Week 7 Mon Oct 11 - WEEK 7 Mon Oct 11 Chapter 5 Completion...

Info icon This preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
1 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
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
2 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 5-22 . 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,
Image of page 2
3 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. Exercise 5-28 modified . Suppose that there are 187 boutiques in Milan from which the random sample of 60 boutiques is chosen. Uuse the normal distribution as the distribution of to calculate Ans. We know that with simple random sampling and Thus, using Table 2. 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.
Image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
4 Simple Random Sampling. Select n units at random from the population Stratified Random Sampling. First stratify the populations into strata and then use simple random sampling on each stratum.
Image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern