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Sampling Distributions

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Parameter & Statistic Parameter Summary measure about population Sample Statistic Summary measure about sample P P PP in Population & Parameter S S SS in Sample & Statistic

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Common Statistics & Parameters Sample StatisticPopulation Parameter Variance S2S2 Standard Deviation S Mean X Binomial Proportion pp ^

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1. Theoretical probability distribution 2. Random variable is sample statistic Sample mean, sample proportion, etc. 3. Results from drawing all possible samples of a fixed size 4. List of all possible [x, p(x)] pairs Sampling distribution of the sample mean Sampling Distribution

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Sampling from Normal Populations

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Let Y 1,Y 2,…,Y n be a random sample of size n from a normal distribution with mean μand varianceσ 2. Then is normally distribution with mean And variance

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Proof:

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Properties of the Sampling Distribution of x

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3. Formula (sampling with replacement) 2. Less than population standard deviation 1. Standard deviation of all possible sample means, x Measures scatter in all sample means, x Standard Error of the Mean

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Central Tendency Dispersion Sampling with replacement = 50 = 10 X n =16 X = 2.5 n = 4 X = 5 X = 50 - X Sampling Distribution Population Distribution Sampling from Normal Populations

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Standardizing the Sampling Distribution of x Standardized Normal Distribution = 0 = 1 Z Sampling Distribution X X X

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Youre an operations analyst for AT&T. Long-distance telephone calls are normally distribution with = 8 min. and = 2 min. If you select random samples of 25 calls, what percentage of the sample means would be between 7.8 & 8.2 minutes? © T/Maker Co. Thinking Challenge

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Sampling Distribution 8 X = X 0 = 1 –.50 Z Standardized Normal Distribution.1915 Sampling Distribution Solution*

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Sampling from Non-Normal Populations

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Population size, N = 4 Random variable, x Values of x: 1, 2, 3, 4 Uniform distribution © T/Maker Co. Suppose Theres a Population... Developing Sampling Distributions

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Population DistributionSummary Measures P(x) x Population Characteristics

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Sample with replacement Samples 1st Obs 1,11,21,31,4 2,12,22,32,4 3,13,23,33,4 4,14,24,34,4 2nd Observation st Obs 16 Sample Means All Possible Samples of Size n = 2

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nd Observation st Obs 16 Sample MeansSampling Distribution of the Sample Mean P(x) x Sampling Distribution of All Sample Means

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Summary Measures of All Sample Means

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PopulationSampling Distribution P(x) x x Comparison

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A fair die is thrown infinitely many times, with the random variable X = # of spots on any throw. The probability distribution of X is: …and the mean and variance are calculated as well: 9.25 x P(x)1/6 Sampling Distribution of the Mean…

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Sampling Distribution of Two Dice A sampling distribution is created by looking at all samples of size n=2 (i.e. two dice) and their means… While there are 36 possible samples of size 2, there are only 11 values for, and some (e.g. =3.5) occur more frequently than others (e.g. =1). 9.26

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9.27 P( ) 6/36 5/36 4/36 3/36 2/36 1/36 P( ) Sampling Distribution of Two Dice…

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9.28 Compare… Compare the distribution of X… …with the sampling distribution of. As well, note that:

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Sampling from Non-Normal Populations

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Law of Large Numbers The law of large numbers states that, under general conditions, will be near with very high probability when n is large. The conditions for the law of large numbers are Y i, i=1, …, n, are i.i.d. The variance of Y i,, is finite.

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9.32 Central Limit Theorem… The sampling distribution of the mean of a random sample drawn from any population is approximately normal for a sufficiently large sample size. The larger the sample size, the more closely the sampling distribution of X will resemble a normal distribution.

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9.33 Central Limit Theorem… If the population is normal, then X is normally distributed for all values of n. If the population is non-normal, then X is approximately normal only for larger values of n. In most practical situations, a sample size of 30 may be sufficiently large to allow us to use the normal distribution as an approximation for the sampling distribution of X.

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Central Tendency Dispersion Sampling with replacement Population Distribution Sampling Distribution n =30 X = 1.8 n = 4 X = 5 m = 50 s = 10 X m X = 50 - X Sampling from Non-Normal Populations

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X As sample size gets large enough (n 30)... sampling distribution becomes almost normal. Central Limit Theorem

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Central Limit Theorem Example The amount of soda in cans of a particular brand has a mean of 12 oz and a standard deviation of.2 oz. If you select random samples of 50 cans, what percentage of the sample means would be less than oz? SODA

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Sampling Distribution 12 X = X 0 = 1 –1.77 Z.0384 Standardized Normal Distribution.4616 Shaded area exaggerated Central Limit Theorem Solution*

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9.40 Example One survey interviewed 25 people who graduated one year ago and determines their weekly salary. The sample mean to be $750. To interpret the finding one needs to calculate the probability that a sample of 25 graduates would have a mean of $750 or less when the population mean is $800 and the standard deviation is $100. After calculating the probability, he needs to draw some conclusions.

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We want to find the probability that the sample mean is less than $750. Thus, we seek The distribution of X, the weekly income, is likely to be positively skewed, but not sufficiently so to make the distribution of nonnormal. As a result, we may assume that is normal with mean and standard deviation 9.41 Example

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9.42 Example Thus, The probability of observing a sample mean as low as $750 when the population mean is $800 is extremely small. Because this event is quite unlikely, we would have to conclude that the dean's claim is not justified.

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9.43 Using the Sampling Distribution for Inference Heres another way of expressing the probability calculated from a sampling distribution. P(-1.96 < Z < 1.96) =.95 Substituting the formula for the sampling distribution With a little algebra

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9.44 Using the Sampling Distribution for Inference Returning to the chapter-opening example where µ = 800, σ = 100, and n = 25, we compute or This tells us that there is a 95% probability that a sample mean will fall between and Because the sample mean was computed to be $750, we would have to conclude that the dean's claim is not supported by the statistic.

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9.45 Using the Sampling Distribution for Inference For example, with µ = 800, σ = 100, n = 25 and α=.01, we produce

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Sampling Distributions The Proportion The proportion of the population having some characteristic is denoted π.

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Standard error for the proportion: Z value for the proportion: Sampling Distributions The Proportion

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If the true proportion of voters who support Proposition A is π =.4, what is the probability that a sample of size 200 yields a sample proportion between.40 and.45? In other words, if π =.4 and n = 200, what is P(.40 p.45) ? Sampling Distributions The Proportion: Example

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Find : Convert to standardize d normal:

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Sampling Distributions The Proportion: Example Use cumulative normal table: P(0 Z 1.44) = P(Z 1.44) – 0.5 =.4251 Z Standardize Sampling Distribution Standardized Normal Distribution.400 p

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