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## Chapter 8 Section 1

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**Chapter 8Section 1**Distributions of the Sample Mean**Learning objectives**Understand the concept of a sampling distribution Describe the distribution of the sample mean for samples obtained from normal populations Describe the distribution of the sample mean for samples obtained from a population that is not normal 1 2 3 Chapter 8 – Section 1**Learning objectives**Understand the concept of a sampling distribution Describe the distribution of the sample mean for samples obtained from normal populations Describe the distribution of the sample mean for samples obtained from a population that is not normal 1 2 3 Chapter 8 – Section 1**Lets look at a small population.**2,4,6 8,4,3,7 Find the mean. Is this mu or x bar? Now lets take all the possible samples of 2 from this population. Chapter 8 – Section 1**2,4,6 8,4,3,7**2,2 4,4 6,6 8,8 4,4 2,4 4,6 6,8 8,4 4,3 2,6 4,8 6,4 8,3 4,7 2,8 4,4 6,3 8,7 2,4 4,3 6,7 3.3 2,3 4,7 3,7 2,7 7,7**2,4,6 8,4,3,7**2,2 4,4 6,6 8,8 4,4 2,4 4,6 6,8 8,4 4,3 2,6 4,8 6,4 8,3 4,7 2,8 4,4 6,3 8,7 2,4 4,3 6,7 3.3 2,3 4,7 3,7 2,7 7,7**Often the population is too large to perform a census … so**we take a sample How do the results of the sample apply to the population? What’s the relationship between the sample mean and the population mean? What’s the relationship between the sample standard deviation and the population standard deviation? This is statisticalinference Chapter 8 – Section 1**We want to use the sample mean x to estimate the population**mean μ If we want to estimate the heights of eight year old girls, we can proceed as follows Randomly select 100 eight year old girls Compute the sample mean of the 100 heights Use that as our estimate This is using the sample mean to estimate the population mean Chapter 8 – Section 1**Chapter 8 – Section 1**• However, if we take a series of different random samples • Sample 1 – we compute sample mean x1 • Sample 2 – we compute sample mean x2 • Sample 3 – we compute sample mean x3 • Etc. • Each time we sample, we may get a different result • The sample mean x is a random variable!**Because the sample mean is a random variable**The sample mean has a mean The sample mean has a standard deviation The sample mean has a probability distribution This is called the samplingdistributionofthesamplemean Chapter 8 – Section 1**The Central Limit Theorem is about the distribution of**sample means. x CENTRAL LIMIT THEOREM • The Central Limit Theorem is what we will base most or the statistical concepts on for the rest of this course.**Definition**Sampling Distribution of the mean is the probability distribution of sample means, with all samples having the same sample size n.**1. The random variable xhas a distribution (which may or may**not be normal) with mean µ and standard deviation . 2. Samples all of the same size n are randomly selected from the population of xvalues. http://onlinestatbook.com/stat_sim/sampling_dist/index.html Central Limit Theorem Given:**Central Limit Theorem**Conclusions:**Central Limit Theorem**Conclusions: 1. The distribution of sample x will, as the sample size increases, approaches a normal distribution.**Central Limit Theorem**Conclusions: 1. The distribution of sample x will, as the sample size increases, approach a normal distribution. 2. The mean of the sample means will be the population mean µ.**Central Limit Theorem**Conclusions: 1. The distribution of sample x will, as the sample size increases, approach a normal distribution. 2. The mean of the sample means will be the population mean µ. 3. The standard deviation of the sample means will approach n**1. For samples of size n larger than 30, the distribution of**the sample means can be approximated reasonably well by a normal distribution. The approximation gets better as the sample size n becomes larger. 2. If the original population is itself normally distributed, then the sample means will be normally distributed for any sample sizen (not just the values of n larger than 30).http://onlinestatbook.com/stat_sim/sampling_dist/index.html Practical Rules Commonly Used:**the mean of the sample means** Notation µx= µ**the mean of the sample means**the standard deviation of sample mean Notation µx= µ x= n**the mean of the sample means**the standard deviation of sample mean (often called standard error of the mean) Notation µx= µ x= n**As the sample size increases, the sampling distribution of**sample means approaches a normal distribution.**Example: Given the population of women has normally**distributed weights with a mean of 143 lb and a standard deviation of 29 lb, a.) if one woman is randomly selected, find the probability that her weight is greater than 150 lb.b.) if 36 different women are randomly selected, find the probability that their mean weight is greater than 150 lb.**Example: Given the population of women has normally**distributed weights with a mean of 143 lb and a standard deviation of 29 lb, a.) if one woman is randomly selected, find the probability that her weight is greater than 150 lb. z = 150-143 = 0.24 29 P(x>150) = P(z>.24) =1 - 0.5948 = 0.4052 P(z>.24)= 1 - 0.5948 = 0.4052 0.5948 = 143 150 = 29 0 0.24**Example: Given the population of women has normally**distributed weights with a mean of 143 lb and a standard deviation of 29 lb, b.) if 36 different women are randomly selected, find the probability that their mean weight is greater than 150 lb.**Example: Given the population of women has normally**distributed weights with a mean of 143 lb and a standard deviation of 29 lb, b.) if 36 different women are randomly selected, find the probability that their mean weight is greater than 150 lb. x = 143 150 x= 29 = 4.83333 36**Example: Given the population of women has normally**distributed weights with a mean of 143 lb and a standard deviation of 29 lb, b.) if 36 different women are randomly selected, find the probability that their mean weight is greater than 150 lb. z = 150-143 = 1.45 29 36 0.9265 x = 143 150 x= 4.83333 0 1.45**Example: Given the population of women has normally**distributed weights with a mean of 143 lb and a standard deviation of 29 lb, b.) if 36 different women are randomly selected, find the probability that their mean weight is greater than 150 lb. z = 150-143 = 1.45 29 36 = 0.0735 0.9265 x = 143 150 x= 4.83333 0 1.45**Example: Given the population of women has normally**distributed weights with a mean of 143 lb and a standard deviation of 29 lb, b.) if 36 different women are randomly selected, the probability that their mean weight is greater than 150 lb is 0.0735. z = 150-143 = 1.45 29 36 1 - 0.9265 = 0.0735 0.4265 x = 143 150 x= 4.83333 0 1.45**Example: Given the population of women has normally**distributed weights with a mean of 143 lb and a standard deviation of 29 lb,**a.) if one woman is randomly selected, find the probability**that her weight is greater than 150 lb.P(x > 150) = 0.4052 Example: Given the population of women has normally distributed weights with a mean of 143 lb and a standard deviation of 29 lb,**a.) if one woman is randomly selected, find the probability**that her weight is greater than 150 lb.P(x > 150) = 0.4052 b.) if 36 different women are randomly selected, their mean weight is greater than 150 lb.P(x> 150) = 0.0735 Example: Given the population of women has normally distributed weights with a mean of 143 lb and a standard deviation of 29 lb,**a.) if one woman is randomly selected, find the probability**that her weight is greater than 150 lb.P(x > 150) = 0.4052 b.) if 36 different women are randomly selected, their mean weight is greater than 150 lb.P(x36 > 150) = 0.0735It is much easier for an individual to deviate from the mean than it is for a group of 36 to deviate from the mean. Example: Given the population of women has normally distributed weights with a mean of 143 lb and a standard deviation of 29 lb,**If n > 0.05 N**Sampling WithoutReplacement**If n > 0.05 N**Sampling WithoutReplacement N - n x = n N - 1**If n > 0.05 N**Sampling WithoutReplacement N - n x = n N - 1 finite population correction factor**0**Distribution of 50 Sample Means for 50 Students 15 Frequency 10 5 0 1 2 3 4 5 6 7 8 9 Figure 5-20