Week 10 Mon Nov 1 - Posted after class WEEK 10 – Mon Nov 1 Large Sample CI for p(Chap 6 Sec 4 Dichotomous Population The point estimate of p is

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Unformatted text preview: Posted after class WEEK 10 – Mon, Nov 1 Large Sample CI for p (Chap 6, Sec 4) Dichotomous Population. The point estimate of p is the sample proportion p ˆ . Ignoring the fpcf , a large n, (1 – α)100% confidence interval estimator (CI) for p is n p p z p ) ˆ 1 ( ˆ ˆ 2 /- ± α . (6-7) This interval estimator is delivers approximate coverage probability (1 – α)100% for the population p provided n is large. The textbook suggests that if both np > 5 and n(1 – p) > 5, then the interval is reliable, that is, it delivers close to the (1 – α)100% coverage probability it is designed to deliver. Example, Suppose p = .10, n > 50. Then np > 5 and n(1 - p) > 5, so the book suggests that the (random) interval estimate (6-7) will capture p with probability close to the nominal 1 – α. The margin of error is n p p z ) ˆ 1 ( ˆ 2 /- α . It is the half-width of the (1 – α)100% confidence interval (6-7). Example. Exercise 6-46 . That random sampling was used is stated. Sample size n = 100 given. Population size N not given. Dictomous population. Sample proportion 85 . 100 / 85 ˆ = = p given. Confidence level (1 – α)100% = 90% given. Large n, so we use the z-interval estimate of the population proportion p. Here α = .10, α/2 = .05, z .05 = 1.645. Thus, the z-interval (6-7) evaluates to be 100 ) 85 . 1 ( 85 . 645 . 1 85 .- ± There are two common ways to present this: .85 ± .059 and .791 < p < .909. Discuss interpretation of a confidence interval estimate. 1 Summary of the CI’s as presented so far and in the textbook. n z X / 2 / σ α ± (6-3) n s t X / 2 / α ± (6-5) n p p z p ) ˆ 1 ( ˆ ˆ 2 /- ± α (6-7) Review Exercises Exercise 6-5 . The variable of interest is x = value (in $), a numerical characteristic. We are told that x is approximately normally distributed in the population. Population size N not given. Population standard deviation is given as σ = $5,500. Random sampling was used with sample size n = 16 with x =$89,673. Find 95% CI for µ. Ans. Using (6-3) with α = 0.05 gives $89,673 ± 1.960 ( 16 / $5,500) , which is $89,673 ± $2,695. Can be written [$86,978, $92,368] or $86,978 ≤ µ ≤ $92,368....
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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 10 Mon Nov 1 - Posted after class WEEK 10 – Mon Nov 1 Large Sample CI for p(Chap 6 Sec 4 Dichotomous Population The point estimate of p is

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