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Section 7-4: Confidence Intervals And Sample Means for Proportions

Section 7-4: Confidence Intervals And Sample Means for Proportions. Symbols Used in Proportion Notation. p =. population proportion. ^. p =. sample proportion. For a sample proportion,. _ X _ n. ^. ^. ^. 1 - p. q =. p =. ^. * Sometimes p is given. Where:

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Section 7-4: Confidence Intervals And Sample Means for Proportions

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  1. Section 7-4: Confidence Intervals And Sample Means for Proportions

  2. Symbols Used in Proportion Notation p = population proportion ^ p = sample proportion For a sample proportion, _X_ n ^ ^ ^ 1 - p q = p = ^ * Sometimes p is given Where: X = number of sample units that possess the characteristic of interest n = sample size

  3. ^ ^ If necessary, round p and q to three decimal places. Example: In a survey of 200 workers, 168 said they were interrupted three or more times an hour by phone messages, faxes, etc.Find p and q. ^ ^ X = 168 n = 200 _X_ n 168 200 ^ = 0.84 (or p will be given to you ) p = = ^ ^ 1 - p = 1 – 0.84 = 0.16 q =

  4. 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

  5. (Set up each formula and round to THREE DECIMAL PLACES!!) Example 1: In a sample of 100 teenage girls, 30% used hair coloring. Find the 95% confidence interval of the true proportion of teenage girls who use hair coloring. ^ ^ C.I. = n = p = q = 95% zα/2 = 100 0.30 0.70 1.96 √ √ ^ ^ ^ ^ ( ) ( ) p · q n p · q n ^ ^ p– zα/2 < p < p+ zα/2 √ √ ( ) ( ) .3 · .7 100 .3 · .7 100 .3– 1.96 < p < .3+ 1.96 .3 – 0.090 < p < .3 + 0.090 .210 < p < .390

  6. (Set up each formula and round to THREE DECIMAL PLACES!!) Example 2: A survey of 120 female freshman college students showed that 18 did not wish to work after marriage. Find the 95% confidence interval of the true proportion of females who do not wish to work after marriage. ^ _18 120 ^ C.I. = n = p = = .15 q = 95% zα/2 = 120 0.85 1.96 √ √ ^ ^ ^ ^ ( ) ( ) p · q n p · q n ^ ^ p– zα/2 < p < p+ zα/2 √ √ ( ) ( ) .15 · .85 120 .15 · .85 120 .15– 1.96 < p < .15+ 1.96 .15 – 0.064 < p < .15 + 0.064 .086 < p < .214

  7. (Set up each formula and round to THREE DECIMAL PLACES!!) Example 3: A survey of 120 female freshman college students showed that 18 did not wish to work after marriage. Find the 95% confidence interval of the true proportion of females who do not wish to work after marriage. ^ 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 < .15+ 1.96 .13 – 0.021 < p < .13 + 0.021 .109 < p < .151

  8. 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

  9. There are two situations that must be considered: ^ • If some approximation of p is known, that • value can be used in the formula 2) If no approximation of p is known, usep = 0.5 ^

  10. (Set up each formula and round to THREE DECIMAL PLACES!!) Example 4: A researcher wishes to be 95% confident that her estimate of the true proportion of individuals who travel overseas is within 0.04 of the true proportion. Find the sample size necessary. In a prior study, a sample of 200 people showed that 80 traveled overseas last year. _80 200 ^ ^ q = p = = .4 1 – .4 = .6 E = 0.04 C.I. = 95% zα/2 = 1.96 2 ( ) 2 ( ) zα/2 E 1.96 0.04 ^^ n = p ∙ q = .4 ∙ .6 = 576.24 = 577  ROUND UP!!

  11. 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!!

  12. 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!!

  13. p350 – 351 # 4, 8 – 14 even, 15, 16, 19, 20 n = ___ < p < ___

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