Lecture 11

# Lecture 11 - LECTURE 11 Interval Estimation This lecture...

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LECTURE 11 Interval Estimation This lecture covers material on interval estimation of a population mean and proportion -- large sample and small sample cases, confidence intervals, t- distribution, and determination of the sample size. Microsoft Excel is used for some applications. Read: Chapter 8, Sections 8.1 through 8.4. 1

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This lecture builds further on the issue of statistical inference. The manager knows that point estimates of population parameters that are based on sample statistics derived from a single sample cannot be expected to be equal to the population parameter values. Therefore it is important to know how precise are these estimates. That is where interval estimation plays a role. The objective is to say with some degree of confidence that the estimate of a population parameter (e.g., mean, proportion etc.,) lies within an interval. Interval Estimation of Population Means In statistical inference, we need to make estimates of population parameters. So we take a random sample and use its sample statistics to make these point estimates. From our discussion in Lecture 9: The sample mean is a point estimate of the population mean μ . may not be exactly equal to μ . The interval estimate for the population mean is obtained by: ± Margin of Error Large Sample Case - Population Standard Deviation known: In this case sample size is large or n 30. Population Mean = μ is not known. Population standard deviation = σ is known. Sample mean = Sample standard deviation = s . The sampling distribution of sample means can be used to make probability statements about the margin of error. Calculating Probability Estimates of Margin of Error: By the central limit theorem, the sampling distribution of sample means will be normally distributed with a mean μ and standard deviation σ x = σ / n. Suppose we want to know that given a sample mean of , the probability that the margin of error is 6. 2
Note: This means that the maximum difference between and μ or ( - μ ) = ± 6. If is > μ , then ( - μ ) + 6; If is < μ , then ( - μ ) - 6. We want to find P( | - μ | 6). P( | - μ | 6) = P(0 < 6) + P(0 > -6). Converting to the standard normal table for the positive value of : z = ( - μ )/ σ x = 6/ σ x . (Note: Therefore ‘6’ = z * σ x ). We need to find P(0 < z 6/ σ x ). As the normal curve is symmetrical, this is also the same as P(0> z -6/ σ x ) Therefore P( | - μ | 6) = P(0 < z 6/ σ x ) + P(0 > z -6/ σ x ). 3

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EXAMPLE 1 We take a sample of size 64 from a population having a known standard deviation of 16. In this case sample size n = 64 is large or n 30. Population Mean = μ is unknown. Population standard deviation
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## This note was uploaded on 09/21/2011 for the course OM 210 taught by Professor Singer during the Summer '08 term at George Mason.

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Lecture 11 - LECTURE 11 Interval Estimation This lecture...

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