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Module 22: Proportions: One Sample

This module explores hypothesis testing and confidence intervals for proportions using one random sample from a population. It includes examples and calculations for males keeping appointments and asthma prevalence.

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Module 22: Proportions: One Sample

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  1. Module 22: Proportions: One Sample This module presents confidence intervals and tests of hypotheses for proportions for the situation with one random sample from a population. Reviewed 06 June 05 /MODULE 22 22 - 1

  2. Proportions P = Population parameter, the proportion of population with characteristic x = Number in sample with the characteristic n = Total number in sample p = x/n, the sample estimate of the proportion with the characteristic 22 - 2

  3. Hypothesis tests and confidence intervals are based on the normal approximation to the binomial distribution. For the hypothesis: H0: P = P0 vs. H1: P  P0 The options for the test statistic are: where Q0 = 1- P0 22 - 3

  4. Example: AJPH, Nov. 1977;67:1033 - 1036 22 - 4

  5. 22 - 5

  6. One Sample Hypothesis test Table 1 on the previous page indicates that 85.39% of the 1,109 males kept their appointments. Suppose that you had worked in this area previously and the percent of appointments kept before had always been 90% or higher. A question this raises is whether this observed rate is sufficiently different from 90% for you to think your population was different. That is, you would like to test the hypothesis: H0: PM = 0.90 vs. H1: PM 0.90 22 - 6

  7. Hypothesis test for males keeping appointments • The hypothesis: H0: PM = 0.90 vs. H1: PM 0.90 • 2.The assumptions: Independence • 3.The  level:  = 0.05 • 4. The test statistic: • 5. The rejection region: Reject H0: PM = 0.90 if z is not between 1.96 22 - 7

  8. 6. The result: 7. The conclusion: Reject H0: PM = 0.90 since z is not between ± 1.96 22 - 8

  9. Confidence Interval for P Confidence Interval for P, with  = 0.05 22 - 9

  10. Appointment keeping behavior for males 22 - 10

  11. 22 - 11

  12. 22 - 12

  13. Social Class I, Asthma or Wheezy bronchitis n = 191 p = 0.047 or 4.7% 1. The hypothesis: H0: P = 0.10 vs. H1: P 0.10 2. The assumptions: Independence 3. The -level :  = 0.05 4. The test statistic: 5. The rejection region: Reject H0: P = 0.10, if z is not between ±1.96 22 - 13

  14. 6. The result: 7. The conclusion: Reject H0: P = 0.10, since z is not between ± 1.96 22 - 14

  15. Confidence Interval for P with  = 0.05 n = 191, p = 0.047, p(1-p) = 0.047(0.953) = 0.0448 22 - 15

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