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1. Biostatistics Unit 5 Samples Needs to be completed. 12/24/13

2. Sampling distributions • Sampling distributions are important in the understanding of statistical inference.  • Probability distributions permit us to answer questions about sampling and they provide the foundation for statistical inference procedures.

3. Definition • The sampling distribution of a statistic is the distribution of all possible values of the statistic, computed from samples of the same size randomly drawn from the same population.  • When sampling a discrete, finite population, a sampling distribution can be constructed.  • Note that this construction is difficult with a large population and impossible with an infinite population.

4. Construction of sampling distributions 1.  From a population of size N, randomly draw all possible samples of size n. 2.  Compute the statistic of interest for each sample.3.  Create a frequency distribution of the statistic.

5. Properties of sampling distributions We are interested in the • mean, • standard deviation, and • appearance of the graph (functional form) of a sampling distribution.

6. Types of sampling distributions We will study the following types of sampling distributions. • Distribution of the sample mean • Distribution of the difference between two means • Distribution of the sample proportion • Distribution of the difference between two proportions

7. (A) Sampling distribution of Given a finite population with mean (m) and variance (s2).  When sampling from a normally distributed population, it can be shown that the distribution of the sample mean will have the following properties.

8. Properties of the sampling distribution •  The distribution of will be normal. • The mean , of the distribution of the values of will be the same as the mean of the population from which the samples were drawn; = m. 3.  The variance, , of the distribution of will be equal to the variance of the population divided by the sample size; = .

9. Standard error The square root of the variance of the sampling distribution is called the standard error of the mean which is also called the standard error.

10. Nonnormally distributed populations When the sampling is done from a nonnormally distributed population, the central limit theorem is used.

11. The central limit theorem Given a population of any nonnormal functional form with mean (m) and variance (s2) , the sampling distribution of , computed from samples of size n from this population will have mean, m, and variance, s2/n, and will be approximately normally distributed when the sample is large (30 or higher).

12. The central limit theorem Note that the standard deviation of the sampling distribution is used in calculations of z scores and is equal to:

13. Sampling distribution of the mean andCentral Limit Theorem We do in class together

14. Data • A small apartment building has 3 apartments. • How many people live in each apartment?

15. Find m and s • Use the TI to obtain the values for the population. The values are: m = s =

16. Form samples of size 2 • We need to form all samples of size 2, using replacement since the population is very small. • Then we find the sample mean for each sample of 2 apartments.

17. Find m and s • Use the TI to obtain the values for the means of the samples. The values are: m = s =

18. ResultsMean of Sample means • Mean of population equals mean of the sample means

19. Results of Standard deviation of the sample means • S.D. equals the population standard deviation divided by the square root of the sample size

20. Distribution of the sample means • If the population is normally distributed, then the sample means will be normally distributed. • If the population is not normally distributed, then the sample means will be normally distributed if the sample size is at least 30.

21. Important Consequence • If we take samples of size n from some population, under the previous conditions, then we can determine the probability of the sample means fulfilling some condition. We use:

22. Example #1 • The heights of kindergarten children are approximately normally distributed with a mean of 39 and a standard deviation of 2. If one child is randomly selected, what is the probability that the child is taller than 41 inches? • This is 1 child – Not the Central Limit Theorem!

23. Example #2 • Suppose we have a class of 30 kindergarten children. What is the probability that the mean height of these children exceeds 41 inches? • This is the Central Limit Theorem as it is asking about the probability of a sample mean!

24. Conclusion • It is not unusual for one child, selected at random from a kindergarten class, to be taller than 41 inches. • It is highly unlikely that the mean height for 30 kindergarten students exceeds 41 inches.

25. An analogy • It would not be unusual for a student to get an A on a statistics test. • It would be unusual if the class average for a statistics class was an A!

26. Demonstration that Central Limit Theorem Really Works (1) We start with a dwelling that has 3 apartments. Here is the list of occupancies. Apt A = 3 Apt B = 4 Apt C = 2 This is the entire population. It is entered into a list on the TI-83

27. Demonstration that Central Limit Theorem Really Works (2) We calculate 1-Var Stats to obtain the population parameters for this population. Mean: m = 3 Standard Deviation: s = .8164965809 Note:we do not use s = 3 because this is the entire population, not a sample.

28. Demonstration that Central Limit Theorem Really Works (3) • Knowing the population parameters of m and s, we now determine them using a sampling distribution. • We can find the population parameters because it is a very small population. • Normally, populations are too large to determine m and s directly from the population.

29. Demonstration that Central Limit Theorem Really Works (4) • We need to form all samples of size 2, using replacement since the population is very small. • Then we find the sample mean for each sample of 2 apartments.

30. Demonstration that Central Limit Theorem Really Works (5) We calculate 1-Var Stats to obtain the population parameters for the sampling distribution. Mean: m = 3 Standard Deviation: s = .5773502692

31. Demonstration that Central Limit Theorem Really Works (6)

32. Example Given the information below, what is the probability that x is greater than 53? (1) Write the given information.m = 50s = 16     n = 64 x = 53

33. Example (2) Sketch a normal curve.

34. Example (3) Convert x to a z score.

35. Example (4) Find the appropriate value(s) in the table.A value of z = 1.5 gives an area of .9332.  This is subtracted from 1 to givethe probability P (z > 1.5) = .0668

36. Example (5) Complete the answer.The probability that x is greater than 53 is .0668.

37. (B) Distribution of the difference between two means • It often becomes important to compare two population means.  • Knowledge of the sampling distribution of the difference between two means is useful in studies of this type.  • It is generally assumed that the two populations are normally distributed.

38. Sampling distribution of Plotting mean sample differences against frequency gives a normal distribution with mean equal to which is the difference between the two population means.

39. Variance The variance of the distribution of the sample differences is equal to Therefore, the standard error of the differences between two means would be equal to

40. Converting to a z score To convert to the standard normal distribution, we use the formula  We find the z score by assuming that there is no difference between the population means.

41. Sampling from normal populations This procedure is valid even when Sampling from normal populations the population variances are different or when the sample sizes are different.  Given two normally distributed populations with means,  and , and variances,  and , respectively. (continued)

42. Sampling from normal populations The sampling distribution of the difference, , between the means of independent samples of size n1 and n2 drawn from these populations is normally distributed with mean, , and variance,

43. Example In a study of annual family expenditures for general health care, two populations were surveyed with the following results:Population 1: n1 = 40,  = \$346 Population 2: n2 = 35,  = \$300

44. Example If the variances of the populations are   = 2800 and  = 3250, what is the probability of obtaining sample results as large as those shown if there is no difference in the means of the two populations?

45. Solution (1) Write the given informationn1 = 40,  = \$346, = 2800 n2 = 35,  = \$300,  = 3250

46. Solution (2) Sketch a normal curve

47.  Solution (3) Find the z score

48. Solution (4) Find the appropriate value(s) in the table    A value of z = 3.6 gives an area of .9998.  This is subtracted from 1 to give the probability        P (z > 3.6) = .0002

49. Solution (5) Complete the answer    The probability that  is as large as given is .0002.

50. C) Distribution of the sample proportion ( ) While statistics such as the sample mean are derived from measured variables, the sample proportion is derived from counts or frequency data.