  39. Applying the Central Limit Theorem What is the probability that their mean weight will be more than 530 pounds? Consider the distribution of sample means: P( x > 530): z = 530 – 510 = 20 = 2.22 9 9 P(z > 2.22) = _______ .0132

  40. Applying the Central Limit Theorem What is the probability that their mean weight will be less than 500 pounds? P( x < 500): z = 500 – 510 = –10 = – 1.11 9 9 P(z < – 1.11) = _______ .1335

  41. Applying the Central Limit Theorem What is the probability that their mean weight will be between 495 and 515 pounds? P(495 < x < 515) : for 495: z = 495 – 510 = 15 =  1.67 9 9 for 515: z = 515 – 510 = 5 = 0.56 9 9 P(  1.67 < z < 0.56) = _______ .6648

  42. Sampling Distributions for Proportions Allow us to work with the proportion of successes rather than the actual number of successes in binomial experiments.

  43. Sampling Distribution of the Proportion • n= number of binomial trials • r = number of successes • p = probability of success on each trial • q = 1 - p = probability of failure on each trial

  44. Sampling Distribution of the Proportion If np > 5 and nq > 5 then p-hat = r/n can be approximated by a normal random variable (x) with:

  45. The Standard Error for

  46. Continuity Correction • When using the normal distribution (which is continuous) to approximate p-hat, a discrete distribution, always use the continuity correction. • Add or subtract 0.5/n to the endpoints of a (discrete) p-hat interval to convert it to a (continuous) normal interval.

  47. If n = 20, convert a p-hat interval from 5/8 to 6/8 to a normal interval. Note: 5/8 = 0.625 6/8 = 0.75 So p-hat interval is 0.625 to 0.75. Since n = 20, .5/n = 0.025 5/8 - 0.025 = 0.6 6/8 + 0.025 = 0.775 Required x interval is 0.6 to 0.775 Continuity Correction

  48. Suppose 12% of the population is in favor of a new park. • Two hundred citizen are surveyed. • What is the probability that between 10 % and 15% of them will be in favor of the new park?

  49. Is it appropriate to the normal distribution? • 12% of the population is in favor of a new park. p = 0.12, q= 0.88 • Two hundred citizen are surveyed. n = 200 • Both np and nq are greater than five.

  50. Find the mean and the standard deviation

More Related