Sampling Distributions and the Central Limit Theorem. Definition

Transcription

1 Sampling Distributions and the Central Limit Theorem We have been studying the relationship between the mean of a population and the values of a random variable. Now we will study the relationship between a population mean and the means of samples taken from the population. Definition A sampling distribution is the probability distribution of a sample statistic that is formed when samples of size n are repeatedly taken from a population. If the sample statistic is the sample mean, then the distribution is the sampling distribution of sample means. Every sample statistic has a sampling distribution. Page 26, look at the Venn diagram. the rectangle represents a large population and each circle represents a sample of size n. Because the sample entries can differ, so can the sample means. Properties of Sampling Distributions of Sample Means. The mean of the sample means is equal to the population mean μ 2. The standard deviation of the sample mean is equal to the population standard deviation σ divided by the square root of the sample size n. The standard deviation of the sampling distribution of the sample means is called the standard error of the mean.

2 You write the population values {,,,7} on slips of paper and put them in a box. Then you randomly choose two slips of paper, with replacement. List all possible samples of size n =2 and calculate the mean of each of each. These means form the sampling distributions of the sample means. Find the mean, variance, and standard deviation of the sample means. Compare your results with the mean μ =, variance σ 2 =, and standard deviation σ = of the population. Sample sample mean, x Sample sample mean, x,,..7,,,,7 2 2,,,,7 7, 7, 7, 7, Make a probability distribution of the sample means, draw a histogram of the probabilities. x f probability /6 = /6 = 0.20 /6 = 0.87 /6 = /6 = /6 = 0.20 /6 =

3 Using the properties of sampling distributions of sample means, we get: List all possible samples of size n =, with replacement, from the population {,,}. Calculate the mean of each sample. Find the mean, variance, and standard deviation of the sample means. Compare your results with the mean μ =, variance σ 2 = 8/, and standard deviation σ = 8/.6 of the population. Sample Sample mean, x Sample Sample mean, x,,,,,,,,,,,,,,,,,,,,,,,,,,,,,, ,,,,,,,,,,,,,,,,,,,,,,,, Probability /27=0.070 /27=0. 6/27= /27=0.29 6/27= /27=0. /27=0.070

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5 THE CENTRAL LIMIT THEOREM. If samples of size n, where n 0, are drawn from any population with a mean μ and a standards deviation σ, then the sampling distribution of the sample means approximates a normal distribution. The greater the sample size, the better the approximation. (See figures for "Any Population Distribution on next page) 2. If the population itself is normally distributed, then the sampling distribution of sample means is normally distributed for any sample size n. (See figures for "Normal Population Distribution" on next page) In either case, the sampling distribution of sample means has a mean equal to the population mean. mean of the sample means The sampling distribution of the sample means has a value equal to /n times the variance of the population and a standard deviation equal to the population standard deviation divided by the square root of n. Standard deviation of the Variance of the sample means sample means Remember that the standard deviation of the sampling distribution of the sample means,, is also called the standard error of the mean.

6 . Any Population Distribution Standard deviation 2. Normal Population Distribution Standard deviatio Mean Distribution of Sample means, n 0 standard deviation of the sample means Mean Distribution of Sample Means (any n) standard deviation of the sample means mean mean 6

7 EXAMPLE: Cell phone bills for residents of a city have a mean of $7 of $9, as shown in the figure. Random samples of 00 cell phone bills are drawn from this population, and the mean of each sample is determined. Find the mean and standard deviation of the sampling distribution of the sample means. Then sketch a graph of the sampling distribution. The mean of the sampling distribution is equal to the population mean. So, The standard deviation of the sample means is equal to the population standard deviation divided by n. So, 7

8 Because the sample size is greater than 0, the sampling distribution can be approximated by a normal distribution with a mean of $7 and a standard deviation of $0.90. Assume the training heart rates of all 20 year old athletes are normally distributes, with a mean of beats per minute and a standard deviation of 8 beats per minute. Random samples of size are drawn from the population, and the mean sample is determined. Find the mean and standard deviation of the sampling distribution of the sampling means. Then sketch a graph of the sampling distribution. The sampling mean is and the standard deviation of the sample means is 9. So the graph should look like: 8

9 PROBABILITY AND THE CENTRAL LIMIT THEOREM In section.2, you learned how to find the probability that a random variable x will lie in a given interval of population values. We can find the probability that a sample mean x will lie in a given interval of the x sampling distribution. To transform x to a z score, you can use the formula EXAMPLE P. 266 The figure on P. 266 show the lengths of time people spend driving each day. You randomly select 0 drivers ages to 9. What is the probability that the mean time they spend driving each day is between 2.7 and 2. minutes. Assume σ =.. (a) Since we need to find the probability between 2.7 and 2., we must find both z scores. 9

10 We are trying to find the probability that the drivers are between 2.7 and 2. minutes. This means that of the samples of 0 drivers ages to 9, about 9% will have a drive time that is between 2.7 and 2. minutes. You randomly select 00 drivers ages to 9 from the chart on page 266. What is the probability that the mean time they spend driving each day is between 2.7 and 2. minutes? Use μ = 2 and σ =.. 0

11 The mean room and board expense per year at four year colleges is $926. You randomly select 9 four year colleges. What is the probability that the mean room and board is less than $900? Assume that the room and board expenses are normally distributed with a standard deviation of $00.

12 The average credit card debt carried by undergraduates is normally distributed, with a mean of $7 and a standard deviation of $20. (a) What is the probability that a randomly selected undergraduate, who is a credit card holder, has a credit card balance less than $2700? μ = 7 σ = 20 P(x < 2700) = 0.6 (b) You randomly select 2 undergraduates who are credit card holders. What is the probability that their mean credit card balance is less than $2700. P(x < 2700) = 0.07 There is a % chance that an undergraduate will have a balance less than $2700, but there is only about a 2% chance that a sample of 2 graduates will have a balance less than $2700. Try It Yourself 6 P. 268 P odds 2