Chapter 8: Sampling Distributions - U of L Class Index
Chapter 9: Sampling Distributions 9.1
Introduction
This chapter connects the material in Chapters 4 through 8 (numerical descriptive statistics, sampling, and probability distributions, in particular) with statistical inference, which is introduced in Chapter 10. At the completion of this chapter, you are expected to know the following: 1. How the sampling distribution of the mean is created and the shape and parameters of the distribution. 2. How to calculate probabilities using the sampling distribution of the mean. 3. Understand how the normal distribution can be used to approximate the binomial distribution. 4. How to calculate probabilities associated with a sample proportion. 5. How to calculate probabilities associated with the difference between two sample means.
9.2
Sampling Distribution of the Mean
The most important thing to learn from this section is that if we repeatedly draw samples from any population, the values of x calculated in each sample will vary. This new random variable created by sampling will have three important characteristics: 1. x is approximately normally. 2. The mean of x will equal the mean of the original random variable. That is, µ x = µ x . 3. The variance of x will equal the variance of the original random variable divided by n. That is, σ 2x = σ 2x / n . The sampling distribution of x allows us to make probability statements about x based on knowing the values of the sample size n and the population parameters µ and σ 2 .
Example 9.1 A random variable possesses the following probability distribution: x
p(x)
1 2 3
.2 .5 .3
106
a) Find all possible samples of size 2 that can be drawn from this population. b) Using the results in part a), find the sampling distribution of x . c) Confirm that µ x = µ x and σ 2x = σ 2x / n .
Solution a) There are nine possible samples of size 2. They are (1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), and (3,3). b) The samples, the values of x , and the probability of each sample outcome are shown below: Sample
x
Probability
(1,1) (1,2) (1,3) (2,1) (2,2) (2,3) (3,1) (3,2) (3,3)
1.0 1.5 2.0 1.5 2.0 2.5 2.0 2.5 3.0
(.2)(.2) = .04 (.2)(.5) = .10 (.2)(.3) = .06 (.5)(.2) = .10 (.5)(.5) = .25 (.5)(.3) = .15 (.3)(.2) = .06 (.3)(.5) = .15 (.3)(.3) = .09
The sampling distribution of x follows: x
p( x )
1.0 1.5 2.0 2.5 3.0
.04 .20 .37 .30 .09
c) Using our definitions of expected value and variance, we find the mean and variance of the random variable x: µ x = E( x) = σ 2x
=
∑ x p(x)
∑ (x −
µ) 2 p(x)
= 1(.2) + 2(.5) + 3(.3) = 2.1 = (1 − 2.1) 2 (.2) + (2 − 2.1)2 (.5) + (3 − 2.1) 2 (.3) = 0.49
107
The mean and variance of the random variable x are computed as follows: µ x = E(x ) =
∑ x p(x ) = 1.0(.04) + 1.5(.20) + 2.0(.37) + 2.5(.30) + 3.0(.09) = 2.1
σ 2x =
∑ (x − µx )2 p(x )
= (1.0 − 2.1)2 (.04) + (1.5 − 2.1)2 (.20) + (2.0 − 2.1)2 (.37) + (2.5 − 2.1) 2 (.30) + (3.0 − 2.1)2 (.09) = 0.245
As you can see, µ x = µ x = 2.1
and σ 2x = σ 2x / n = 0.49 / 2 = 0.245
Example 9.2 Suppose a random sample of 100 observations is drawn from a normal population whose mean is 600 and whose variance is 2,500. Find the following probabilities: a) b) c) d)
P(590 < x < 610) P(590 < x < 610) P(x > 650) P( x > 650)
Solution a) X is normally distributed with mean µ x = 600 and variance σ 2x = 2,500. We standardize x by subtracting µ x = 600 and dividing by σ x = 50. Therefore, x − µx ⎛ 590 − 600 610 − 600 ⎞ P(590 < x < 610) = P⎜ < < ⎟ σ 50 50 ⎠ ⎝ x
= P(–.2 < z < .2) = .1586 b) We know that x is normally distributed with µ x = µ x = 600 and σ 2x = σ 2x / n = 2, 500 /100 = 25. Thus, σ x = 5. Hence, ⎛ 590 − 600 x − µx 610 − 600 ⎞ P(590 < x < 610) = P⎜ < < ⎟ σx 5 5 ⎠ ⎝
= P(–2 < z < 2) = .9544
108
c) P(x > 650)
⎛ x − µx 650 − 600⎞ = P⎜ > ⎟ 50 ⎠ ⎝ σx
= P(z > 1) = .1587 ⎛ x − µx 650 − 600 ⎞ > d) P( x > 650) = P⎜ ⎟ 5 ⎠ ⎝ σx
= P(z > 10) = 0
Example 9.3 Refer to Example 9.2. Suppose a random sample of 100 observations produced a mean of x = 650. What does this imply about the statement that µ = 600 and σ 2 = 2,500?
Solution From Example 9.2 part d), we found that P( x > 650) = 0 Therefore, it is quite unlikely that we could observe a sample mean of 650 in a sample of 100 observations drawn from a population whose mean is 600 and whose variance is 2,500.
