1. Chapter 10: Estimating With Confidence

2. 10.1 – Confidence Intervals: The basics

3. Statistical Inference: Using sample data to draw conclusions about a population Note:Each sample may vary, but the population parameter doesn’t!

4. Sampling Distribution: If population is approximately normal, so is the sample distribution If population is skewed, the sample distribution is approximately normal if n  30 by the central limit theorem If given sample data, look at the distribution to assess normality if needed. (Normal Prob. Plot)

5. Confidence Interval: Uses the sample distribution to predict population parameter It is an interval of numbers above and below the sample statistic

6. Confidence Level: The probability the interval will capture the true parameter value in repeated samples Critical Value: The probability p lying to its right under the standard Normal curve. ( Z* )

7. Margin of Error: How accurate our estimate is based on the variability of the sample distribution. We add and subtract this from our estimate. estimate  margin of error Caution! Margin of error is only from random sampling errors. This does not include errors in collecting the data!

8. Most Common Critical Values Confidence Level (C)Upper tail prob.Z* Value 90% Z=? 0.05  1.645 0.90

10. Most Common Critical Values Confidence Level (C)Upper tail prob.Z* Value 95% Z=? 0.025  1.96 0.95

11. Most Common Critical Values Confidence Level (C)Upper tail prob.Z* Value 99% Z=? 0.005  2.576 0.99

12. Calculator Tip:Critical Values 2 nd Dist – invNorm( (1 + C)/2 ) OR: Look at the T-Tables for the most common ones! (You will learn more about them later)

13. (Z-Interval) estimate  margin of error Confidence Interval for a Population mean (  known)  estimate  critical value  standard error

14. Properties of Confidence Intervals The interval is always centered around the statistic The higher the confidence level, the wider the interval becomes If you increase n, then the margin of error decreases

15. Calculator Tip:Z-Interval Stat – Tests – ZInterval Data: If given actual values Stats: If given summary of values

16. Interpreting a Confidence Interval: I am C% confident that the true parameter is captured in the interval What you will say: If we took many, many, SRS from a population and calculated a confidence interval for each sample, C% of the confidence intervals will contain the true mean What it means:

17. CAUTION! Never Say: The interval will capture the true mean C% of the time. It either does or does not!

19. Conditions for a Z-Interval: 1.SRS (CLT or population approx normal) (should say) (Population 10x sample size) 2. Normality 3. Independence

20. Steps to Construct ANY Confidence Interval: PANIC P:Parameter of Interest (what are you looking for?) A:Assumptions (what are the conditions?) N:Name the type of interval (what type of data do we have?) I:Interval (Finally! You can calculate!) C:Conclusion in context (I am ___% confident the true parameter lies between ________ and _________)

21. Example #1 Serum Cholesterol-Dr. Paul Oswick wants to estimate the true mean serum HDL cholesterol for all of his 20-29 year old female patients. He randomly selects 30 patients and computes the sample mean to be 50.67. Assume from past records, the population standard deviation for the serum HDL cholesterol for 20-29 year old female patients is  =13.4. a.Construct a 95% confidence interval for the mean serum HDL cholesterol for all of Dr. Oswick’s 20-29 year old female patients. P:The true mean serum HDL cholesterol for all of Dr. Oswick’s 20-29 year old female patients.

22. A: SRS: Normality: Independence: N:One sample Z-Interval Says randomly selected Approximately normal by the CLT (n  30) I am assuming that Dr. Oswick has 300 patients or more.

23. I:

24. C:I am 95% confident the true mean serum HDL cholesterol for all of Dr. Oswick’s 20-29 year old female patients is between 45.875 and 55.465

25. Example #1 Serum Cholesterol-Dr. Paul Oswick wants to estimate the true mean serum HDL cholesterol for all of his 20-29 year old female patients. He randomly selects 30 patients and computes the sample mean to be 50.67. Assume from past records, the population standard deviation for the serum HDL cholesterol for 20-29 year old female patients is  =13.4. b. If the US National Center for Health Statistics reports the mean serum HDL cholesterol for females between 20-29 years old to be  = 53, do Dr. Oswick’s patients appear to have a different serum level compared to the general population? Explain. No, 53 is contained in the interval.

26. Example #1 Serum Cholesterol-Dr. Paul Oswick wants to estimate the true mean serum HDL cholesterol for all of his 20-29 year old female patients. He randomly selects 30 patients and computes the sample mean to be 50.67. Assume from past records, the population standard deviation for the serum HDL cholesterol for 20-29 year old female patients is  =13.4. c. What two things could you do to decrease your margin of error? Increase n Lower confidence level

27. Example #2 Suppose your class is investigating the weights of Snickers 1- ounce Fun-Size candy bars to see if customers are getting full value for their money. Assume that the weights are Normally distributed with standard deviation = 0.005 ounces. Several candy bars are randomly selected and weighed with sensitive balances borrowed from the physics lab. The weights are 0.95 1.020.980.971.051.010.981.00 ounces. Determine a 90% confidence interval for the true mean, µ. Can you say that the bars weigh 1oz on average? P:The true mean weight of Snickers 1-oz Fun-size candy bars

28. A: SRS: Normality: Independence: N:One sample Z-Interval Says randomly selected Approximately normal because the population is approximately normal I am assuming that Snickers has 80 bars or more in the 1-oz size

29. I:

30. C:I am 90% confident the true mean weight of Snickers 1-oz Fun-size candy bars is between.9921 and.9979 ounces. I am not confident that the candy bars weigh as advertised at the 90% level.

