Presentation on theme: "Sampling Distributions. Parameter & Statistic Parameter Summary measure about population Sample Statistic Summary measure about sample P P PP in Population."— Presentation transcript:
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
Common Statistics & Parameters Sample StatisticPopulation Parameter Variance S2S2 Standard Deviation S Mean X Binomial Proportion pp ^
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
Sampling from Normal Populations
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
Properties of the Sampling Distribution of x
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
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
Standardizing the Sampling Distribution of x Standardized Normal Distribution = 0 = 1 Z Sampling Distribution X X X
Population DistributionSummary Measures P(x) x Population Characteristics
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
nd Observation st Obs 16 Sample MeansSampling Distribution of the Sample Mean P(x) x Sampling Distribution of All Sample Means
Summary Measures of All Sample Means
PopulationSampling Distribution P(x) x x Comparison
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…
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
9.27 P( ) 6/36 5/36 4/36 3/36 2/36 1/36 P( ) Sampling Distribution of Two Dice…
9.28 Compare… Compare the distribution of X… …with the sampling distribution of. As well, note that:
Sampling from Non-Normal Populations
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.
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.
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.
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
X As sample size gets large enough (n 30)... sampling distribution becomes almost normal. Central Limit Theorem
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
Sampling Distribution 12 X = X 0 = 1 –1.77 Z.0384 Standardized Normal Distribution.4616 Shaded area exaggerated Central Limit Theorem Solution*
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.
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
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.
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
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.
9.45 Using the Sampling Distribution for Inference For example, with µ = 800, σ = 100, n = 25 and α=.01, we produce
Sampling Distributions The Proportion The proportion of the population having some characteristic is denoted π.
Standard error for the proportion: Z value for the proportion: Sampling Distributions The Proportion
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
Find : Convert to standardize d normal:
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