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Chapter 8 1 Inferences for proportions • sample proportion ( pˆ ) fraction of measurements in the sample that are “successes”: € pˆ = (# successes in sample) / n • population proportion ( p ) € fraction of the measurements in the population that are “successes” Sampling distribution of pˆ From an SRS of size n selected from a large population € (sample should be no more than 10% of the population) with population proportion p, we compute the sample proportion pˆ . • mean of the sampling distribution is p. € • standard deviation of the sampling distribution, also called the standard error for pˆ , is p(1− p) € n • distribution becomes more nearly normal as the € sample size increases Chapter 8 2 Procedures for population proportions From an SRS selected from a large population (and at most 10% of the size of the population), determine the sample statistic pˆ . An approximate level C confidence interval for € p is pˆ(1− pˆ ) pˆ ± z* n € where z* is the upper € € 1−C normal critical value. 2 [TI-83: STAT TESTS 1-PropZInt… ] Chapter 8 3 One-sample z test for pˆ Assumptions: SRS selected from a population with € population proportion pˆ , large enough that both np0 (number of successes) and n(1 – p0) (number of failures) are sufficiently large (10 or more). € • State hypotheses: Null hypothesis Alternative hypothesis H0 : p = p 0 Ha: p > p0, or p < p0, or p ≠ p0 • Calculate test statistic: z-statistic based on H0: z= pˆ − p0 p0 (1− p0 ) n • Find P-value: Probability associated with appropriate H0: € P = P( Z ≥ z ), or P = P( Z ≤ z ), or P = 2P( Z ≥ z ) Conclusion: assess evidence against H0 in favor of Ha depending on how small P is. [TI-83: STAT TESTS 1-PropZTest… ] Chapter 8 4 Choosing the sample size The margin of error in using pˆ as an estimate for p is m =€ z * pˆ (1− pˆ ) n To solve for the sample size n, we would need to know € the value of pˆ before we even know the size of the same we will select to measure pˆ ! We can, however, estimate it sufficiently well. If we use the estimate p*, then € solving for n yields € z* 2 n = p*(1− m p*) Using the conservative estimate p* = 0.5 produces a € larger value of n in most circumstances than may be necessary (but larger samples produce even more accurate estimates, so this is not a big problem). Chapter 8 5 Sampling distribution of pˆ 1 – pˆ 2 An SRS of size n1 is selected from a large population € € with population proportion p1, and an independent SRS of size n2 is selected from a large population with population proportion p2. (Neither sample should contain more than 10% of the population.) Compute the sample proportions pˆ 1 and pˆ 2. • mean of the sampling distribution is p1 – p2. € € • standard deviation of the sampling distribution, also called the standard error of the statistic pˆ 1 – pˆ 2, is € € p1(1− p1) n1 + p2 (1− p2 ) n2 • distribution becomes more nearly normal as the € sample size increases Chapter 8 6 Confidence level for comparing two proportions An SRS of size n1 is selected from a large population with population proportion p1, and an independent SRS of size n2 is selected from a large population with population proportion p2. An approximate level C confidence interval for p1 – p2 is pˆ (1 − pˆ1 ) pˆ 2 (1 − pˆ 2 ) ( pˆ 1 – pˆ 2) ± z* 1 + n1 n2 € € where z* is the upper € 1−C normal critical value. 2 € [TI-83: STAT TESTS 2-PropZInt… ] • pooled proportion if H0: p1 = p2 is assumed true, then both samples come from the same population; the samples can then be combined to compute a pooled proportion pˆ = (combined count of successes in both samples) n1 + n 2 € Chapter 8 7 Two-sample z test for comparing two proportions Assumptions: Two independent SRSs are selected from large populations, large enough that n1p1 and n2p2 (number of successes) and n1(1 – p1) and n2(1 – p2) (number of failures) are 10 or more. • State hypotheses: Null hypothesis Alternative hypothesis H0: p1 = p2 Ha: p1 > p2, or p1 < p2, or p1 ≠ p2 • Calculate test statistic: z-statistic (here, pˆ is the pooled proportion): pˆ1 − pˆ 2 z= € 1 1 pˆ(1− pˆ ) + n n2 1 • Find P-value: Probability associated with appropriate H0: € P = P( Z ≥ z ), or P = P( Z ≤ z ), or P = 2P( Z ≥ z ) Conclusion: assess evidence against H0 in favor of Ha depending on how small P is. [TI-83: STAT TESTS 2-PropZTest… ]