Week 11 Mon Nov 8

# Week 11 Mon Nov 8 - Posted before class WEEK 11 Mon Nov 8...

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1 Posted before class WEEK 11 – Mon, Nov 8 Lower Estimates - (1 – α)100% Confidence Level One-Sided Confidence Intervals Guidelines from the Centers for Medicare & Medicaid Services state “In most situations, the lower limit of a one-sided 90 percent confidence interval should be used as the amount of overpayment to be demanded for recovery from the physician or supplier.” CI for μ - Case σ is known Book (N indefinitely large) N specified – Correct Formula Discuss limitations. CI for μ - Case σ is unknown Book (N indefinitely large) N specified – Correct Formula Use with n – 1 df (degrees of freedom). If you use the t-table in the book and the df value is not in the table, go to the row above the missing value. Discuss limitations. CI for p (appropriate for large n) Book (N indefinitely large) N specified – Correct Formula Discuss limitations. Example of Lower Estimate . The x-characteristic of a population of size 900 has unknown mean and unknown standard deviation. A 90% lower estimate is needed for the population mean. A simple random sample of 48 units is selected. The sample has average 36.54 and standard deviation 22.45. Calculate the lower estimate. Ans. 32.34

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2 Upper Estimates - (1 – α)100% Confidence Level One-Sided Confidence Intervals CI for μ - Case σ is known Book (N indefinitely large) N specified – Correct Formula Discuss limitations. CI for μ - Case σ is unknown Book (N indefinitely large) N specified – Correct Formula Use with n – 1 df (degrees of freedom). If you use the t-table in the book and the df value is not in the table, go to the row above the missing value. Discuss limitations. CI for p (appropriate for large n) Book (N indefinitely large) N specified – Correct Formula Discuss limitations. Example . The 600 parts in a bin has an unknown proportion p of parts that will not pass inspection. A simple random sample of 150 parts from the bin is inspected and it is found that 12 of the 150 will not pass inspection. A 95% upper estimate is needed for the number of parts in the bin that will not pass inspection. Calculate the upper estimate. Ans. The upper estimate for p is 0.1116 which multiplied by 600 gives the answer 66 or 67 parts.
Review of Sampling Distributions based on Simple Random Sampling Sampling from a dichotomous population of size N . We are interested in the sampling distribution of S N - S Simple Random Sampling n from N x n - x Successes Others n N Sample Population Then, with the sample proportion , X has a Hypergeometric distribution H( n , S , N ). Example.

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Week 11 Mon Nov 8 - Posted before class WEEK 11 Mon Nov 8...

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