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 and Central Limit Theorem
We do in class together

14. Data A small apartment building has 3 apartments.
How many people live in each apartment?
Apartment
People A B C

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
Samples
Sample mean
A, A
A, B
A, C
B, A
B, B
B, C
C, A
C, B
C, C
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. Results Mean 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 =
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.
Samples
Sample mean
A, A
3.0
A, B
3.5
A, C
2.5
B, A
B, B
4.0
B, C
C, A
C, B
C, C
2.0

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 =

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 = 50      s = 16      n = 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  
This is subtracted from 1 to give the probability P (z > 1.5) = .0668

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

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 information n1 = 40, = $346, = 2800

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   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

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.

51. Properties of the sample proportion
Construction of the sampling distribution of the sample proportion is done in a manner similar to that of the mean and the difference between two means.  When the sample size is large, the distribution of the sample proportion is approximately normally distributed because of the central limit theorem.

52. Mean and variance
The mean of the distribution,  , will be equal to the true population proportion, p, and the variance of the distribution, , will be equal to p(1-p)/n.  

53. The z-score
The z-score for the sample proportion is

54. Example
In the mid seventies, according to a report by the National Center for Health Statistics, 19.4 percent of the adult U.S. male population was obese.  What is the probability that in a simple random sample of size 150 from this population fewer than 15 percent will be obese?

55. Solution
(1) Write the given information       n = 150       p =      Find P( < .15)

56. Solution
(2) Sketch a normal curve  

57. Solution
(3) Find the z score    

58. Solution
(4) Find the appropriate value(s) in the table A value of z = gives an area of which is the probability         P (z < -1.36) = .0869

59. Solution (5) Complete the answer
The probability that < .15 is

60. D) Distribution of the difference between two proportions
This is for situations with two population proportions.  We assess the probability associated with a difference in proportions computed from samples drawn from each of these populations.  The appropriate distribution is the distribution of the difference between two sample proportions.

61. Sampling distribution of
The sampling distribution of the difference between two sample proportions is constructed in a manner similar to the difference between two means. 
(continued)

62. Sampling distribution of
Independent random samples of size n1 and n2 are drawn from two populations of dichotomous variables where the proportions of observations with the character of interest in the two populations are p1 and p2 , respectively.

63. Mean and variance The distribution of the difference between two
sample proportions, , is approximately normal.   The mean is
The variance is
These are true when n1 and n2 are large.

64. The z score
The z score for the difference between two proportions is given by the formula

65. Example
In a certain area of a large city it is hypothesized that 40 percent of the houses are in a dilapidated condition.  A random sample of 75 houses from this section and 90 houses from another section yielded difference, , of .09.  If there is no difference between the two areas in the proportion of dilapidated houses, what is the probability of observing a difference this large or larger?

66. Solution (1) Write the given information n1 = 75, p1 = .40
  = .09 Find P( )

67. Solution
(2) Sketch a normal curve

68. Solution
(3) Find the z score                

69. Solution
(4) Find the appropriate value(s) in the table     A value of z = 1.17 gives an area of which is subtracted from 1 to give the probability         P (z > 1.17) = .121

70. Solution (5) Complete the answer The probability of observing
of .09 or greater is .121.

71. fin