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Chapter 8 -- Confidence Intervals

Chapter 8 -- Confidence Intervals. m or p is unknown (another course: s ) We want to estimate the parameter Confidence Intervals are accurate estimates of the true value of the parameter. Decisions to make. How many to survey? the entire population? just one object or person?

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Chapter 8 -- Confidence Intervals

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  1. Chapter 8 -- Confidence Intervals • m or p is unknown (another course: s) • We want to estimate the parameter • Confidence Intervals are accurate estimates of the true value of the parameter

  2. Decisions to make • How many to survey? the entire population? just one object or person? • How confident do we want to be? • What error can we live with?

  3. Confidence • CL = 1 - a = confidence level • CI = confidence interval ex. Let’s build a 95% Confidence Interval • 1 - a = .95 (confidence level) • a = “unconfidence” = 0.05 • a/2 = .025 z a/2 = z.025 = 1.96

  4. Confidence Interval

  5. Error Bounds • EBP = error bound for a proportion • EBM = error bound for a mean

  6. Error Bound for Proportions • P = X/n • P = proportion of successes • X ~ N(np, npq) • P = X/n ~ N(np/n, npq/n2) • P ~ N(p, pq/n) • z = (p’ - p)/(pq/n)1/2 • EBP = p’ - p = z*(p’q’/n)1/2

  7. Error Bounds for Means • X ~ N(m, s/n)

  8. Error Bound For Means • Student-t Distribution

  9. Confidence Interval (C. I.) • Means: (sample mean - EBM, sample mean + EBM) • Proportions: (p’ - EBP, p’ + EBP)

  10. C. I. For a Mean - Population Standard Deviation Known • For her project, Maria bought 50 small bags of jelly beans. The average number of jelly beans in a bag was 25. It is known that the standard deviation is 3 jelly beans. Maria wants to construct a 90% confidence interval for the true average number of jelly beans in a small bag.

  11. C.L. = 0.90 implies z = 1.645 C.I. = (sample mean + EBM, sample mean - EBM ) = (25 - 0.70, 25 + 0.70) = (24.30 , 25.70) We are 90% confident that the true average number of jelly beans is in the range 24.30 to 25.70. C. I. For a Mean - Population Standard Deviation Known

  12. C. I. For a Mean - Population Standard Deviation Unknown • Uses a new distribution called the Student-t (invented by William Gossett) • Notation: t degrees of freedom degrees of freedom is abbreviated df. • Use Student-t when the population standard deviation is not known, the sample is “small” (less than 30), and the population from which the sample comes is normal.

  13. C. I. For a Mean - Population Standard Deviation Unknown • Seven fast food restaurants were surveyed concerning the number of calories in four ounces of french fries. The data is 296, 329, 306, 324, 292, 310, 350 calories. Construct a 95% confidence interval for the true average number of calories in a four ounce serving of french fries. Assume the number of calories in french fries follows a normal distribution.

  14. Sample mean = 315.3 Sample stdev = 20.4 Use Student-t with 6 df. (df = n - 1 = 7 - 1) t=2.45 comes from a Student-t table where df = 6 and C.I. = 0.95 C. I. = (sample mean - EBM, sample mean + EBM) = (315.3-18.9, 315.3+18.9) =(296.4, 334.2) C. I. For a Mean - Population Standard Deviation Unknown

  15. C. I. For a Binomial Proportion • At a local cabana club, 102 of the 450 families who are members have children who swam on the swim team in 1998. Construct an 80% confidence interval for the true proportion of families with children who swim on the swim team in any year.

  16. C.L. = 0.80 implies z = 1.28; n = 450 p’ = 102/450 q’ = 348/450 C. I. For a Binomial Proportion

  17. C. I. For a Binomial Proportion • C. I. = (p’ - EBP, p’ + EBP) ((102/450)-0.0253, (102/450)+0.0253) = (0.2014, 0.2520) • We are 80% confident that the true proportion of families that have children on the swim team in any year is between 20% and 25%.

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