2 Parameter & Statistic Parameter Sample Statistic Summary measure about populationSample StatisticSummary measure about sampleP in Population & ParameterS in Sample & Statistic
3 Common Statistics & Parameters Sample StatisticPopulation ParameterMeanXStandard DeviationSVarianceS22Binomial Proportionp^
4 Sampling Distribution Theoretical probability distributionRandom variable is sample statisticSample mean, sample proportion, etc.Results from drawing all possible samples of a fixed sizeList of all possible [x, p(x)] pairsSampling distribution of the sample mean
13 Standard Error of the Mean 1. Standard deviation of all possible sample means, x ● Measures scatter in all sample means, xLess than population standard deviation3. Formula (sampling with replacement)
14 Sampling from Normal Populations Central TendencyDispersionSampling with replacementPopulation Distributionm= 50s= 10XSampling Distributionn =16 X = 2.5n = 4 X = 5mX= 50-
15 Standardizing the Sampling Distribution of x Standardized Normal Distributionm= 0s= 1Z
20 Population Characteristics Summary MeasuresPopulation DistributionP(x).3.2Have students verify these numbers..1.0x1234
21 All Possible Samples of Size n = 2 2nd Observation12341stObs16 Sample Means2nd Observation12341stObs1,11,21,31,41.01.52.02.52,12,22,32,41.52.02.53.03,13,23,33,42.02.53.03.54,14,24,34,42.53.03.54.0Sample with replacement
22 Sampling Distribution of All Sample Means 2nd Observation12341stObs16 Sample MeansSampling Distribution of the Sample Mean.0.1.2.31.01.52.02.53.03.54.0P(x)x1.01.52.02.51.52.02.53.02.02.53.03.52.53.03.54.0
23 Summary Measures of All Sample Means Have students verify these numbers.
24 Sampling Distribution ComparisonPopulationSampling DistributionP(x).0.1.2.31126.96.36.199.31.01.52.02.53.03.54.0P(x)xx
25 Sampling Distribution of the Mean… 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:x123456P(x)1/69.25
26 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
27 Sampling Distribution of Two Dice… 6/365/364/363/362/361/36P( )9.279.27
28 Compare… Compare the distribution of X… …with the sampling distribution of .As well, note that:9.28
30 Law of Large NumbersThe 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 areYi , i=1, …, n, are i.i.d.The variance of Yi , , is finite.
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.9.32
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.9.33
36 Sampling from Non-Normal Populations Central TendencyDispersionSampling with replacementPopulation Distributions= 10m= 50XSampling Distributionn = 4 X = 5n =30 X = 1.8m-= 50XX
37 X Central Limit Theorem As sample size gets large enough (n 30) ... sampling distribution becomes almost normal.X
38 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
39 Central Limit Theorem Solution* Shaded area exaggeratedSampling Distribution12s`X= .0311.95s= 1–1.77Z.0384Standardized Normal Distribution.4616
40 ExampleOne 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.9.40
41 ExampleWe want to find the probability that the sample mean is less than $750. Thus, we seekThe 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 meanand standard deviation9.41
42 ExampleThus,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.9.42
43 Using the Sampling Distribution for Inference Here’s another way of expressing the probability calculated from a sampling distribution.P(-1.96 < Z < 1.96) = .95Substituting the formula for the sampling distributionWith a little algebra9.43
44 Using the Sampling Distribution for Inference Returning to the chapter-opening example where µ = 800, σ = 100, and n = 25, we computeorThis 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.9.44
45 Using the Sampling Distribution for Inference For example, with µ = 800, σ = 100, n = 25 and α= .01, we produce9.45
46 Sampling Distributions The Proportion The proportion of the population having some characteristic is denoted π.
47 Sampling Distributions The Proportion Standard error for the proportion:Z value for the proportion:
48 Sampling Distributions The Proportion: Example 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 isP(.40 ≤ p ≤ .45) ?
49 Sampling Distributions The Proportion: Example Find :Convert to standardized normal:
50 Sampling Distributions The Proportion: Example Use cumulative normal table:P(0 ≤ Z ≤ 1.44) = P(Z ≤ 1.44) – 0.5 = .4251Standardized Normal DistributionSampling Distribution.4251Standardize.40.451.44pZ