31. Choosing a Sample Size for a specific margin of error Note: Always round up! You can’t have part of a person! Ex: 163.2 rounds up to 164.

32. Example #3 A statistician calculates a 95% confidence interval for the mean income of the depositors at Bank of America, located in a poverty stricken area. The confidence interval is $18,201 to $21,799. a.What is the sample mean income?

33. Example #3 A statistician calculates a 95% confidence interval for the mean income of the depositors at Bank of America, located in a poverty stricken area. The confidence interval is $18,201 to $21,799. b. What is the margin of error? m = 21,799 – 20,000 m = 1,799 m

34. Example #4 A researcher wishes to estimate the mean number of miles on four-year-old Saturn SCI’s. How many cars should be in a sample in order to estimate the mean number of miles within a margin of error of  1000 miles with 99% confidence assuming  =19,700.

35. 10.2 – Estimating a Population Mean

36. In the 10.1 we made an unrealistic assumption that the population standard deviation was known and could be used to calculate confidence intervals.

37. When the standard deviation of a statistic is estimated from the data Standard Error:

38. 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!

39. 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

40. 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

44. 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

46. 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:

47. 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

48. 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

50. 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

51. 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

52. 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

54. .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

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

56. One-Sample t-interval:

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

58. 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

59. 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.

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

62. 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

64. 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

66. 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

68. 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

69. 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.

70. 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.

71. I: df = 20 – 1 =19

73. I: df = 20 – 1 = 19

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

75. 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

76. Confidence Intervals for Matched Pairs

77. 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

78. 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

79. 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

80. I: df = 5 – 1 =4

82. I: df = 20 – 1 = 19

83. 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.

84. 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?

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

86. 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

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

89. 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

90. I: df = 20 – 1 = 19

92. I: df = 20 – 1 = 19

93. 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.

94. 10.3 – Estimating a Population Proportion

95. Properties of :

96. Confidence Interval for a Population Proportion: Notice! We use Z* and not t*

97. Conditions for a p-interval: 1.SRS (should say) (Population 10x sample size) 2. Normality 3. Independence

98. Calculator Tip:One sample p-Interval Stat – Tests – 1–PropZInt x = # of successes in the sample

99. Example #1 A news release by the IRS reported 90% of all Americans fill out their tax forms correctly. A random sample of 1500 returns revealed that 1200 of them were correctly filled out. Calculate a 92% confidence interval for the proportion of Americans who correctly fill out their tax forms. Is the IRS correct in their report? P:The true percent of Americans who fill out their tax forms correctly

100. A: SRS: Normality: Independence: Says randomly selected It is safe to assume that there are more than 15,000 people who file their taxes 0.80 Yes, safe to assume an approximately normally distribution

101. N:One Sample Proportion Interval I: Z* = ?

102. ?

103. Confidence Level (C)Upper tail prob.Z* Value 92% Z=? 0.04 0.92

105. Confidence Level (C)Upper tail prob.Z* Value 92% Z=? 0.04  1.75 0.92

106. N:One Sample Proportion Interval I:

107. C:I am 92% confident the true percent of Americans who fill out their tax forms correctly is between 78.19% and 81.8% Is the IRS correct in their report? No, 90% is not in the interval!

108. Sample size for a Desired Margin of Error If we want the margin of error in a level C confidence interval for p to be m, then we need n subjects in the sample, where: n  p* = An estimate for

109. Note: If p is unknown use the most conservative value of p = 0.5. Since n is the sample size, it must be a whole number!!! Round up!

110. Example #2 You wish to estimate with 95% confidence; the proportion of computers that need repairs or have problems by the time the product is three years old. Your estimate must be accurate within 3.5% of the true proportion. a. Find the sample size needed if a prior study found that 19% of computers needed repairs or had problems by the time the product as three years old.

111. Example #2 You wish to estimate with 95% confidence; the proportion of computers that need repairs or have problems by the time the product is three years old. Your estimate must be accurate within 3.5% of the true proportion. b. If no preliminary estimate is available, find the most conservative sample size required.

112. Example #2 You wish to estimate with 95% confidence; the proportion of computers that need repairs or have problems by the time the product is three years old. Your estimate must be accurate within 3.5% of the true proportion. c. Compare the results from a and b. Using 0.5 makes the sample size very large, ensuring that enough people will be surveyed.