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ELEMENTARY STATISTICS 3E William Navidi and Barry Monk ยฉMcGraw-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 ยฉMcGraw-Hill Education. Objectives 1. Construct the sampling distribution of a sample mean 2. Use the Central Limit Theorem to compute probabilities involving sample means ยฉMcGraw-Hill Education. Objective 1 Construct the sampling distribution of a sample mean ยฉMcGraw-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, ๐‘ฅ,าง is computed. If several samples are drawn from a population, they are likely to have different values for ๐‘ฅ.าง Because the value of ๐‘ฅาง varies each time a เดฅ is a random variable. For each value of the sample is drawn, ๐’™ random variable, ๐‘ฅ,าง we can compute a probability. The probability distribution of ๐‘ฅาง is called the sampling distribution of ๐‘ฅ.าง ยฉMcGraw-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, ๐œ‡ = 2.5, and the population standard deviation ๐œŽ = 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. ยฉMcGraw-Hill Education. An Example of a Sampling Distribution (Continued) The table displays all possible samples of size 3 and the sample mean ๐‘ฅาง of each. The mean of all of values of ๐‘ฅาง is ๐œ‡๐‘ฅาง = 2.5 and the standard deviation of all values of ๐‘ฅาง is ๐œŽ๐‘ฅาง = 0.6455. Next, we compare these values to the population mean (2.5) and population standard deviation (1.118). ยฉMcGraw-Hill Education. Mean and Standard Deviation of a Sampling Distribution The mean of the sampling distribution is ๐œ‡๐‘ฅาง = 2.5, which is the same as the mean of the population, ๐œ‡ = 2.5. This relation always holds. The mean of the sampling distribution is denoted by ๐๐’™ and equals the mean of the population: ๐๐’™ = ๐ The standard deviation of the sampling distribution is ๐œŽ๐‘ฅาง = 0.6455, which is less than the population standard deviation ๐œŽ = 1.118. It is not obvious how 1.118 ๐œŽ these two quantities are related. Note, that ๐œŽ๐‘ฅาง = 0.6455 = = . Recall that the sample size is ๐‘› = 3, which suggests that ๐œŽ๐‘ฅาง = ๐œŽ . ๐‘› 3 3 The standard deviation of the sampling distribution, sometimes called the standard error, is denoted by ๐ˆ๐’™ and equals the standard deviation of the population divided by the square root of the sample size: ๐ˆ๐’™ = ยฉMcGraw-Hill Education. ๐ˆ ๐’ Example: Sampling Distribution Among students at a certain college, the mean number of hours of television watched per week is ๐œ‡ = 10.5, and the standard deviation is ๐œŽ = 3.6. A simple random sample of 16 students is chosen for a study of viewing habits. Let ๐‘ฅาง be the mean number of hours of TV watched by the sampled students. Find the mean ๐œ‡๐‘ฅ and the standard deviation ๐œŽ๐‘ฅ of ๐‘ฅาง . Solution: The mean of ๐‘ฅาง is: ๐œ‡๐‘ฅ = ๐œ‡ = 10.5 The sample size is ๐‘› = 16. Therefore, the standard deviation of ๐‘ฅาง is: ๐œŽ๐‘ฅ = ยฉMcGraw-Hill Education. ๐œŽ ๐‘› = 3.6 16 = 0.9 Sampling Distribution for Sample of Size 3 Consider again the tetrahedral die example. The sampling distribution for ๐‘ฅาง can be determined from the table of all possible values of ๐‘ฅ.าง 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 ๐‘ฅาง =1.33 is , 64 or 0.047, because there are 3 samples whose sample mean is 1.33. The remaining probabilities are as follows. เดฅ ๐’™ 1.00 1.33 1.67 2.00 2.33 2.67 3.00 3.33 3.67 4.00 ๐‘ท(เดฅ ๐’™) 0.016 0.047 0.094 0.156 0.188 0.188 0.156 0.094 0.047 0.016 ยฉMcGraw-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 ๐‘ฅาง 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 ๐‘ฅาง 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 ๐‘› = 3. ยฉMcGraw-Hill Education. Sampling Distribution from a Skewed Population If a population is skewed, a larger sample size is necessary for the sampling distribution of ๐‘ฅาง to be approximately normal. Consider the following probability distribution. Below are the probability histograms for the sampling distribution of ๐‘ฅาง 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 ๐‘› > 30 is large enough. ยฉMcGraw-Hill Education. The Central Limit Theorem The remarkable fact that the sampling distribution of ๐‘ฅาง 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 ๐‘ฅาง be the mean of a large (๐‘› > 30) simple random sample from a population with mean ๐œ‡ and standard deviation ๐œŽ. Then ๐‘ฅาง has an approximately normal distribution, with mean ๐œ‡๐‘ฅ = ๐œ‡ and standard ๐œŽ deviation ๐œŽ๐‘ฅ = . ๐‘› The Central Limit Theorem applies for all populations. However, for symmetric populations, a smaller sample size may suffice. If the เดฅ will be normal for population itself is normal, the sample mean ๐’™ any sample size. ยฉMcGraw-Hill Education. Example 1: The Central Limit Theorem A sample of size 45 will be drawn from a population with mean ๐œ‡ = 15 and standard deviation ๐œŽ = 3.5. Is it appropriate to use the normal distribution to find probabilities for ๐‘ฅ? าง Solution: Yes, by The Central Limit Theorem, since ๐‘› > 30, ๐‘ฅาง has an approximately normal distribution. ยฉMcGraw-Hill Education. Example 2: The Central Limit Theorem A sample of size 8 will be drawn from a normal population with mean ๐œ‡ = โ€“60 and standard deviation ๐œŽ = 5. Is it appropriate to use the normal distribution to find probabilities for ๐‘ฅ? าง Solution: Yes, since the population itself is approximately normal, ๐‘ฅาง has an approximately normal distribution. ยฉMcGraw-Hill Education. Example 3: The Central Limit Theorem A sample of size 24 will be drawn from a population with mean ๐œ‡ = 35 and standard deviation ๐œŽ = 1.2. Is it appropriate to use the normal distribution to find probabilities for ๐‘ฅ? าง Solution: No, since the population is not known to be normal and ๐‘› is not greater than 30, we cannot be certain that ๐‘ฅาง has an approximately normal distribution. ยฉMcGraw-Hill Education. Objective 2 Use the Central Limit Theorem to compute probabilities involving sample means ยฉMcGraw-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 ๐œ‡ = 25 years, with a standard deviation of ๐œŽ = 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 ๐œ‡๐‘ฅาง and ๐œŽ๐‘ฅาง ๐œ‡๐‘ฅ = ๐œ‡ = 25 and ๐œŽ๐‘ฅ = ๐œŽ ๐‘› = 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. ยฉMcGraw-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 ๐œ‡๐‘ฅาง and ๐œŽ๐‘ฅาง ๐œŽ 97 ๐œ‡๐‘ฅ = ๐œ‡ = 1135 and ๐œŽ๐‘ฅ = = = 9.7 ๐‘› 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. ยฉMcGraw-Hill Education. You Should Know . . . โ€ข How to construct the sampling distribution of a sample mean โ€ข How to find the mean and standard deviation of a sampling distribution of ๐‘ฅาง โ€ข The Central Limit Theorem โ€ข How to use the Central Limit Theorem to compute probabilities involving sample means (Calculator) ยฉMcGraw-Hill Education.
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