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Test about a Population Proportion Let p denote the proportion of individuals or objects in a population who possess a specified property (labeled as “S”). Let X be the number of S’s in the sample. Then p̂ = Xn is the sample proportion. X is a binomial random variable with parameters p and n, i.e. X ∼ Bin(n, p). Furthermore, when the sample size n itself is large, both X and p̂ are approximately normally distributed, i.e. X ∼ N(np, np(1 − p)) ). and p̂ ∼ N(p, p(1−p) n Test about population proportion p will depend on the sample size. Test about a Population Proportion Large-Sample Tests When the sample size is large (n ≥ 30), p̂ is approximately normally distributed with mean p and variance p(1 − p)/n. In particular, under the null hypothesis H0 : p = p0 , p̂ is approximately normally distributed with mean p0 and variance p0 (1 − p0 )/n, i.e. p̂ ∼ N(p0 , p0 (1 − p0 )/n). Therefore the test statistic p̂ − p0 Z=p p0 (1 − p0 )/n has approximately a standard normal distribution. Test about a Population Proportion Example: Problem 35 State DMV records indicate that of all vehicles undergoing emissions testing during the previous year, 70% passed on the first try. A random sample of 200 cars tested in a particular county during the current year yields 124 that passed on the initial test. Does this suggest that the true proportion for this county during the current year differs from the previous statewide proportion? Test the relevant hypotheses using significance level α = 0.5 In this case, p̂ = 124/200 = 0.62. The null hypothesis is H0 : p = 0.70 and the alternative hypothesis is Ha : p 6= 0.70. Test about a Population Proportion 1. Parameter of interest: population proportion p. 2. Null hypothesis: H0 : p = 0.70. 3. Alternative hypothesis: Ha : p 6= 0.70. 4. Test statistic value: 0.62 − 0.70 p̂ − p0 =p = −2.469 z=p p0 (1 − p0 )/n 0.70(1 − 0.70)/200 5. Rejection region: z ≥ z0.025 or z ≤ −z0.025 , where z0.025 = 1.96. 6. Conclusion: since −2.469 < −1.96, we reject H0 . Test about a Population Proportion Summary for large-sample tests for population proportion p Null hypothesis: H0 : p = p0 Test statistic value: z = √ p̂−p0 p0 (1−p0 )/n Alternative Hypothesis Ha : p > p0 Ha : p < p0 Ha : p 6= p0 Rejection Region z ≥ zα (upper-tailed) z ≤ −zα (lower-tailed) either z ≥ zα/2 or z ≤ −zα/2 (two-tailed) Remark: These test procedures are valid provided that np0 ≥ 10 and n(1 − p0 ) ≥ 10. Test about a Population Proportion Type II Error β for large-sample tests Alternative Hypothesis β(p 0 ) Ha : p > p0 Φ p0 −p 0 +zα √ Ha : p < p0 1−Φ Ha : p > p0 Φ Φ p0 −p 0 −zα √ √ p0 (1−p0 )/n p 0 (1−p 0 )/n p0 −p 0 +zα/2 √ √ p0 (1−p0 )/n p 0 (1−p 0 )/n √ p0 (1−p0 )/n p 0 (1−p 0 )/n p0 −p 0 −zα/2 √ √ p0 (1−p0 )/n p 0 (1−p 0 )/n − Test about a Population Proportion Small-Sample Tests When the sample size n is small (n ≤ 30), we test the hypotheses based directly on the binomial distribution. For example, if the null hypothesis is H0 : p = p0 and the alternative hypothesis is Ha : p > p0 , then the rejection region is of the form X ≥ c, where X ∼ Bin(n, p). P(type I error) = P(reject H0 | H0 ) = P(X ≥ c | p = p0 ) = 1 − P(X < c | p = p0 ) = 1 − P(X ≤ c − 1 | p = p0 ) = 1 − B(c − 1; n, p0 ) And P(type II error) = P(fail to reject H0 | p = p 0 ) = P(X < c | p = p 0 ) = P(X ≤ c − 1 | p = p 0 ) = B(c − 1; n, p 0 ) Test about a Population Proportion Small-Sample Tests Remark: in the samll-sample case, it is usually not possible to find a vale c for which P(type I error) is exactly the desired significance level α. Therefore we choose the largest rejection region which statisfying P(type I error) < α. Test about a Population Proportion Example: A coin is tossed 10 times and x = 6 heads are observed. Let p = P(head). Do you believe the coin prefers head with significance level 0.10? Test about a Population Proportion 1. Parameter of interest: proportion for “Heads” p. 2. Null hypothesis: H0 : p = 0.5. 3. Alternative hypothesis: Ha : p > 0.5. 4. Test statistic: X = number of heads. 5. Rjection region: X ≥ c for some c. 6. Significance level: 0.10 ≥ P(type I error) = P(reject H0 | H0 ) = P(X ≥ c | p = 0.5) = 1 − P(X < c | p = 0.5) = 1 − P(X ≤ c − 1 | p = 0.5) = 1 − B(c − 1; 10, 0.5) Therefore c − 1 = 7 and c = 8. 7. Conclusion: since 6 < 7, we fail to reject H0 .