6/7/10 Lecture 18 1STOR 155 Introductory StatisticsLecture 18: Inference for a single proportionSection 8.1The UNIVERSITY of NORTH CAROLINAat CHAPEL HILL6/7/10 Lecture 18 2Normal Approximation for Counts and Proportions• Let X ~ B(n, p) and• If n is large, then• Rule of Thumb: np 10, n(1 - p) 10../ˆnXp )./)1( ,( approx. is ˆn-pppNp6/7/10 Lecture 18 3Confidence Interval for p• Expression:where the margin of error – Assumption: n is large– Confidence level C determines m = zãpê(1 àpê)=np[ pê à m ; pê + m ]zã6/7/10 Lecture 18 4Hypothesis Testing for p• For a hypothesized value , we want to test versus some alternative (1-sided or 2-sided). Recall the 4 steps …• Step 1: only need to specify • Step 2: Test statistic where p0H0: p = p0Haz = (pê à p0)=û0û0= p0(1 à p0)=nq6/7/10 Lecture 18 5Hypothesis Testing for p (continued)• Step 3: The P-value will be equal to P(Z > z) for 1-sided (upper tail) P(Z < z) for 1-sided (lower tail) 2 P(Z > |z|) for 2-sided • Step 4: Compare the P-value with the significance level and draw your conclusion.Ha: p > p0Ha: p < p0Ha: p6=p0ë6/7/10 Lecture 18 6“Biased” one-Euro Coin?• A group of statistics students spun the Belgian one-Euro coin 250 times, and heads came up 140 times.• p = P(H) in each spin• Claim: the coin is biased (more specifically, p is greater than 0.5)6/7/10 Lecture 18 7“Biased” one-Euro Coin? (continued)• Sample: 140 heads among 250 spins of a Belgian one-Euro coin (a hint)• p = P(H) in each spin• H0: p = 0.5 vs Ha: p > 0.5 (one-sided upper)• P-value = P(Z > 1.897) = 1 – 0.9713 = 0.0287• … Conclude based on a givenz = (140=250 à0:5) ä 0:5(1 à0:5)=250p= 1:897ë6/7/10 Lecture 18 8“Biased” one-Euro Coin? (continued)What about a 95% CI for p ?Note: Margin of error 95% CI = [0.56 – 0.06, 0.56 + 0.06] = [0.5, 0.62]ëm = zãpê(1 àpê)=np= 1:96 0:56(1 à0:56)=250pù 0:066/7/10 Lecture 18 9Take Home Message• CI for a single proportion p• Margin of error m in CI• Hypothesis testing for p: 4 steps• Assumption: large n How large ? np0õ 10 and n(1 à p0) õ
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