Question:
What purpose does the sampling distribution serve? In particular, why do we need to calculate probabilities associated with the sample mean?
Answer:
(in reverse order) We are not terribly interested in making probability statements about x . Since knowledge of µ and σ 2 is required in order to compute the probability that x falls into some specific interval, we acknowledge that this procedure is quite unrealistic. However, the sampling distribution will eventually allow us to infer something about an unknown population mean from a sample mean. This process, called statistical inference, will be the main topic throughout the rest of the textbook.
EXERCISES 9.1
If 64 observations are taken from a population with µ = 100 and σ = 40, find P(102 < x < 112).
109
9.2
A normally distributed random variable has a mean of 20 and a standard deviation of 10. If a random sample of 25 is drawn from this population, find P( x > 23).
9.3
Given the probability distribution of x below, find all samples of size 3, the sampling distribution of x , the mean, and the variance of x . x
p(x)
0 1
.7 .3
110
9.3 Creating the Sampling Distribution by Computer Simulation In Section 9.2, we described how the sampling distribution was created theoretically. We also pointed out that sampling distributions can be created empirically, but that the effort can be extremely time-consuming. In this section, we used the computer and our two software packages to create several sampling distributions empirically. The concept of the sampling distribution is critical to the development of statistical inference. It is important that you understand that the sampling distribution of any statistic is created theoretically or empirically by repeated sampling from a population. In each sample we calculate the statistic and thus create the distribution of that statistic. Throughout the book we will introduce about 20 different sampling distributions.
9.4 Sampling Distribution of a Proportion The sampling distribution of a sample proportion is actually based on the binomial distribution. However, the primary purpose of creating the sampling distribution is for inference and the binomial distribution, which is discrete, makes inference somewhat difficult. Consequently, we use the normal approximation to the binomial distribution. The details are not particularly important to the applied statistician (that's you). What is important is that you understand how the sampling distribution is used.
ˆ is approximately normal with mean p and variance np(1-p). Thus The sampling distribution of p z=
ˆp − p p( 1 − p ) / n
is approximately standard normally distributed.
Example 9.4 A fair coin is flipped 400 times. Find the probability that the proportion of heads falls between .48 and .52.
Solution ˆ < .52). We employ the approximate normal sampling distribution. Because We wish to find P(.48 < p the coin is fair, p = .5.
111
⎛ ˆ < .52) = P⎜ .48 − .5 P(.48 < p < ⎜ (.5 )(.5 ) / 400 ⎝
ˆp − p p( 1 − p ) / n
Introduction
This chapter connects the material in Chapters 4 through 8 (numerical descriptive statistics, sampling, and probability distributions, in particular) with statistical inference, which is introduced in Chapter 10. At the completion of this chapter, you are expected to know the following: 1. How the sampling distribution of the mean is created and the shape and parameters of the distribution. 2. How to calculate probabilities using the sampling distribution of the mean. 3. Understand how the normal distribution can be used to approximate the binomial distribution. 4. How to calculate probabilities associated with a sample proportion. 5. How to calculate probabilities associated with the difference between two sample means.
9.2
Sampling Distribution of the Mean
The most important thing to learn from this section is that if we repeatedly draw samples from any population, the values of x calculated in each sample will vary. This new random variable created by sampling will have three important characteristics: 1. x is approximately normally. 2. The mean of x will equal the mean of the original random variable. That is, µ x = µ x . 3. The variance of x will equal the variance of the original random variable divided by n. That is, σ 2x = σ 2x / n . The sampling distribution of x allows us to make probability statements about x based on knowing the values of the sample size n and the population parameters µ and σ 2 .
Example 9.1 A random variable possesses the following probability distribution: x
p(x)
1 2 3
.2 .5 .3
106
a) Find all possible samples of size 2 that can be drawn from this population. b) Using the results in part a), find the sampling distribution of x . c) Confirm that µ x = µ x and σ 2x = σ 2x / n .
