1 / 34

Chapter 8: Estimation

Understandable Statistics Eighth Edition By Brase and Brase Prepared by: Lynn Smith Gloucester County College Edited by: Jeff, Yann, Julie, and Olivia. Chapter 8: Estimation. Section 8.3. Estimating p in the Binomial Distribution. Focus Points.

nancy
Download Presentation

Chapter 8: Estimation

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Understandable StatisticsEighth Edition By Brase and BrasePrepared by: Lynn SmithGloucester County CollegeEdited by: Jeff, Yann, Julie, and Olivia Chapter 8: Estimation

  2. Section 8.3 Estimating p in the Binomial Distribution

  3. Focus Points • Compute the maximal margin of error for proportions using a given level of confidence. • Compute confidence intervals for p and interpret the results. • Interpret poll results.

  4. Statistics Quote “There are two kinds of statistics, the kind you look up and the kind you make up.”  —Rex Stout

  5. Review of the Binomial Distribution • Completely determined by the number of trials n and the probability of success p in a single trial. • q = 1 – p • If np > 5 and nq > 5, the binomial distribution can be approximated by the normal distribution.

  6. A Point Estimate for p, the Population Proportion of Successes

  7. Point Estimate for q (Population Proportion of Failures)

  8. For a sample of 500 airplane departures, 370 departed on time. Use this information to estimate the probability that an airplane from the entire population departs on time. We estimate that there is a 74% chance that any given flight will depart on time.

  9. Margin of error for “p hat” as a Point Estimate for p

  10. A c Confidence Interval for p for Large Samples (np > 5 and nq > 5) • zc = critical value for confidence level c taken from a normal distribution

  11. For a sample of 500 airplane departures, 370 departed on time. Find a 99% confidence interval for the proportion of airplanes that depart on time. • Is the use of the normal distribution justified?

  12. For a sample of 500 airplane departures, 370 departed on time. Find a 99% confidence interval for the proportion of airplanes that depart on time. • Can we use the normal distribution?

  13. For a sample of 500 airplane departures, 370 departed on time. Find a 99% confidence interval for the proportion of airplanes that depart on time. so the use of the normal distribution is justified.

  14. Out of 500 departures, 370 departed on time. Find a 99% confidence interval.

  15. 99% confidence interval for the proportion of airplanes that depart on time: E = 0.0506 Confidence interval is:

  16. 99% confidence interval for the proportion of airplanes that depart on time Confidence interval is 0.6894 < p < 0.7906 We are 99% confident that between 69% and 79% of the planes depart on time.

  17. The point estimate and the confidence interval do not depend on the size of the population. The sample size, however, does affect the accuracy of the statistical estimate.

  18. Margin of Error The margin of error is the maximal error of estimate E for a confidence interval. Usually, a 95% confidence interval is assumed.

  19. Interpretation of Poll Results The proportion responding in a certain way is

  20. General interpretation of poll results • p-hat: the sample estimate of the population proportion • E, margin of error: maximal error of a 95% confidence interval for p

  21. A 95% confidence interval for population proportion p is:

  22. Interpret the following poll results: “A recent survey of 400 households indicated that 84% of the households surveyed preferred a new breakfast cereal to their previous brand. Chances are 19 out of 20 that if all households had been surveyed, the results would differ by no more than 3.5 percentage points in either direction.”

  23. “Chances are 19 out of 20 …” 19/20 = 0.95 A 95% confidence interval is being used.

  24. “... 84% of the households surveyed preferred…” 84% represents the percentage of households who preferred the new cereal.

  25. “... the results would differ by no more than 3.5 percentage points in either direction.” 3.5% represents the margin of error, E. The confidence interval is: 84% - 3.5% < p < 84% + 3.5% 80.5% < p < 87.5%

  26. The poll indicates (with 95% confidence): between 80.5% and 87.5% of the population prefer the new cereal.

  27. Calculator Instructions CONFIDENCE INTERVALS FOR THE PROBABILITY OF SUCCESS p IN A BINOMIAL DISTRIBUTION To find a confidence interval for a proportion, press the Stat key and use option A:1-PropZlnt under TESTS. Notice that the normal distribution will be used.

  28. Example • The public television station BPBS wants to find the percent of its viewing population who give donations to the station. 300 randomly selected viewers were surveyed, and it was found that 123 made contributions to the station. Find a 95% confidence interval for the probability that a viewer of BPBS selected at random contributes to the station.

  29. Example • The letter x is used to count the number of successes (the letter r is used in the text). • Enter 123 for x and 300 for n. Use 0.95 for the C-level.

  30. Example • Highlight Calculate and press Enter. • The result is the interval from 0.35 to 0.47.

  31. Section 8.3, Problem 8 Law Enforcement: Escaped Convicts Case studies showed that out of 10,351 convicts who escaped from u.s. prisons, only 7867 were recaptured (The Book of Odds, by Shook and Shook, Signet). • Let p represent the proportion of all escaped convicts who will eventually be recaptured. Find a point estimate for p. (b) Find a 99% confidence interval for p. Give a brief statement of the meaning of the confidence interval. (c) Is use of the normal approximation to the binomial justified in this problem? Explain.

  32. Solution

  33. Section 8.3, Problem 17 Lifestyle: Smoking In a survey of 1000 large corporations, 250 said that, given a choice between a job candidate who smokes and an equally qualified nonsmoker, the nonsmoker would get the job (USA Today). • Let p represent the proportion of all corporations preferring a nonsmoking candidate. Find a point estimate for p. (b) Find a 0.95 confidence interval for p. (c) As a new writer, how would you report the survey results regarding the proportion of corporations that would hire the equally qualified nonsmoker? What is the margin of error based on a 95% confidence interval?

  34. Solution

More Related