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Sec7.3:Estimating p in the Binomial Distribution

Sec7.3:Estimating p in the Binomial Distribution. Understandable Statistics, 10 th ed. By Brase and Brase. 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

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Sec7.3:Estimating p in the Binomial Distribution

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  1. Sec7.3:Estimating p in the Binomial Distribution Understandable Statistics, 10th ed. By Brase and Brase

  2. 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 and nq are both > 5, the binomial distribution can be approximated by the normal distribution.

  3. Point Estimates • Point estimate for p: • Point estimate for q:

  4. 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.

  5. A c Confidence Interval for p ( ) • zc = critical value for confidence level c taken from a normal distribution

  6. 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?

  7. Can we use the normal distribution? 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.

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

  9. 99% confidence interval for the proportion of airplanes that depart on time: E = 0.0506 Confidence interval is:

  10. 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.

  11. Margin of Error • The margin of error is the maximal error of estimate E for a confidence interval. • When a poll states the result of a survey, the proportion reported to respond in the designated manner is • A 95% confidence interval for the population proportion is

  12. 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.”

  13. “Chances are 19 out of 20 …” • 19/20 = 0.95 • A 95% confidence interval is being used. • “... 84% of the households surveyed preferred…” 84% represents the percentage of households who preferred the new cereal

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

  15. Finding Sample Size n • if you have a preliminary estimate for p • if you don’t have a preliminary estimate for p Where E = specified maximal error of estimate zc = critical value from the normal distribution for the desired confidence level

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