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Week 10 Mon Nov 1

# Week 10 Mon Nov 1 - Posted before class WEEK 10 Mon Nov 1...

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1 Posted before 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 / . (6-7) This interval estimator is delivers approximate coverage probability (1 α) for the population p provided n is large. May not be reliable if p is too near to 0 or too near to 1. The margin of error is n p p z ) ˆ 1 ( ˆ 2 / . 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.

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2 Summary of the CI’s as presented so far and in the textbook. n z X / 2 / (6-3) n s t X / 2 / (6-5) n p p z p ) ˆ 1 ( ˆ ˆ 2 / (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. Exercise 6-6 . Find 99% CI for μ. Ans. Using (6- 3) with α = 0.0 1 gives \$89,673 ± 2.576 ( 16 / \$5,500) , which is \$89,673 ± \$3,542. Can be written [\$86,131, \$93,215] or \$86,131 ≤ µ ≤ \$93,215. Exercise 6-22 . Find 95% CI for μ. Ans. Using (6-5 ) with α = 0.0 5, n = 100, x =3.2 and s = 2.1 yields 3.2 ± 2.000 ( 100 / 2.1) , 3.2 ± 0.42. Can be written [2.78, 3.62] or 2.78 ≤ µ ≤ 3.62. Exercise 6-46 . Find 90% CI for p. Ans. Using (6-7 ) with α = 0. 10, n = 100, p ˆ = 0.85 yields 100 ) 15 . 0 )( 85 . 0 ( 645 . 1 85 . 0 , 0.85 ± 0.059. Can be written [0.791, 0.909] or 0.791 ≤ p ≤ 0.909 or 79.1% ≤ p% ≤ 90.9%.
3 Please note that the CI’s corrected for the finiteness of the population are: n N n N z

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