Solution a) There are nine possible samples of size 2. They are (1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), and (3,3). b) The samples, the values of x , and the probability of each sample outcome are shown below: Sample
x
Probability
(1,1) (1,2) (1,3) (2,1) (2,2) (2,3) (3,1) (3,2) (3,3)
1.0 1.5 2.0 1.5 2.0 2.5 2.0 2.5 3.0
(.2)(.2) = .04 (.2)(.5) = .10 (.2)(.3) = .06 (.5)(.2) = .10 (.5)(.5) = .25 (.5)(.3) = .15 (.3)(.2) = .06 (.3)(.5) = .15 (.3)(.3) = .09
The sampling distribution of x follows: x
p( x )
1.0 1.5 2.0 2.5 3.0
.04 .20 .37 .30 .09
c) Using our definitions of expected value and variance, we find the mean and variance of the random variable x: µ x = E( x) = σ 2x
=
∑ x p(x)
∑ (x −
µ) 2 p(x)
= 1(.2) + 2(.5) + 3(.3) = 2.1 = (1 − 2.1) 2 (.2) + (2 − 2.1)2 (.5) + (3 − 2.1) 2 (.3) = 0.49
107
The mean and variance of the random variable x are computed as follows: µ x = E(x ) =
∑ x p(x ) = 1.0(.04) + 1.5(.20) + 2.0(.37) + 2.5(.30) + 3.0(.09) = 2.1
σ 2x =
∑ (x − µx )2 p(x )
= (1.0 − 2.1)2 (.04) + (1.5 − 2.1)2 (.20) + (2.0 − 2.1)2 (.37) + (2.5 − 2.1) 2 (.30) + (3.0 − 2.1)2 (.09) = 0.245
As you can see, µ x = µ x = 2.1
and σ 2x = σ 2x / n = 0.49 / 2 = 0.245
Example 9.2 Suppose a random sample of 100 observations is drawn from a normal population whose mean is 600 and whose variance is 2,500. Find the following probabilities: a) b) c) d)
P(590 < x < 610) P(590 < x < 610) P(x > 650) P( x > 650)
Solution a) X is normally distributed with mean µ x = 600 and variance σ 2x = 2,500. We standardize x by subtracting µ x = 600 and dividing by σ x = 50. Therefore, x − µx ⎛ 590 − 600 610 − 600 ⎞ P(590 < x < 610) = P⎜ < < ⎟ σ 50 50 ⎠ ⎝ x
= P(–.2 < z < .2) = .1586 b) We know that x is normally distributed with µ x = µ x = 600 and σ 2x = σ 2x / n = 2, 500 /100 = 25. Thus, σ x = 5. Hence, ⎛ 590 − 600 x − µx 610 − 600 ⎞ P(590 < x < 610) = P⎜ < < ⎟ σx 5 5 ⎠ ⎝
= P(–2 < z < 2) = .9544
108
c) P(x > 650)
⎛ x − µx 650 − 600⎞ = P⎜ > ⎟ 50 ⎠ ⎝ σx
= P(z > 1) = .1587 ⎛ x − µx 650 − 600 ⎞ > d) P( x > 650) = P⎜ ⎟ 5 ⎠ ⎝ σx
= P(z > 10) = 0
Example 9.3 Refer to Example 9.2. Suppose a random sample of 100 observations produced a mean of x = 650. What does this imply about the statement that µ = 600 and σ 2 = 2,500?
Solution From Example 9.2 part d), we found that P( x > 650) = 0 Therefore, it is quite unlikely that we could observe a sample mean of 650 in a sample of 100 observations drawn from a population whose mean is 600 and whose variance is 2,500.
Question:
What purpose does the sampling distribution serve? In particular, why do we need to calculate probabilities associated with the sample mean?
Answer:
(in reverse order) We are not terribly interested in making probability statements about x . Since knowledge of µ and σ 2 is required in order to compute the probability that x falls into some specific interval, we acknowledge that this procedure is quite unrealistic. However, the sampling distribution will eventually allow us to infer something about an unknown population mean from a sample mean. This process, called statistical inference, will be the main topic throughout the rest of the textbook.
EXERCISES 9.1
If 64 observations are taken from a population with µ = 100 and σ = 40, find P(102 < x < 112).
109
9.2
A normally distributed random variable has a mean of 20 and a standard deviation of 10. If a random sample of 25 is drawn from this population, find P( x > 23).
9.3
Given the probability distribution of x below, find all samples of size 3, the sampling distribution of x , the mean, and the variance of x . x
p(x)
0 1
.7 .3
110
9.3 Creating the Sampling Distribution by Computer Simulation In Section 9.2, we described how the sampling distribution was created theoretically. We also pointed out that sampling distributions can be created empirically, but that the effort can be extremely time-consuming. In this section, we used the computer and our two software packages to create several sampling distributions empirically. The concept of the sampling distribution is critical to the development of statistical inference. It is important that you understand that the sampling distribution of any statistic is created theoretically or empirically by repeated sampling from a population. In each sample we calculate the statistic and thus create the distribution of that statistic. Throughout the book we will introduce about 20 different sampling distributions.
9.4 Sampling Distribution of a Proportion The sampling distribution of a sample proportion is actually based on the binomial distribution. However, the primary purpose of creating the sampling distribution is for inference and the binomial distribution, which is discrete, makes inference somewhat difficult. Consequently, we use the normal approximation to the binomial distribution. The details are not particularly important to the applied statistician (that's you). What is important is that you understand how the sampling distribution is used.
ˆ is approximately normal with mean p and variance np(1-p). Thus The sampling distribution of p z=
ˆp − p p( 1 − p ) / n
is approximately standard normally distributed.
Example 9.4 A fair coin is flipped 400 times. Find the probability that the proportion of heads falls between .48 and .52.
Solution ˆ < .52). We employ the approximate normal sampling distribution. Because We wish to find P(.48 < p the coin is fair, p = .5.
111
⎛ ˆ < .52) = P⎜ .48 − .5 P(.48 < p < ⎜ (.5 )(.5 ) / 400 ⎝
ˆp − p p( 1 − p ) / n
