Week 7 Mon Oct 11

# Week 7 Mon Oct 11 - This version posted after class. WEEK 7...

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This version posted after class. WEEK 7 – Mon, Oct 11 Chapter 5. Completion of Coverage of Chapter 5 Material. I will start by going over the textbook’s example that runs from page 190 to page 193 concerning sample average SUMMARY Sampling from a Population with Numerical Characteristic x Simple Random Sampling n from N Population x-values x i , i = 1, 2, …, N Sample x-values x i , i = 1, 2, …, n n N Sample Statistics Population Parameters_______ µ ( Sample average (mean) ) ( Population average (mean) ) ( Sample variance ) ( Population variance ) ( Sample standard deviation ) ( Population standard deviation ) Important: At this point you are not expected to calculate the sample variance (in Exer 5-2) or the sample standard deviation s (in Exer 5-5). Facts : With simple random sampling ( SRS ), 1. is a random variable fpcf 2. 3. so that 1

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Finite population correction factor ( fpcf ). When the ratio is small the fpcf is practically 1, that is, . The textbook usually does not include the fpcf . Additional Properties (p. 194). Property 1. When n is large (and not too close to N ), then the distribution of is approximately . The book ignores the fpcf and takes the distribution of to be approximately . These facts are established under the name Central Limit Theorem ( CLT ). Property 2. If the population distribution is then the distribution of is whatever the sample size n . Exercise 5-28 . Population size N is not given. Value of population mean µ is given, µ = \$700. The population standard deviation is given, σ = \$100. Sample size is given, n = 60. We are asked to determine . We will use the normal distribution as the distribution of Ans. We know that with simple random sampling and Thus, using Table 2. 2
Exercise 5-28 modified . Suppose that there are 187 boutiques in Milan from which the random sample of 60 boutiques is chosen. Use the normal distribution as the distribution of to calculate Ans. We know that with simple random sampling and Thus, using Table 2. Exercise 5-22 . Population size N is not given. Value of population mean µ is not given. The population standard deviation is given, σ = \$4500. Sample size is given, n = 225. We will calculate based on simple random sampling and normal distribution for ( CLT ). Sketch probability density for Ans. and Thus, FOR CALCULATOR USERS . Since your TI-83 cannot understand the command normalcdf ( µ - 800, µ + 800, µ , 300), you have to enter a number for µ in order to make the calculation. Take any value for µ . For µ = 1000, we get normalcdf (200, 1800, 1000, 300) = .9923. Added Note (10/15/10) : Someone pointed out that my old TI-83 can handle this situation using the key X,T, θ , n to enter the variable µ . One-Stage Probability-Based Sampling Plans (pp. 187-188) . In the practice of sampling many different sampling designs are available to the researcher. We have only considered estimators based on simple random sampling. We have formulas that give the expectations and the standard deviations of the estimators based on simple random sampling. The estimators and formulas are different for the different sampling plans. Below we illustrate three probability-based sampling plans and estimators for population mean for each plan.

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## This note was uploaded on 12/11/2010 for the course STT 315 taught by Professor Anderson during the Spring '08 term at Michigan State University.

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Week 7 Mon Oct 11 - This version posted after class. WEEK 7...

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