ELEMENTARY STATISTICS 3E William Navidi and Barry Monk ©McGraw-Hill Education. All rights reserved. Authorized only for instructor use in the classroom. No reproduction or further distribution permitted without the prior written consent of McGraw-Hill Education. Sampling Distributions and The Central Limit Theorem Section 7.3 ©McGraw-Hill Education. Objectives 1. Construct the sampling distribution of a sample mean 2. Use the Central Limit Theorem to compute probabilities involving sample means ©McGraw-Hill Education. Objective 1 Construct the sampling distribution of a sample mean ©McGraw-Hill Education. Sampling Distribution of the Sample Mean In real situations, statistical studies involve sampling several individuals then computing numerical summaries of the samples. Most often the sample mean, 𝑥,ҧ is computed. If several samples are drawn from a population, they are likely to have different values for 𝑥.ҧ Because the value of 𝑥ҧ varies each time a ഥ is a random variable. For each value of the sample is drawn, 𝒙 random variable, 𝑥,ҧ we can compute a probability. The probability distribution of 𝑥ҧ is called the sampling distribution of 𝑥.ҧ ©McGraw-Hill Education. An Example of a Sampling Distribution Tetrahedral dice are shaped like a pyramid with four faces. Each face corresponds to a number between 1 and 4. Tossing a tetrahedral die is like sampling a value from the population 1, 2, 3, 4 . We can easily find the population mean, 𝜇 = 2.5, and the population standard deviation 𝜎 = 1.118. Suppose that a tetrahedral die is tossed three times. The sequence of three numbers observed may be thought of as a sample of size 3. We may get samples such as [1, 1, 1], [1, 1, 2], [1, 1, 3], and so on. ©McGraw-Hill Education. An Example of a Sampling Distribution (Continued) The table displays all possible samples of size 3 and the sample mean 𝑥ҧ of each. The mean of all of values of 𝑥ҧ is 𝜇𝑥ҧ = 2.5 and the standard deviation of all values of 𝑥ҧ is 𝜎𝑥ҧ = 0.6455. Next, we compare these values to the population mean (2.5) and population standard deviation (1.118). ©McGraw-Hill Education. Mean and Standard Deviation of a Sampling Distribution The mean of the sampling distribution is 𝜇𝑥ҧ = 2.5, which is the same as the mean of the population, 𝜇 = 2.5. This relation always holds. The mean of the sampling distribution is denoted by 𝝁𝒙 and equals the mean of the population: 𝝁𝒙 = 𝝁 The standard deviation of the sampling distribution is 𝜎𝑥ҧ = 0.6455, which is less than the population standard deviation 𝜎 = 1.118. It is not obvious how 1.118 𝜎 these two quantities are related. Note, that 𝜎𝑥ҧ = 0.6455 = = . Recall that the sample size is 𝑛 = 3, which suggests that 𝜎𝑥ҧ = 𝜎 . 𝑛 3 3 The standard deviation of the sampling distribution, sometimes called the standard error, is denoted by 𝝈𝒙 and equals the standard deviation of the population divided by the square root of the sample size: 𝝈𝒙 = ©McGraw-Hill Education. 𝝈 𝒏 Example: Sampling Distribution Among students at a certain college, the mean number of hours of television watched per week is 𝜇 = 10.5, and the standard deviation is 𝜎 = 3.6. A simple random sample of 16 students is chosen for a study of viewing habits. Let 𝑥ҧ be the mean number of hours of TV watched by the sampled students. Find the mean 𝜇𝑥 and the standard deviation 𝜎𝑥 of 𝑥ҧ . Solution: The mean of 𝑥ҧ is: 𝜇𝑥 = 𝜇 = 10.5 The sample size is 𝑛 = 16. Therefore, the standard deviation of 𝑥ҧ is: 𝜎𝑥 = ©McGraw-Hill Education. 𝜎 𝑛 = 3.6 16 = 0.9 Sampling Distribution for Sample of Size 3 Consider again the tetrahedral die example. The sampling distribution for 𝑥ҧ can be determined from the table of all possible values of 𝑥.ҧ The probability that the sample mean is 1.00 is 1 , or 0.016, because out of the 64 possible 64 samples, only 1 has a sample mean equal to 3 1.00. Similarly, the probability that 𝑥ҧ =1.33 is , 64 or 0.047, because there are 3 samples whose sample mean is 1.33. The remaining probabilities are as follows. ഥ 𝒙 1.00 1.33 1.67 2.00 2.33 2.67 3.00 3.33 3.67 4.00 𝑷(ഥ 𝒙) 0.016 0.047 0.094 0.156 0.188 0.188 0.156 0.094 0.047 0.016 ©McGraw-Hill Education. Probability Histogram for a Sampling Distribution In the tetrahedral die example, the population is 1, 2, 3, 4 . When a die is rolled, each number has the 1 same chance of appearing, or 0.25. 