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Chapter Eight Part 1 (Sections 8.1 to 8.3) Estimation

Understandable Statistics S eventh Edition By Brase and Brase Prepared by: Lynn Smith Gloucester County College. Chapter Eight Part 1 (Sections 8.1 to 8.3) Estimation. Point Estimate. an estimate of a population parameter given by a single number. Examples of Point Estimates.

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Chapter Eight Part 1 (Sections 8.1 to 8.3) Estimation

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  1. Understandable StatisticsSeventh EditionBy Brase and BrasePrepared by: Lynn SmithGloucester County College Chapter Eight Part 1 (Sections 8.1 to 8.3) Estimation

  2. Point Estimate an estimate of a population parameter given by a single number

  3. Examples of Point Estimates

  4. Examples of Point Estimates • is used as a point estimate for  .

  5. Examples of Point Estimates • is used as a point estimate for  . • s is used as a point estimate for .

  6. Error of Estimate the magnitude of the difference between the point estimate and the true parameter value

  7. The error of estimate using as a point estimate for  is

  8. Confidence Level • A confidence level, c, is a measure of the degree of assurance we have in our results. • The value of c may be any number between zero and one. • Typical values for c include 0.90, 0.95, and 0.99.

  9. Critical Value for a Confidence Level, c the value zc such that the area under the standard normal curve falling between – zc and zc is equal to c.

  10. – zc 0 zc Critical Value for a Confidence Level, c P(– zc < z < zc ) = c This area = c.

  11. – z.90 0 z.90 Find z0.90 such that 90% of the area under the normal curve lies between z-0.90 and z0.90. P(-z0.90 < z < z0.90 ) = 0.90 .90

  12. – z.90 0 z.90 Find z0.90 such that 90% of the area under the normal curve lies between z-0.90 and z0.90. • P(0< z < z0.90 ) = 0.90/2 = 0.4500 .4500

  13. – z.90 0 z.90 Find z0.90 such that 90% of the area under the normal curve lies between z-0.90 and z0.90. • P( z < z0.90 ) = .5 + 0.4500 = .9500 .9500

  14. Find z0.90 such that 90% of the area under the normal curve lies between z-0.90 and z0.90. • According to Table 5a in Appendix II, 0.9500 lies exactly halfway between two area values in the table (.9495 and .9505). • Averaging the z values associated with these areas gives z0.90 = 1.645.

  15. Common Levels of Confidence and Their Corresponding Critical Values

  16. Confidence Interval for the Mean of Large Samples (n  30)

  17. Confidence Interval for the Mean of Large Samples (n  30)

  18. Confidence Interval for the Mean of Large Samples (n  30)

  19. Confidence Interval for the Mean of Large Samples (n  30)

  20. Confidence Interval for the Mean of Large Samples (n  30)

  21. Confidence Interval for the Mean of Large Samples (n  30)

  22. Create a 95% confidence interval for the mean driving time between Philadelphia and Boston. Assume that the mean driving time of 64 trips was 6.4 hours with a standard deviation of 0.9 hours.

  23. = 6.4 hourss = 0.9 hoursc = 95%, so zc = 1.96n = 64

  24. = 6.4 hourss = 0.9 hoursApproximate  as s = 0.9 hours. 95% Confidence interval will be from

  25. = 6.4 hourss = 0.9 hoursc = 95%, so zc = 1.96n = 64

  26. 95% Confidence Interval: 6.4 – .2205 <  < 6.4 + .2205 6.1795 <  < 6.6205 We are 95% sure that the true time is between 6.18 and 6.62 hours.

  27. What if it is impossible or impractical to use a large sample? Apply the Student’s t distribution.

  28. Student’s t Variable

  29. The shape of the t distribution depends only only the sample size, n, if the basic variable x has a normal distribution. When using the t distribution, we will assume that the x distribution is normal.

  30. Table 6 in Appendix II gives values of the variable t corresponding to the number of degrees of freedom (d.f.)

  31. Degrees of Freedom d.f. = n – 1 where n = sample size

  32. The t Distribution has a Shape Similar to that of the the Normal Distribution A Normal distribution A “t” distribution

  33. d.f.=7 Find the critical value tc for a 95% confidence interval if n = 8.

  34. Confidence Interval for the Mean of Small Samples (n < 30) from Normal Populations • c = confidence level (0 < c < 1) • tc = critical value for confidence level c, and degrees of freedom = n - 1

  35. The mean weight of eight fish caught in a local lake is 15.7 ounces with a standard deviation of 2.3 ounces. Construct a 90% confidence interval for the mean weight of the population of fish in the lake.

  36. Mean = 15.7 ounces Standard deviation = 2.3 ounces. • n = 8, so d.f. = n – 1 = 7 • For c = 0.90, Table 6 in Appendix II gives t0.90 = 1.895.

  37. Mean = 15.7 ounces Standard deviation = 2.3 ounces. • E = 1.54 • The 90% confidence interval is: • 15.7 - 1.54 <  < 15.7 + 1.54 • 14.16 <  < 17.24

  38. The 90% Confidence Interval:14.16 <  < 17.24 We are 90% sure that the true mean weight of the fish in the lake is between 14.16 and 17.24 ounces.

  39. 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.

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

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

  42. 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.

  43. Error of Estimate for “p hat” as a Point Estimate for p

  44. 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

  45. 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?

  46. 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?

  47. 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.

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

  49. 99% confidence interval for the proportion of airplanes that depart on time: E = 0.0506 Confidence interval is:

  50. 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.

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