Chapter 9: Sampling Distributions 9. 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 0. At the completion of this chapter, you are expected to know the following:. How the sampling distribution of the mean is created and the shape and parameters of the distribution.. 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. 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 sampling will have three important characteristics:. x is approximately normally. calculated in each sample will vary. This new random variable created by. 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 values of the sample size n and the population parameters µ and σ. Example 9. A random variable possesses the following probability distribution: x p(x)..5 3.3 06
a) Find all possible samples of size 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 x = σ x / n. a) There are nine possible samples of size. They are (,), (,), (,3), (,), (,), (,3), (3,), (3,), and (3,3). b) The samples, the values of x, and the probability of each sample outcome are shown below: Sample x Probability (,).0 (.)(.) =.04 (,).5 (.)(.5) =.0 (,3).0 (.)(.3) =.06 (,).5 (.5)(.) =.0 (,).0 (.5)(.5) =.5 (,3).5 (.5)(.3) =.5 (3,).0 (.3)(.) =.06 (3,).5 (.3)(.5) =.5 (3,3) 3.0 (.3)(.3) =.09 The sampling distribution of x follows: x p( x ).0.04.5.0.0.37.5.30 3.0.09 c) Using our definitions of expected value and variance, we find the mean and variance of the random variable x: µ x = E ( x ) = x p ( x ) = (. ) + (. 5) + 3(. 3 ) =. σ x = ( x µ) p ( x ) = (.) (. ) + (.) (. 5 ) + (3.) (. 3) = 0.49 07
The mean and variance of the random variable x are computed as follows: µ x = E ( x ) = x p( x ) =.0 (. 04 ) +.5 (. 0 ) +.0 (. 37 ) +.5(. 30 ) + 3.0 (. 09 ) =. σ x ( ) p( x ) = x µ x = (.0. ) (. 04 ) + (.5.) (. 0 ) + (.0.) (. 37 ) = 0.45 As you can see, and µ x = µ x =. + (.5. ) (.30 ) + ( 3.0.) (. 09 ) σ x = σ x / n = 0.49 / = 0.45 Example 9. Suppose a random sample of 00 observations is drawn from a normal population whose mean is 600 and whose variance is,500. Find the following probabilities: a) P(590 x 60) b) P(590 x 60) c) P(x > 650) d) P( x > 650) a) X is normally distributed with mean µ x = 600 and variance σ x =,500. We standardize x by subtracting µ x = 600 and dividing by σ x = 50. Therefore, P(590 x 60) = P 590 600 50 x µ x σ x 60 600 50 = P(. z.) =.586 b) We know that x is normally distributed with µ x = µ x = 600 and σ x = σ x / n =, 500 /00 = 5. Thus, σ x = 5. Hence, P(590 x 60) = P 590 600 5 x µ x 60 600 σ x 5 = P( z ) =.9544 08
c) P(x > 650) = P x µ x > σ x 650 600 50 = P(z > ) =.587 d) P( x > 650) = P x µ x > σ x = P(z > 0) = 0 650 600 5 Example 9.3 Refer to Example 9.. Suppose a random sample of 00 observations produced a mean of x = 650. What does this imply about the statement that µ = 600 and σ =,500? From Example 9. 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 00 observations drawn from a population whose mean is 600 and whose variance is,500. Question: Answer: What purpose does the sampling distribution serve? In particular, why do we need to calculate probabilities associated with the sample mean? (in reverse order) We are not terribly interested in making probability statements about x. Since knowledge of µ and σ 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. If 64 observations are taken from a population with µ = 00 and σ = 40, find P(0 x ). 09
9. A normally distributed random variable has a mean of 0 and a standard deviation of 0. If a random sample of 5 is drawn from this population, find P( x > 3). 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) 0.7.3 0
9.3 Creating the Sampling Distribution by Computer Simulation In Section 9., 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 timeconsuming. 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 0 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. The sampling distribution of pˆ is approximately normal with mean p and variance np(-p). Thus z = pˆ p p( 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.5. We wish to find P(.48 pˆ.5). We employ the approximate normal sampling distribution. Because the coin is fair, p =.5.
P(.48 pˆ.5) = P. 48. 5 (. 5 )(. 5) / 400 pˆ p p( p ) / n. 5. 5 (. 5 )(. 5) / 400 = P(-.8 z.8) = (.5 -.88) =.438 Example 9.5 Repeat Exercise 9.4 changing the number of flips to 000. P(.48 pˆ.5) = P. 48. 5 (. 5 )(. 5) / 000 pˆ p p( p ) / n. 5. 5 (. 5)(. 5 ) / 000 = P(-.6 z.6) = (.5 -.396) =.076 Question: Answer: Why don't we use the /-correction factor? When the sample size is large the effect of the correction factor is negligible. Omitting it only slightly affects the approximation but simplifies our calculation. EXERCISES 9.4 The proportion of defective units coming off a production line is 5%. Find the probability that in a random sample of 00 units more than 0% are defective? 9.5 Repeat Exercise 9.4 with a sample of 400 units.
9.6 In the last election a local counselor received 5% of the vote. If her popularity level is unchanged what is the probability that in a random sample of 00 voters less than 50% would vote for her? 9.5 Sampling Distribution of the Difference between Two Means The sampling distribution of the difference between two means is developed by extending the Central Limit Theorem. That is, the sampling distribution of is approximately normally distributed (If the two variables are normal then x x is normally distributed as well.) with mean µ µ and standard deviation Thus, σ σ + n n ( x x ) ( µ µ z = σ σ + n n ) x x is either normally distributed or approximately normally distributed. We can use this sampling distribution in the same way we employed the sampling distribution of the sample mean, to make probability statements about the difference between two sample means. Example 9.6 Suppose that we draw random samples of size 5 from two normal populations. The mean and standard deviation of population are 00 and 5. The mean and standard deviation of population are 90 and 40. Find the probability that the mean of sample exceeds the mean of sample. 3
We want to determine P[( x x > 0]. The mean and standard deviation of the sampling distribution are Thus, µ µ = 00-90 = 0 σ σ 5 40 + = + =. n n 5 5 P[( x x ) > 0] = EXERCISES ( x x ) ( µ µ P σ σ + n n ) 0 0 > = (P(z > -.47) =.5+.808 =.6808.. 9.7 The assistant dean of a business school claims that the number of job offers received by MBA's whose major is finance is normally distributed with a mean of and a standard deviation of.5. Furthermore he states that job offers to marketing majors is normally distributed with a mean of 0 and a standard deviation of 3. Find the probability that in a random sample of 0 finance and 0 marketing majors the average finance major receives more job offers than the average marketing major. 9.8 Repeat Exercise 9.7 using sample sizes of 5. 4