1. Chapter 8: Estimating With Confidence

2. 8.3 – Estimating a Population Mean In the previous examples, we made an unrealistic assumption that the population standard deviation was known and could be used to calculate confidence intervals.

3. When the standard deviation of a statistic is estimated from the data Standard Error: When we know  we can use the Z-table to make a confidence interval. But, when we don’t know it, then we have to use something else!

4. Properties of the t-distribution: σ is unknown Degrees of Freedom = n – 1 More variable than the normal distribution (it has fatter tails than the normal curve) Approaches the normal distribution when the degrees of freedom are large (sample size is large). Area is found to the right of the t-value

5. Properties of the t-distribution: If n < 15, if population is approx normal, then so is the sample distribution. If the data are clearly non-Normal or if outliers are present, don’t use! If n > 15, sample distribution is normal, except if population has outliers or strong skewness If n  30, sample distribution is normal, even if population has outliers or strong skewness

9. Example #1 Determine the degrees of freedom and use the t-table to find probabilities for each of the following: DFPictureProbability P(t > 1.093)n = 11 10 1.093

11. DFPictureProbability P(t > 1.093)n = 11 10 1.093 0.15 Example #1 Determine the degrees of freedom and use the t-table to find probabilities for each of the following:

12. Example #1 Determine the degrees of freedom and use the t-table to find probabilities for each of the following: DFPictureProbability P(t < 1.093)n = 11 10 1.093 0.85

13. Example #1 Determine the degrees of freedom and use the t-table to find probabilities for each of the following: DF Picture Probability P(t > 0.685)n = 24 23 0.685

15. Example #1 Determine the degrees of freedom and use the t-table to find probabilities for each of the following: DF Picture Probability P(t > 0.685)n = 24 23 0.685 0.25

16. Example #1 Determine the degrees of freedom and use the t-table to find probabilities for each of the following: DF Picture Probability P(t > 0.685)n = 24 23 -0.685 0.25

17. Example #1 Determine the degrees of freedom and use the t-table to find probabilities for each of the following: DF Picture Probability P(0.70<t<1.093) n = 11 10 1.0930.70

19. .25 –.15 = 0.1 Example #1 Determine the degrees of freedom and use the t-table to find probabilities for each of the following: DF Picture Probability P(0.70<t<1.093) n = 11 10 1.0930.70 0.1

20. Calculator Tip:Finding P(t) 2 nd – Dist – tcdf( lower bound, upper bound, degrees of freedom)

21. One-Sample t-interval:

22. Calculator Tip:One sample t-Interval Stat – Tests – TInterval Data: If given actual values Stats: If given summary of values

23. Conditions for a t-interval: 1.SRS (population approx normal and n<15, or moderate size (15≤ n < 30) with moderate skewness or outliers, or large sample size n ≥ 30) (should say) (Population 10x sample size) 2. Normality 3. Independence

24. Robustness: The probability calculations remain fairly accurate when a condition for use of the procedure is violated The t-distribution is robust for large n values, mostly because as n increases, the t-distribution approaches the Z-distribution. And by the CLT, it is approx normal.

25. Example #2 Practice finding t* nDegrees of Freedom Confidence Interval t* n = 1099% CI n = 2090% CI n = 4095% CI n = 3099% CI 9

27. Example #2 Practice finding t* nDegrees of Freedom Confidence Interval t* n = 1099% CI n = 2090% CI n = 4095% CI n = 3099% CI 93.250 19

29. Example #2 Practice finding t* nDegrees of Freedom Confidence Interval t* n = 1099% CI n = 2090% CI n = 4095% CI n = 3099% CI 93.250 191.729 39

31. Example #2 Practice finding t* nDegrees of Freedom Confidence Interval t* n = 1099% CI n = 2090% CI n = 4095% CI n = 3099% CI 93.250 191.729 39 2.042 29

33. Example #2 Practice finding t* nDegrees of Freedom Confidence Interval t* n = 1099% CI n = 2090% CI n = 4095% CI n = 3099% CI 93.250 191.729 39 2.042 292.756

34. Example #3 As part of your work in an environmental awareness group, you want to estimate the mean waste generated by American adults. In a random sample of 20 American adults, you find that the mean waste generated per person per day is 4.3 pounds with a standard deviation of 1.2 pounds. Calculate a 99% confidence interval for  and explain it’s meaning to someone who doesn’t know statistics. P:The true mean waste generated per person per day.

