Week 9 Wed Oct 27

# Week 9 Wed Oct 27 - Posted after class WEEK 9 – Wed Oct...

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Unformatted text preview: Posted after class WEEK 9 – Wed, Oct 27 Review of Sampling Distributions RANDOM SAMPLING, SAMPLE STATISTICS, SAMPLING DISTRIBUTION (Background material is in Chapter 5) Sampling Distributions of Sample Statistics (Chap 5, Sec 3) Henceforth, unless otherwise stated, the term random sampling will refer to simple random sampling . Simulation study to illustrate using x as a point estimator of µ and s as a point estimator of σ. We have created a population that has numerical characteristic x ranging from 600 to 1000 that has mean µ = 797.6 and standard deviation σ = 116.6. Example 4 . Population of N = 1,000 items with x in the interval [600, 1000]. 960 900 840 780 720 660 600 x Dotplot of x μ σ Variable N Mean StDev Min Max x 1000 797.56 116.57 600.25 999.44 1 Sample Estimation of µ – A simulation with 100 trials of simple random samples of size n = 10 drawn from the Population, in each case computing the sample average x and the sample standard deviation s. Here is a dotplot the 100 x ’s shown below the dotplot of the population values. 960 900 840 780 720 660 600 x xbar Data Dotplot of x, xbar Each symbol represents up to 2 observations. Variable n Mean StDev Min Max xbar 100 790.93 37.94 697.85 905.93 Discussion point: From the dotplot of x values we see that about 90 of the 100 simple random samples yield x ’s that are in the interval 718 to 878, that is, within 80 of the population mean μ = 798. The advantage to random sampling and averaging is: The random samples of size n = 10 produce averages x that vary less about the population mean µ than do the individual values in the population. As we will see, the larger the sample size the less is that variability. 2 Did not go over the review material on pages 3 – 7 in class. Sample Estimation of σ – A simulation with 100 trials of simple random samples of size n = 10 drawn from the population, in each case computing the sample average x and the sample standard deviation s. Here is a histogram of the 100 sample standard deviations s. 156 144 132 120 108 96 84 72 s Dotplot of s Variable n Mean s 100 112.30 Discussion point: From the dotplot of s values we see that about 84 of the 100 simple random samples yield s values between 87 and 147, that is, within 30 of the population standard deviation is σ = 117. The sample standard deviation serves as a point estimator of σ in applications. 3 Simulation study to illustrate using p ˆ as a point estimator of p for a dichotomous population. We have created a population of N = 1000 elements falling into Categories 1 and 0. There are S = 740 in Category 1 and N – S = 240 in Category 0. The population proportion p = .74....
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Week 9 Wed Oct 27 - Posted after class WEEK 9 – Wed Oct...

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