stat2 test - Statistics and Sampling Distributions...

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Statistics and Sampling Distributions Ultimately we would like to know the population characteristics (parameters), but we must estimate them since we do not or can not examine the entire population. We must use the sample information, the statistics. There are 2 basic criteria for a ‘good’ statistic: unbiasedness —in the long run the statistic averages to the parameter, and minimum variance —all of the values will be ‘close’ to the parameter. Since a statistic changes values every time you take a new sample, statistics also have distributions. These distributions, called sampling distributions , are dependent on 3 things: 1. the original (parent) population, including its mean and variance 2. the size of the sample, n —the bigger the sample the less variability the statistic has 3. the sampling method, we will always use simple random samples altho there are others—random samples give us unbiased statistics The Sampling Distribution of X 1. μ ( x ) = μ ( x ) __ unbiasedness 2. σ ( x ) = σ ( x )/ n allows for minimum variance 3. If the data is normally distributed, then the sampling distribution of the sample mean, x , will also be normally distributed, no matter what the sample size, n . Realize that if you took a sample of size 1, you would still have normal data since it would just be a sample from a normal population. 4. The Central Limit Theorem : When n is large enough, then the sampling distribution of the sample mean, x , is approximately normally distributed, no matter how the original population is distributed. “large enough” is usually thought to be 30 more for the CLT to hold. This means we can find a normal distribution for ANY data, as long we look at the sample mean from a large sample! __ _ Let X n be the random average of X’s from a sample of size n where X ~ N( μ x , σ 2 x ), then X n ~N( μ x , σ 2 x / n ) _ z = ( x - μ x )/ σ x / n ~ N( 0, 1) The Central Limit Theorem can even be applied to categorical data . Remember, we must use proportions since we can not calculate a mean. The Sampling Distribution of p 1. μ ( p ) = π _______ 2. σ ( p ) = √π (1 )/ n 3. For n π , n (1 - π ) 10 , the sample proportion, p (= x/n ) is approximately N( π , π (1- π )/ n ) and z = ( p - π )/ √π (1- π )/ n
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This note was uploaded on 03/25/2008 for the course STAT 501 taught by Professor Dongling during the Spring '08 term at Texas A&M.

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