1. Virtual University of Pakistan
Lecture No. 32 of the course on Statistics and Probability by
Miss Saleha Naghmi Habibullah

2. IN THE LAST LECTURE, YOU LEARNT

3. TOPICS FOR TODAY

4. You will recall that, in the last lecture, we discussed the sampling distribution of
We discussed the mean and the standard deviation of the sampling distribution, and, towards the end of the lecture, we consider the very important theorem known as the Central Limit Theorem.

5. Let us now consider the real-life application of this concept with the help of an example:

6. EXAMPLE
A construction company has 310 employees who have an average annual salary of Rs.24,000.
The standard deviation of annual salaries is Rs.5,000.

7. Suppose that the employees of this company launch a demand that the government should institute a law by which their average salary should be at least Rs , and, suppose that the government decides to check the validity of this demand by drawing a random sample of 100 employees of this company, and acquiring information regarding their present salaries.

8. What is the probability that, in a random sample of 100 employees, the average salary will exceed Rs.24,500 (so that the government decides that the demand of the employees of this company is unfounded, and hence does not pay attention to the demand(although, in reality, it was justified))?

9. SOLUTION
The sample size (n = 100) is large enough to assume that the sampling distribution ofX is approximately normally distributed with the following mean and standard deviation:

10. and standard deviation

11. NOTE:
Here we have used finite population correction factor (fpc), because the sample size n = 100 is greater than 5 percent of the population size N = 310.

12. Since X is approximately N(24000, 412.20), therefore

13. We are required to evaluate P(X > 24,500).
Atx = 24,500, we find that

14. 24000
24500
1.21 Z

15. Using the table of areas under the standard normal curve, we find that the area between z = 0 and z = 1.21 is

16. 24000
24500
1.21 Z
0.3869

17. Hence,
P(X > 24,500)
= P(Z > 1.21)
= 0.5 – P(0 < Z < 1.21)
= 0.5 – =

18. 24000
24500
1.21 Z
0.3869
0.1131

19. Hence, the chances are only 11% that in a random sample of 100 employees from this particular construction company , the average salary will exceed Rs.24,500.
In other words, the chances are 89% that, in such a sample, the average salary will not exceed Rs.24,500.

20. Hence, the chances are considerably high that the government might pay attention to the employees’ demand.

21. Next, we consider the SAMPLING DISTRIBUTION OF THE SAMPLE PROPORTION:

22. In this regard, the first point to be noted is that, whenever the elements of a population can be classified into two categories, technically called “success” and “failure”, we may be interested in the proportion of “successes” in the population.

23. If X denotes the number of successes in the population, then the proportion of successes in the population is given by

24. where X represents the number of successes in the sample.
Similarly, if we draw a sample of size n from the population, the proportion of successes in the sample is given by
where X represents the number of successes in the sample.

25. It is interesting to note that X is a binomial random variable and the binomial parameter p is being called a proportion of successes here.

26. The sample proportion has different values in different samples.
It is obviously a random variable and has a probability distribution.

27. This probability distribution of the proportions of successes in all possible random samples of size n, is called the sampling distribution of

28. We illustrate this sampling distribution with the help of the following examples:

29. A population consists of six values 1, 3, 6, 8, 9 and 12.
EXAMPLE-1
A population consists of six values 1, 3, 6, 8, 9 and 12.
Draw all possible samples of size n = 3 without replacement from the population and find the proportion of even numbers in each sample.

30. are sample and population proportions respectively.
Construct the sampling distribution of sample proportions and verify that
are sample and population proportions respectively.

31. SOLUTION
The number of possible samples of size n = 3 that could be selected without replacement from a population of size N is

32. Then the 20 possible samples and the proportion of even numbers are given as follows:

34. The sampling distribution of sample proportion is given below:

35. Sampling Distribution of

36. As n  , the sampling distribution of
approaches normality:

38. To verify the given relations, we first calculate the population proportion p.
Thus :

41. The sampling distribution of has the following important properties:

42. PROPERTIES OF THE SAMPLING DISTRIBUTION OF
Property No. 1:
The mean of the sampling distribution of proportions, denoted by is equal to the population proportion p, that is

43. Property No. 2:
The standard deviation of the sampling distribution of proportions, called the standard error of and denoted by
is given as:

44. a)
when the sampling is performed with replacement

45. b)
when sampling is done without replacement from a finite population.

