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Math 221: Conﬁdence Intervals for a Population Proportion S. K. Hyde Chapter 19, (Moore, 5th Ed.) The tobacco industry closely monitors all surveys that involve smoking. One survey showed that among 785 randomly selected subjects who completed four years of college, 144 smoke (based on data from the American Medical Association). 1. Large-Sample Conﬁdence Interval for a Population Proportion An approximate level C conﬁdence interval for p is x p(1 − p) ˆ ˆ p±m ˆ where p= ˆ , m = z ∗ SEp , ˆ and SEp = ˆ n n Use this interval only when the number of successes and failures in the sample are both at least 15. Find a 99% conﬁdence interval for the proportion of smokers among the population of people who have completed four years of college. 2. The “Plus-Four” Conﬁdence Interval for a Population Proportion A more accurate level C conﬁdence interval for p is x+2 p(1 − p) ˜ ˜ p±m ˜ where p= ˜ , m = z ∗ SEp , ˜ and SEp = ˜ n+4 n+4 This is the same as the above method, with the addition of adding two imaginary successes and two imaginary failures (four overall) to your sample. Hence, the x + 2 and n + 4 in the ˜ deﬁnition of p. Use this interval when the conﬁdence level is at least 90% and the sample size n is at least 10. Find a 99% “plus four” conﬁdence interval for the proportion of smokers among the population of people who have completed four years of college. Conﬁdence Interval for a Population Proportion, page 2 3. Sample Size for a Desired Margin of Error The level C conﬁdence interval for a population proportion p will have margin of error approx- imately equal to a speciﬁed value m when the sample size is z∗ 2 n= p∗ (1 − p∗ ), m where p∗ is a guessed value for the sample proportion. The margin of error will be less than or equal to m if you take the guess p∗ to be 0.5. Suppose the American Medical Association wants to determine the sample size needed to have a margin of error of ±1%? 4. Suppose the American Medical Association wants to determine the sample size needed to have a margin of error of ±1%? What would be the sample size required if you assume that an estimate for p is not known? Conﬁdence Interval for a Population Proportion, page 3 Solution 1. Find a 99% conﬁdence interval for the proportion of smokers among the population of people who have completed four years of college. 144 ˆ The estimate for p is p = 785 = 0.1834395. Thus, the standard error is 144 p(1 − p) ˆ ˆ 785 1 − 144 785 SEp = ˆ = = 0.01381357. n 785 Thus, the margin of error is m = z ∗ SEp = 2.576(0.01381357) = 0.03558374. ˆ So a 99% conﬁdence interval for the proportion of smokers among the population of people who have completed four years of college is p ± m =⇒ 0.1834395 ± 0.03558374 =⇒ (0.148, 0.219) ˆ Note: When using a TI-83 or TI-84 calculator, select STAT −→ TESTS −→ 1-PropZInt , and enter the appropriate data or statistics. 2. Find a 99% “plus four” conﬁdence interval for the proportion of smokers among the population of people who have completed four years of college. Here we repeat the above procedure, but use x + 2 for x and n + 4 for n. 144+2 146 ˜ The estimate for p is p = 785+4 = 789 = 0.1850444. Thus, the standard error is 146 p(1 − p) ˜ ˜ 789 1 − 146 789 SEp = ˜ = = 0.01382504. n+4 789 Thus, the margin of error is m = z ∗ SEp = 2.576(0.01382504) = 0.03561330. ˆ So a 99% conﬁdence interval for the proportion of smokers among the population of people who have completed four years of college is p ± m =⇒ 0.1850444 ± 0.03561330 =⇒ (0.149, 0.221) ˆ To use the TI-83 or TI-84 calculator for the “plus four” intervals, select STAT −→ TESTS −→ 1-PropZInt , and you should enter x + 2 for “x” and n + 4 for “n”. 3. Suppose the American Medical Association wants to determine the sample size needed to have a margin of error of ±1%? Here, we will choose the estimate for p to be p∗ = p = 144 = 0.1834395. ˆ 785 2 z∗ 2 2.576 144 144 n= p∗ (1 − p∗ ) = 1− = 9939.692 m .01 785 785 The American Medical Association needs to have a sample size of 9,940 people to reduce the margin of error to ±1%. 4. Suppose the American Medical Association wants to determine the sample size needed to have a margin of error of ±1%? Assume that an estimate for p is not known. Here, we will choose the estimate for p to be p∗ = p = 2 . ˆ 1 2 z∗ 2 2.576 1 1 n= p∗ (1 − p∗ ) = 1− = 16589.44 m .01 2 2 If no estimate for p is known, then the American Medical Association needs to have a sample size of 16,590 people to reduce the margin of error to ±1%.