Section 8.3: Tests Concerning a PopulationProportion1Assume the observed count X ∼ Bin(n, p), wherep is unknown. Then, the estimate of p is ˆp = X/nandˆp ∼approxN(p,p(1 − p)n).Let α be the significance level. Then• for testH0: p = p0(or p ≤ p0) ↔ Ha: p > p0we reject H0ifz =ˆp − p0qp0(1 − p0)/n> zα• for testH0: p = p0(or p ≥ p0) ↔ Ha: p < p0we reject H0ifz =ˆp − p0qp0(1 − p0)/n< −zα• for testH0: p = p0↔ Ha: p 6= p0we reject H0if|z| = |ˆp − p0qp0(1 − p0)/n| > zα/22First example of Section 8.3: example 8.11 ontextbook. Suppose 1276 individuals in a sampleof 4115 adults were found obese. Is the rate sig-nificantly higher than 30% at α = 0.1 or α = 0.05.TestH0: p ≤ 0.3 ↔ Ha: p > 0.3.Here n = 4115 and x = 1276. Thus,ˆp =12764115= 0.31andz =0.31 − 0.30q0.30 × 0.70/4115= 1.40Note that z0.1= 1.28 and z0.05= 1.645. Wefail to reject H0at α = 0.05 but we reject H0atα =
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