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Confidence Intervals for Population Means: Understanding the Concept and Calculation, Study notes of Statistics

An overview of confidence intervals for population means, including the concept, formulas, and examples. It covers the relationship between sample mean, population mean, standard deviation, and sample size, as well as the use of the student t distribution for calculating confidence intervals when the population standard deviation is unknown.

Typology: Study notes

Pre 2010

Uploaded on 09/02/2009

koofers-user-knr
koofers-user-knr 🇺🇸

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Download Confidence Intervals for Population Means: Understanding the Concept and Calculation and more Study notes Statistics in PDF only on Docsity! Review If X 1, . . . , X n have mean µ and SD σ, E(X̄ ) = µ no matter what SD(X̄ ) = σ/ √ n if the X ’s are independent If X 1, . . . , X n are iid normal(mean=µ, SD=σ), X̄ ∼ normal(mean = µ, SD = σ/ √ n). If X 1, . . . , X n are iid with mean µ and SD σ and the sample size, n, is large, X̄ ∼ normal(mean = µ, SD = σ/ √ n). 1 A discrepancy Caution: Sometimes the order in which the book covers material is a bit odd. (The authors would probably think I’m odd.) But sometimes it is just wrong. (Or perhaps they are making some simplifications to ease learning.) A case in point: Let X 1, . . . , X n be random draws from a population with mean µ and SD σ, and X̄ the sample average. Book Karl σ population SD population SD s SD of the data our estimate of σ σ/ √ n SD of the sampling distribution of X̄ SD(X̄ ) aka SE(X̄ ) s/ √ n Standard error of the mean our estimate of SE(X̄ ) 2 Confidence intervals Suppose we measure the log10 cytokine response in 100 male mice of a certain strain, and find that the sample average (x̄) is 3.52 and sample SD (s) is 1.61. Our estimate of the SE of the sample mean is 1.61/ √ 100 = 0.161. A 95% confidence interval for the population mean (µ) is 3.52± (2× 0.16) = 3.52± 0.32 = (3.20, 3.84). What does this mean? What is the chance that (3.20, 3.84) contains µ? 3 Suppose that X 1, . . . , X n are iid normal(mean=µ, SD=σ). Suppose that we actually know σ. Then X̄ ∼ normal(mean=µ, SD=σ/ √ n) where σ is known but µ is not. How close is X̄ to µ? Pr ( |X̄ − µ| σ/ √ n ≤ 1.96 ) = 95% Pr ( −1.96 σ√ n ≤ X̄ − µ ≤ 1.96 σ√ n ) = 95% µ σ n Pr ( X̄ − 1.96 σ√ n ≤ µ ≤ X̄ + 1.96 σ√ n ) = 95% 4 But we don’t know the SD Use of X̄ ± 1.96 σ/ √ n as a 95% confidence interval for µ requires knowledge of σ. That the above is a 95% confidence interval for µ is a result of the following: X̄ − µ σ/ √ n ∼ normal(0,1) What if we don’t know σ? We plug in the sample SD (s), but then we need to widen the intervals to account for the uncertainty in s. 9 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 500 BAD confidence intervals for µ (σ unknown) 10 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 500 confidence intervals for µ (σ unknown) 11 The Student t distribution If X 1, X 2, . . . X n are iid normal(mean=µ, SD=σ), X̄ − µ s/ √ n ∼ t(df = n− 1) Discovered by William Gossett (“Student”) who worked for Guiness. In R, use the functions pt(), qt(), and dt(). e.g., qt(0.975,9) returns 2.26 (cf 1.96) pt(1.96,9)-pt(-1.96,9) returns 0.918 (cf 0.95) −4 −2 0 2 4 df=2 df=4 df=14 normal 12 The t interval If X 1, . . . , X n are iid normal(mean=µ, SD=σ), X̄ ± t(α/2, n− 1) s/ √ n is a 1− α confidence interval for µ. t(α/2, n−1) is the 1−α/2 quantile of the t distribution with n− 1 “degrees of freedom.” −4 −2 0 2 4 t(α 2, n − 1) α 2 In R: qt(0.975,9) for the case n=10, α=5%. 13 Example 1 Suppose we have measured the log10 cytokine response of 10 mice, and obtained the following numbers: Data 0.2 1.3 1.4 2.3 4.2 4.7 4.7 5.1 5.9 7.0 x̄ = 3.68 s = 2.24 n = 10 qt(0.975,9) = 2.26 95% confidence interval for µ (the population mean): 3.68 ± 2.26 × 2.24 / √ 10 ≈ 3.68 ± 1.60 = (2.1, 5.3) ● ●●●● ●●● ● 0 1 2 3 4 5 6 7 95% CI s 14
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