Class 9.07 Fall 2004 Handout: One Sample Hypothesis Testing And Inference For the Mean Hypothesis Testing I. A sample from Normal distribution. Suppose X1, ..., Xn is an independent sample from Normal(µ, σ2) distribution. Mean Variance Data Sample size Sample Mean Standard Deviation Distribution σµ Unknown 2 Either X1, ..., Xn Known n Known ¯m = X Known SD Known Normal(µ, σ2) Assumed Reasonable estimate for the population mean µ is ¯m = X. Note that E(m)= µ and � � � √ SE(m − µ0)= Var(m − µ0)= Var(m)(= SE(m)) = σ2/n = σ/ n. For testing H0 : µ =µ0 (1) against H1 :1)µ =µ0 or 2)µ<µ0 or 3)µ>µ0 1use test statistics d = m − µ0 obt SE(m) ∗which follows some distribution d. Theorem. Under the above assumptions about the sample X1, ..., Xn,for test-ing test H0: µ − µ0 =0 at α - significance level vs 1) H1: µ − µ0 ∗ ∗ = 0. Reject H0 if |dobt|≥dcrit(α/2) ∗2) H1: µ − µ0 < 0. Reject H0 if d∗ ≤−dcrit(α)obt ∗3) H1: µ − µ0 > 0. Reject H0 if d∗ ≥ dcrit(α)obt ∗Computation of SE(m) and choice of distribution d: 1. σ is known √ SE(m)= σ/ n ∗ √Test statistics dobt = zobt = m−µ0 follows standard Normal distribution z.σ/ n 2. σ is unknown, but n is large (≥ 30) In this case can omit Normality assumption. √ √ SE(m)= σ/ n ≈ SD/ n and ∗ m−µ0test statistics dobt = zobt = SD/ √ n approximately follows standard Normal distribution z. 3. σ is unknown, and n is not large enough (≤ 30)√ √ SE(m)= σ/ n ≈ SD/ n and ∗ m−µ0test Statistics dobt = tobt = SD/ √ n follows t distribution with df = n − 1 degrees of freedom. II. Proportions For a random variable X drawn from a Binomial(n, p) distribution, let p¯ = X/n.For testing 2� H0 : p =p0 (2) against H1 :1)p =p0 or 2)p<p0 or 3)p>p0 p¯−p0use test statistics d = p−p0).obt SE(¯p − p0)= Var(¯n Since for a Binomial(n, p) random variable X and p¯ = X/n, Var(¯p)= p(1−p),for H0 true (that is, p = p0)have � p0(1 − p0) p − p0)= Var(¯SE(¯p − p0)= n Then ∗ p¯ − p0test statistics dobt = zobt = � p0(1−p0) n has approximately Normal z distribution if np, n(1 − p) ≥ 10. 3Confidence Intervals ∗ m−µ0 ∗ ∗For a test statistics dobt = SE(m) we reject H0 if |dobt|≥dcrit(α/2). If ∗ ∗ ∗ <dcrit(α/2)−dcrit(α/2) <dobt we conclude that evidence against H0 is not statistically significant at α - significance level. Conficence interval for µ is computed by inverting non-rejection region ∗ m − µ0 ∗−dcrit(α/2) < <dcrit(α/2)SE(m) ∗ ∗−dcrit(α/2)SE(m) <m − µ0 <dcrit(α/2)SE(m) with (1 − α)100% confidence interval for µ: ∗ ∗ (m − dcrit(α/2)SE(m); m + dcrit(α/2)SE(m))
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