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STAT 110 - Section 5 Lecture 25

STAT 110 - Section 5 Lecture 25. Professor Hao Wang University of South Carolina Spring 2012. TexPoint fonts used in EMF. Read the TexPoint manual before you delete this box.: A A A A A A A. Roulette. Say we bet $10 on Straight up. There are, 0 to 36, so 37 possible outcomes.

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STAT 110 - Section 5 Lecture 25

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  1. STAT 110 - Section 5 Lecture 25 Professor Hao Wang University of South Carolina Spring 2012 TexPoint fonts used in EMF. Read the TexPoint manual before you delete this box.: AAAAAAA

  2. Roulette. Say we bet $10 on Straight up. There are, 0 to 36, so 37 possible outcomes. • Win $360 if this single number (so net gain is $350 ) • Lose money otherwise (so net “gain” is -$10) What’s the chance of winning ? A 1/37 B 33/37 C 19/37 D 3/37 E 12/37

  3. Roulette. Say we bet $10 on Straight up. There are, 0 to 36, so 37 possible outcomes. • Win $360 if this single number (so net gain is $350 ) • Lose money otherwise (so net “gain” is -$10) What’s expected (average) net gain for a $10 bet ? A $-10/37 B $10/37 C $1/37 D $-1/37 E $18/37

  4. The Law of Large Numbers law of large numbers  If a random phenomenon with numerical outcomes is repeated many times independently, the mean of the actually observed outcomes approaches the expected value. • In many independent repetitions, • - the average outcome obtained will be close to the expected value. • http://bcs.whfreeman.com/ips4e/cat_010/applets/expectedvalue.html

  5. The Law of Large Numbers application • The Law of Large Numbers explains why gambling is a business for a casino – • While a casino may lose money in a single spin of the roulette wheel, • its earnings will tend towards a the expected value over a large number of spins.

  6. Chapter 21: What is a Confidence Interval Class Poll: We’ll estimate the proportion of college students who engaged in binge drinking during the last year. Take the definition of binge drinking to be 5 or more drinks on one occasion for males or 4 or more drinks for one occasion for females. Have you engaged in “binge drinking” in the past year? A. YES B. NO Record the estimate of the proportion of college students who engaged in binge drinking at least once in the last year.

  7. Recall the definition of parameter and statistic: • A parameter is a number used to describe a population. • A statistic is a number calculated from a sample and is used to estimate the parameter. • For the class poll on binge drinking, what is the statistic? What is the parameter?

  8. We want to use our statistic to estimate the parameter – will our statistic be exactly the same as the population parameter? • Recall the notion of sampling variability: Different samples from the same population may yield different values of the sample statistic. • How can you reduce the variability of the statistic? A. Reduce the sample size to decrease variability B. Increase the sample size to decrease variability C. We can do nothing to control sampling variability

  9. DEFINITION • The sampling distribution of a statistic is the distribution of values taken by the statistic in all possible samples from the same population. • CENTRAL LIMIT THEOREM (CLT) • The sampling distribution of a sum or percentage will become approximately normal as the sample size gets larger.

  10. SAMPLING DISTRIBUTION OF THE SAMPLE PROPORTION • Take a Simple Random Sample (SRS) of size n from a population that contains a certain true proportion of successes, p. Let be the sample proportion of successes. • If the sample size is large enough, then • a. The sampling distribution of is approximately normal • b. The mean of this normal distribution is • c. The standard deviation of this distribution is

  11. Take a SRS of 1523 adults from a very large population with true proportion who say “yes” 0.60. • The standard deviation of is: • 0.36 • 0.4 • 0.6 • d) • e)

  12. Take a SRS of 1523 adults from a very large population with true proportion who say “yes” 0.60. The sampling distribution has mean 0.6 and standard deviation 0.013. What’s the probability that the sample proportion of adults who say “yes” will be greater than 0.613?

  13. Take a SRS of 1523 adults from a very large population with true proportion who say “yes” 0.60. • The sampling distribution has mean 0.6 and standard deviation 0.013. • What’s the probability that the sample proportion of adults who say “yes” should be between 0.60 and 0.626? • 2.5% d) 68% • 34% e) 95% • c) 47.5%

  14. Estimating p: (1) Point estimator (2) Confidence interval estimator

  15. But, we do not know p! (If we did, we wouldn’t need to make a confidence interval…) So use in place of . This is an approximate 95% confidence interval for p. • CONFIDENCE INTERVAL FOR • An approximate 95% confidence interval for p is

  16. confidence interval calculation • Construct an approximate 95% confidence interval for the true proportion of college students who engaged in binge drinking in the past year, using the results of our class poll.

  17. confidence interval interpretation • The "95%" in the confidence interval listed above represents a level of certainty about our estimate. • If we were to repeatedly make new estimates using exactly the same procedure (by drawing a new sample, conducting new interviews, calculating new estimates and new confidence intervals), the confidence intervals would contain the true value 95% of the time. • (modified from http://www.census.gov/did/www/saipe/methods/statecounty/ci.html

  18. A confidence interval for p z value Percent Confidence 1 68.3% 1.64 90% 1.96 95% 2.58 99% 3 99.7% 3.29 99.9%

  19. A survey of a random sample of 400 product owners shows that 33 had complaints. The observed percentage with complaints is: • 0.05/400 ≈ 0.0001 d) 33/400 ≈ 0.083 • 0.05/33 ≈ 0.001 • e) • c)

  20. A survey of a random sample of 400 product owners shows that 33 had complaints. The estimated standard deviation of is: • 0.083/400 ≈ 0.0002 d) 33/400 ≈ 0.083 • 0.05/33 ≈ 0.001 • e) • c)

  21. From the previous slides, was 0.083 and its estimated standard deviation was 0.0045. A 99.7% confidence interval for p would be approximately: a) b) c)

  22. Assume your 99.7% confidence interval for p is What will your 68% confidence interval be? a) b) c)

  23. Roulette. There are, 0 to 36, so 37 possible outcomes. You are interested in finding out if the roulette is “biased” so you can gain advantage. If there is no bias, the chance of black is 18/37 = 0.4865 Observe 49 blacks out of 100 spins What’s the sample proportion of blacks ? A 49/100 B 51/100 C 19/37 D 3/37 E 12/37

  24. Roulette. There are, 0 to 36, so 37 possible outcomes. You are interested in testing if the roulette is “biased” so you can gain advantage Observe 49 blacks out of 100 spins What’s the estimated standard deviation of the sample proportion of blacks ? A B C

  25. Roulette. There are, 0 to 36, so 37 possible outcomes. You are interested in testing if the roulette is “biased” so you can gain advantage Observe 49 blacks out of 100 spins What’s the 95% confidence interval the proportion of blacks ? A B C

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