Chapter 09 Textbook Study Guide

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Department
Operations Management and Information System
Course
OMIS 2010
Professor
Alan Marshall
Semester
Fall

Description
Chapter 9: Sampling Distributions 9.1 Introduction This chapt er connect s t he m aterial i n C hapters 4 t hrough 8 (num erical descriptive statistics, sampling, and probability dist ributions, in particular) with statisti cal 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 sam ple 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 x =σ x/ n. The sampling distribution of x allows us to make probability statements about x based on knowing the 2 values of the sample size n and the population parameters µ anσ . Example 9.1 A random variable possesses the following probability distribution: x p(x) 1 .2 2 .5 3 .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 =µ and σ 2 = σ / n . x x x x Solution a) There are nine possible sam ples of size 2. They ar e (1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), and (3,3). b) The samples, the values ofx , and the probability of each sample outcome are shown below: Sam ple x Pro bability (1,1) 1.0 (.2)(.2) = .04 (1,2) 1.5 (.2)(.5) = .10 (1,3) 2.0 (.2)(.3) = .06 (2,1) 1.5 (.5)(.2) = .10 (2,2) 2.0 (.5)(.5) = .25 (2,3) 2.5 (.5)(.3) = .15 (3,1) 2.0 (.3)(.2) = .06 (3,2) 2.5 (.3)(.5) = .15 (3,3) 3.0 (.3)(.3) = .09 The sampling distribution ofx follows: x x p( ) 1.0 .04 1.5 .20 2.0 .37 2.5 .30 3.0 .09 c) Using our definitions of expected value and variance, we fi nd the mean and vari ance of t he random variable x: µ x = E(x) = ∑ xp(x) = 1(.2) + 2(.5) + 3(.3) = 2.1 σ x = ∑ (x −µ ) p(x) = (1 − 2.1) (.2) + (2 − 2.1) (.5) + (3 − 2.1) (.3) = 0.49 107 The mean and variance of the random variablex 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 2 = () −µ 2 p(x) x ∑ x = (1.0 − 2.1)2(.04) + (1.5 − 2.1) (.20) + (2.0 − 2.1) (.37) + (2.5 − 2.1)(.30) + (3.0 − 2.1) (.09) = 0.245 As you can see, µ x =µ x = 2.1 and 2 2 σ x =σ x / n = 0.49/2 = 0.245 Example 9.2 Suppose a random sample of 100 observat ions is drawn from a normal population whose mean is 600 and whose variance is 2,500. Find the following probabilities: a) P(590 < x < 610) x b) P(590 < < 610) c) P(x > 650) x d) P( > 650) Solution 2 a) X is normally distributed with mean x = 600 and variance σx = 2,500. We standardize x by subtractingµ x = 600 and dividing byσx = 50. Therefore, P(590 < x < 610) = P ⎜590 − 600 < x − µ x < 610 − 600⎟ ⎝ 50 σ x 50 ⎠ = P(–.2 < z < .2) = .1586 2 2 b) We know t hat x is norm ally di stributed wi tµ x = µ x = 600 and σ x =σ x/n = 2, 500/100 = 25. Thus, σ = 5. Hence, x P(590 < x < 610) = P ⎜590 − 600 < x − µ x < 610 − 600 ⎟ ⎝ 5 σ x 5 ⎠ = P(–2 < z < 2) = .9544 108 c) P(x > 650) = P ⎛x −µ x > 650 − 600 ⎞ ⎝ σ x 50 ⎠ = P(z > 1) = .1587 ⎛ x −µ x 650 − 600 ⎞ d) P( x > 650) = P ⎜ > ⎟ ⎝ σ x 5 ⎠ = 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. 2 What does this imply about the statement that µ = 600 anσ = 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 observa- tions 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 state- x 2 ments about . Since knowledge of µ and σ is required in order to compute the probability that falls into some specific interval, we acknowledge that this procedure is q uite unrealistic. However, the sampling distribution will eventually allow us t o infer something about an un
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