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Formula for a Specific Confidence Interval For a Proportion

Formula for a Specific Confidence Interval For a Proportion. √. √. ^. ^. ^. ^. ( ). ( ). p · q n. p · q n. ^. ^. p – z α /2 < p < p + z α /2. Hint: Put the entire fraction in parentheses. ^. ^. ( ). p · q n. ^.

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Formula for a Specific Confidence Interval For a Proportion

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  1. Formula for a Specific Confidence Interval For a Proportion √ √ ^ ^ ^ ^ ( ) ( ) p · q n p · q n ^ ^ p – zα/2 < p < p + zα/2 Hint: Put the entire fraction in parentheses ^ ^ ( ) p · q n ^ You’re using p-values which are rounded to 3 decimals, so your answer should be rounded to 3 decimals

  2. (Set up each formula and round to THREE DECIMAL PLACES!!) Example 3: A Today Poll of 1015 adults found that 132 approved of the Job Congress was doing in 1995. Find the 95% confidence interval of the true proportion of adults who felt this way. ^ 132 1015 ^ C.I. = n = p = = .13 q = 95% zα/2 = 1015 0.87 1.96 √ √ ^ ^ ^ ^ ( ) ( ) p · q n p · q n ^ ^ p – zα/2 < p < p + zα/2 √ √ ( ) ( ) .13 · .87 1015 .13 · .87 1015 .13 – 1.96 < p < .13 + 1.96 .13 – 0.021 < p < .13 + 0.021 .109 < p < .151

  3. Formula for the Minimum Sample Size Needed For an interval estimate of population proportion 2 ( ) zα/2 E ^ ^ n = p ∙ q Where E is the maximum error of estimate. ALWAYS ROUND UP TO THE NEXT WHOLE NUMBER

  4. Example 5: An educator desires to estimate, within 0.03, the true proportion of high school students who study at least one hour each school night. He wants to be 98% confident. How large a sample is necessary?A) A previous study showed that 60% surveyed spent at least one hour each school night studying. ^ ^ .6 q = p = .4 E = 0.03 C.I. = 98% zα/2 = 2.33 2 2 ( ) ( ) zα/2 E 2.33 0.03 ^^ n = p ∙ q = .6 ∙ .4 = 1447.71 = 1448  ROUND UP!!

  5. Example 5: An educator desires to estimate, within 0.03, the true proportion of high school students who study at least one hour each school night. He wants to be 98% confident. How large a sample is necessary?B) If no estimate of the sample proportion is available, how large should the sample be? ^ ^ .5 q = p = .5 E = 0.03 C.I. = 98% zα/2 = 2.33 2 2 ( ) ( ) zα/2 E 2.33 0.03 ^^ n = p ∙ q = .6 ∙ .4 = 1508.03 = 1509  ROUND UP!!

  6. p 350 – 351 # 4 – 14 even (omit 6), 15, 16, 19, 20

  7. ^ ^ 1. a) p = .5 q = .5 ^ ^ b) p = .45 q = .55 ^ ^ c) p = .462 q = .538 ^ ^ d) p = .583 q = .417 ^ ^ e) p = .453 q = .547

  8. ^ ^ 2. a) p = .12 q = .88 ^ ^ b) p = .29 q = .71 ^ ^ c) p = .65 q = .35 ^ ^ d) p = .53 q = .47 ^ ^ e) p = .67 q = .33

  9. 4. √ √ ( ) ( ) .27 · .73 100 .27 · .73 100 .27– 1.65 < p < .27+ 1.65 .27 – 0.073 < p < .27 + 0.073 .197 < p < .343

  10. 8. √ √ .431 · .569 763 .431 · .569 763 .431– 1.75 < p <.431+ 1.75 .431 – 0.031 < p < .431 + 0.031 .400 < p < .462

  11. 10. √ √ .575 · .425 80 .575 · .425 80 .575– 1.96 < p <.575+ 1.96 .575 – 0.108 < p < .575 + 0.108 .467 < p < .683

  12. 12. √ √ .45 · .55 500 .45 · .55 500 .45– 1.65 < p < .45+ 1.65 .45 – 0.037 < p < .45 + 0.037 .413 < p < .487

  13. 14. √ √ .56 · .44 1000 .56 · .44 1000 .56– 1.96 < p < .56+ 1.96 .56 – 0.031 < p < .56 + 0.031 .529 < p < .591

  14. 15a. 2 ( ) 2.58 .02 n = .25 · .75 n = 3120.188  3121

  15. 15b. 2 ( ) 2.58 .02 n = .5 · .5 n = 4160.25  4161

  16. 16a. 2 ( ) 1.65 .05 n = .29· .71 n = 224.225  225

  17. 16b. 2 ( ) 1.65 .05 n = .5· .5 n = 272.25  273

  18. 19. 2 ( ) 1.96 .03 n = .5· .5 n = 1067.111  1068

  19. 20. 2 ( ) 1.96 .02 n = .27· .73 n = 1892.948  1893

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