4 The probability histogram for the sampling distribution of 𝑥ҧ with sample size 3 is obtained from the sampling distribution on the previous slide. The probability histogram for the sampling distribution looks a lot like the normal curve, whereas the probability histogram for the population does not. Remarkably, it is true that, for any population, if the sample size is large enough, the sample mean 𝑥ҧ will be approximately normally distributed. For a symmetric population like the tetrahedral die population, the sample mean is approximately normally distributed even for a small sample size like 𝑛 = 3. ©McGraw-Hill Education. Sampling Distribution from a Skewed Population If a population is skewed, a larger sample size is necessary for the sampling distribution of 𝑥ҧ to be approximately normal. Consider the following probability distribution. Below are the probability histograms for the sampling distribution of 𝑥ҧ for samples of size 3, 10, and 30. Note that the shapes of the distributions begin to approximate a normal curve as the sample size increases. The size of the sample needed to obtain approximate normality depends mostly on the skewness of the population. In practice, a sample of size 𝑛 > 30 is large enough. ©McGraw-Hill Education. The Central Limit Theorem The remarkable fact that the sampling distribution of 𝑥ҧ is approximately normal for a large sample from any distribution is part of one of the most used theorems in Statistics, the Central Limit Theorem. Let 𝑥ҧ be the mean of a large (𝑛 > 30) simple random sample from a population with mean 𝜇 and standard deviation 𝜎. Then 𝑥ҧ has an approximately normal distribution, with mean 𝜇𝑥 = 𝜇 and standard 𝜎 deviation 𝜎𝑥 = . 𝑛 The Central Limit Theorem applies for all populations. However, for symmetric populations, a smaller sample size may suffice. If the ഥ will be normal for population itself is normal, the sample mean 𝒙 any sample size. ©McGraw-Hill Education. Example 1: The Central Limit Theorem A sample of size 45 will be drawn from a population with mean 𝜇 = 15 and standard deviation 𝜎 = 3.5. Is it appropriate to use the normal distribution to find probabilities for 𝑥? ҧ Solution: Yes, by The Central Limit Theorem, since 𝑛 > 30, 𝑥ҧ has an approximately normal distribution. ©McGraw-Hill Education. Example 2: The Central Limit Theorem A sample of size 8 will be drawn from a normal population with mean 𝜇 = –60 and standard deviation 𝜎 = 5. Is it appropriate to use the normal distribution to find probabilities for 𝑥? ҧ Solution: Yes, since the population itself is approximately normal, 𝑥ҧ has an approximately normal distribution. ©McGraw-Hill Education. Example 3: The Central Limit Theorem A sample of size 24 will be drawn from a population with mean 𝜇 = 35 and standard deviation 𝜎 = 1.2. Is it appropriate to use the normal distribution to find probabilities for 𝑥? ҧ Solution: No, since the population is not known to be normal and 𝑛 is not greater than 30, we cannot be certain that 𝑥ҧ has an approximately normal distribution. ©McGraw-Hill Education. Objective 2 Use the Central Limit Theorem to compute probabilities involving sample means ©McGraw-Hill Education. Example 1: Using the Central Limit Theorem Recent data from the U.S. Census indicates that the mean age of college students is 𝜇 = 25 years, with a standard deviation of 𝜎 = 9.5 years. A simple random sample of 125 students is drawn. What is the probability that the sample mean age of the students is greater than 26 years? Solution: The sample size is 125, which is greater than 30. Therefore, we may use the normal curve. We compute 𝜇𝑥ҧ and 𝜎𝑥ҧ 𝜇𝑥 = 𝜇 = 25 and 𝜎𝑥 = 𝜎 𝑛 = 9.5 125 = 0.85 Find the area under the normal curve. The probability that the sample mean age of the students is greater than 26 years is approximately 0.1197. ©McGraw-Hill Education. Example 2: Using the Central Limit Theorem Hereford cattle are one of the most popular breeds of cattle. Based on data from the Hereford Cattle Society, the mean weight of a one-yearold Hereford bull is 1135 pounds, with a standard deviation of 97 pounds. Would it be unusual for the mean weight of 100 head of cattle to be less than 1100 pounds? Solution: The sample size is 100, which is greater than 30. Therefore, we may use the normal curve. We compute 𝜇𝑥ҧ and 𝜎𝑥ҧ 𝜎 97 𝜇𝑥 = 𝜇 = 1135 and 𝜎𝑥 = = = 9.7 𝑛 100 Find the area under the normal curve. This probability is less than 0.05, so it would be unusual for the sample mean to be less than 1100. ©McGraw-Hill Education. You Should Know . . . • How to construct the sampling distribution of a sample mean • How to find the mean and standard deviation of a sampling distribution of 𝑥ҧ • The Central Limit Theorem • How to use the Central Limit Theorem to compute probabilities involving sample means (Calculator) ©McGraw-Hill Education.
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