46. (As in the case of the sampling distribution of X,
is known as the finite population correction factor (fpc).)

47. The sampling distribution of
Property No. 3:
SHAPE OF THE DISTRIBUTION:
The sampling distribution of
is the binomial distribution. However, for sufficiently large sample sizes, the sampling distribution of is approximately normal.

48. As n  , the sampling distribution of
approaches normality:

49. As a rule of thumb, the sampling distribution of will be approximately normal whenever both np and nq are equal to or greater than 5.

50. Let us apply this concept to a real-world situation:

51. EXAMPLE-2
Ten percent of the 1-kilogram boxes of sugar in a large warehouse are underweight.
Suppose a retailer buys a random sample of 144 of these boxes.
What is the probability that at least 5 percent of the sample boxes will be underweight?

52. SOLUTION
Here the statistic is the sample proportion
The sample size (n = 144) is large enough to assume that the sample proportion is approximately normally distributed with mean

54. Therefore, the sampling distribution of is approximately N(0. 10, 0
And, hence:
is approximately N(0, 1).

55. We are required to find the probability that the proportion of underweight boxes in the sample is equal to or greater than 5% i.e., we require

56. In this regard, a very important point to be noted is that, just as we use a continuity correction of + ½ whenever we consider the normal approximation to the binomially distributed random variable X, in this situation, since
therefore, we need to use the following continuity correction:

57. correction of We need to use a continuity
in the case of the sampling distribution of

58. (using the area table of the standard normal distribution)

59. 0.10
-2.14 Z
0.4838
0.5

60. Hence, the probability that at least 5% of the sample boxes are under-weight is as high as 98% !

61. The sampling distributions of X and
The sampling distributions of X and pertain to the situation when we are drawing all possible samples of a particular size from one particular population.

62. Next, we will discuss the case when we are dealing with all possible samples drawn from two populations, such that the samples from the two populations are independent.
In this regard, we will consider the sampling distributions of
and

63. We begin with the sampling distribution of

67. We illustrate the sampling distribution of
with the help of the following example:

68. EXAMPLE
Draw all possible random samples of size n1 = 2 with replacement from a finite population consisting of 4, 6, 8.
Similarly, draw all possible random samples of size n = 2 with replacement from another finite population consisting of 1, 2, 3.

69. a) Find the possible differences between the sample means of the two population.
b) Construct the sampling distribution of and compute its mean and variance.

70. c) Verify that

71. SOLUTION:

72. Whenever we are sampling with replacement from a finite population, the total number of possible samples is Nn
(where N is the population size, and n is the sample size).

73. Hence, in this example, there are (3)2 = 9 possible samples which can be drawn with replacement from each population.
These two sets of samples and their means are given below:

75. a) Since there are 9 samples from the first population as well as 9 from the second, hence, there are 81 possible combinations of x1 andx2 .
The 81 possible differences x1 –x2 are presented in the following table:

77. b) The sampling distribution of
is as follows:

79. Thus the mean and the variance are

80. c). In order to verify the properties of the sampling distribution of
c) In order to verify the properties of the sampling distribution of we first need to compute the mean and variance of the first population:

81. The mean and standard deviation of the first population are:

82. and

83. The mean and variance of the second population are:

84. Hence, two properties of the sampling distribution of are satisfied.

89. In case of sampling without replacement from a finite population, the formula for the standard error of
will be suitably modified.

91. is normally distributed with zero mean and unit variance.
In other words, the variable
is normally distributed with zero mean and unit variance.

92. b) If the POPULATIONS are non-normal and if both sample sizes are large, (i.e., greater than or equal to 30),
then the sampling distribution of the differences between means is approximately a normal distribution by the Central Limit Theorem.

93. In this case too, the variable
will be approximately normally distributed with mean zero and variance one.

94. IN TODAY’S LECTURE,
YOU LEARNT

95. IN THE NEXT LECTURE, YOU WILL LEARN
Sampling Distribution of
Point Estimation
Desirable Qualities of a Good Point Estimator
Unbiasedness
Consistency
Efficiency
Methods of Point Estimation:
The Method of Moments,