35. A: SRS: Normality: Independence: N:One Sample t-interval Says randomly selected 15<n<30. We must assume the population doesn’t have strong skewness. Proceeding with caution! It is safe to assume that there are more than 200 Americans that create waste.

36. I: df = 20 – 1 =19

38. I: df = 20 – 1 = 19

39. C:I am 99% confident the true mean waste generated per person per day is between 3.5323 and 5.0677 pounds.

40. Matched Pairs t-procedures: Subjects are matched according to characteristics that affect the response, and then one member is randomly assigned to treatment 1 and the other to treatment 2. Recall that twin studies provide a natural pairing. Before and after studies are examples of matched pairs designs, but they require careful interpretation because random assignment is not used. Apply the one-sample t procedures to the differences

41. Confidence Intervals for Matched Pairs

42. Example #4 Archaeologists use the chemical composition of clay found in pottery artifacts to determine whether different sites were populated by the same ancient people. They collected five random samples from each of two sites in Great Britain and measured the percentage of aluminum oxide in each. Based on these data, do you think the same people used these two kiln sites? Use a 95% confidence interval for the difference in aluminum oxide content of pottery made at the sites and assume the population distribution is approximately normal. Can you say there is no difference between the sites? New Forrest 20.818 15.818.3 Ashley Trails 19.114.816.718.317.7 Difference 1.73.21.3-2.5.6

43. P:μ n = New Forrest percentage of aluminum oxide The true mean difference in aluminum oxide levels between the New Forrest and Ashley Trails. μ a = Ashley Trails percentage of aluminum oxide μ d = μ n - μ a = Difference in aluminum oxide levels

44. A: SRS: Normality: Independence: N:Matched Pairs t-interval Says randomly selected Says population is approx normal It is safe to assume that there are more than 50 samples available

45. I: df = 5 – 1 =4

47. I: df = 20 – 1 = 19

48. C:I am 95% confident the true mean difference in aluminum oxide levels between the New Forrest and Ashley Trails is between –1.754 and 3.4743. Can you say there is no difference between the sites? Yes, zero is in the confidence interval, so it is safe to say there is no difference.

49. Example #5 The National Endowment for the Humanities sponsors summer institutes to improve the skills of high school language teachers. One institute hosted 20 Spanish teachers for four weeks. At the beginning of the period, the teachers took the Modern Language Association’s listening test of understanding of spoken Spanish. After four weeks of immersion in Spanish in and out of class, they took the listening test again. (The actual spoken Spanish in the two tests was different, so that simply taking the first test should not improve the score on the second test.) Below is the pretest and posttest scores. Give a 90% confidence interval for the mean increase in listening score due to attending the summer institute. Can you say the program was successful?

50. SubjectPretestPosttestSubjectPretestPosttest 13029113032 22830122928 33132133134 42630142932 52016153432 63025162027 73431172628 81518 2529 92833193132 102025202932

51. P:μ B = Pretest score The true mean difference in test scores between the Pretest and Posttest μ d = μ B - μ A = Difference in test scores μ A = Posttest score

52. A: SRS: Normality: We must assume the 20 teachers are randomly selected

54. A: SRS: Normality: Independence: N:Matched Pairs t-interval We must assume the 20 teachers are randomly selected 15<n<30 and distribution is approximately normal, so safe to assume It is safe to assume that there are more than 200 Spanish teachers

55. I: df = 20 – 1 = 19

57. I: df = 20 – 1 = 19

58. C:I am 90% confident the true mean difference in test scores between the Pretest and Posttest is between –2.689 and –0.2115. Can you say the program was successful? Yes, zero is not in the confidence interval, so the pretest score is lower than the posttest score.