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Statistics for Business and Economics Chapter 5 Inferences Based on a Single Sample: Estimation with Confidence Intervals Learning Objectives 1. State What Is Estimated 2. Distinguish Point & Interval Estimates 3. Explain Interval Estimates 4. Compute Confidence Interval Estimates for Population Mean & Proportion 5. Compute Sample Size 6. Discuss Finite Population Correction Factor Thinking Challenge Suppose you’re interested in the average amount of money that students in this class (the population) have on them. How would you find out? Introduction to Estimation Statistical Methods Statistical Methods Descriptive Statistics Inferential Statistics Estimation Hypothesis Testing Estimation Process Population Mean, , is unknown Sample Random Sample Mean X = 50 I am 95% confident that is between 40 & 60. Unknown Population Parameters Are Estimated Estimate Population Parameter... Mean Proportion p Variance Differences with Sample Statistic x p^ 2 1 - 2 s 2 x1 -x2 Estimation Methods Estimation Point Estimation Interval Estimation Point Estimation Estimation Methods Estimation Point Estimation Interval Estimation Point Estimation 1. Provides a single value • Based on observations from one sample 2. Gives no information about how close the value is to the unknown population parameter 3. Example: Sample mean x = 3 is point estimate of unknown population mean Interval Estimation Estimation Methods Estimation Point Estimation Interval Estimation Interval Estimation 1. Provides a range of values • Based on observations from one sample 2. Gives information about closeness to unknown population parameter • Stated in terms of probability – Knowing exact closeness requires knowing unknown population parameter 3. Example: Unknown population mean lies between 50 and 70 with 95% confidence Key Elements of Interval Estimation Sample statistic Confidence interval (point estimate) Confidence limit (lower) Confidence limit (upper) A probability that the population parameter falls somewhere within the interval. Confidence Limits for Population Mean Parameter = Statistic ± Error © 1984-1994 T/Maker Co. (1) X Error (2) Error X or X X (3) Z (4) Error Z x (5) X Z x x Error x Many Samples Have Same Interval X = ± Zx x_ -1.65 -2.58x x -1.96x +1.65x +2.58x +1.96x 90% Samples 95% Samples 99% Samples X Confidence Level 1. Probability that the unknown population parameter falls within interval 2. Denoted (1 – • is probability that parameter is not within interval 3. Typical values are 99%, 95%, 90% Intervals & Confidence Level Sampling Distribution of Sample Mean /2 _ 1- x /2 x = _ X (1 – α)% of intervals contain μ Intervals extend from X – ZσX to X + ZσX α% do not Large number of intervals Factors Affecting Interval Width 1. Data dispersion • Measured by Intervals extend from X – ZX toX + ZX 2. Sample size X n 3. Level of confidence (1 – ) • Affects Z © 1984-1994 T/Maker Co. Confidence Interval Estimates Confidence Intervals Mean σ Known σ Unknown Proportion Confidence Interval Estimate Mean ( Known) Confidence Interval Estimates Confidence Intervals Mean σ Known σ Unknown Proportion Confidence Interval Mean ( Known) 1. Assumptions • • • Population standard deviation is known Population is normally distributed If not normal, can be approximated by normal distribution (n 30) 2. Confidence interval estimate X Z / 2 X Z / 2 n n Estimation Example Mean ( Known) The mean of a random sample of n = 25 isX = 50. Set up a 95% confidence interval estimate for if = 10. X Z / 2 X Z / 2 n n 10 10 50 1.96 50 1.96 25 25 46.08 53.92 Thinking Challenge You’re a Q/C inspector for Gallo. The for 2-liter bottles is .05 liters. A random sample of 100 bottles showed x = 1.99 liters. What is the 90% confidence interval estimate of the true mean amount in 2-liter bottles? 22 liter liter © 1984-1994 T/Maker Co. Confidence Interval Solution* X Z / 2 n X Z / 2 n .05 .05 1.99 1.645 1.99 1.645 100 100 1.982 1.998 Confidence Interval Estimate Mean ( Unknown) Confidence Interval Estimates Confidence Intervals Mean σ Known σ Unknown Proportion Confidence Interval Mean ( Unknown) 1. Assumptions • • Population standard deviation is unknown Population must be normally distributed 2. Use Student’s t–distribution Student’s t Distribution Standard Normal Bell-Shaped t (df = 13) Symmetric t (df = 5) ‘Fatter’ Tails 0 Z t Degrees of Freedom (df) 1. Number of observations that are free to vary after sample statistic has been calculated 2. Example – Sum of 3 numbers is 6 X1 = 1 (or any number) X2 = 2 (or any number) X3 = 3 (cannot vary) Sum = 6 degrees of freedom =n-1 =3-1 =2 Student’s t Table /2 v t .10 t .05 t .025 1 3.078 6.314 12.706 Assume: n=3 df = n - 1 = 2 = .10 /2 =.05 2 1.886 2.920 4.303 /2 3 1.638 2.353 3.182 t values 0 2.920 t Confidence Interval Mean ( Unknown) S S X t / 2 X t / 2 n n df n 1 Estimation Example Mean ( Unknown) A random sample of n = 25 has x = 50 and s = 8. Set up a 95% confidence interval estimate for . S S X t / 2 X t / 2 n n 8 8 50 2.064 50 2.064 25 25 46.69 53.30 Thinking Challenge You’re a time study analyst in manufacturing. You’ve recorded the following task times (min.): 3.6, 4.2, 4.0, 3.5, 3.8, 3.1. What is the 90% confidence interval estimate of the population mean task time? Confidence Interval Solution* • x = 3.7 • s = 3.8987 • n = 6, df = n - 1 = 6 - 1 = 5 • t.05 = 2.015 3.8987 3.8987 3.7 2.015 3.7 2.015 6 6 .492 6.908 Confidence Interval Estimate of Proportion Confidence Interval Estimates Confidence Intervals Mean σ Known σ Unknown Proportion Confidence Interval Proportion 1. Assumptions • • Random sample selected Normal approximation can be used if npˆ 15 and nqˆ 15 2. Confidence interval estimate ˆˆ ˆˆ pq pq pˆ z 2 p pˆ z 2 n n Estimation Example Proportion A random sample of 400 graduates showed 32 went to graduate school. Set up a 95% confidence interval estimate for p. ˆˆ ˆˆ pq pq pˆ Z / 2 p pˆ Z / 2 n n .08 .92 .08 .92 .08 1.96 p .08 1.96 400 400 .053 p .107 Thinking Challenge You’re a production manager for a newspaper. You want to find the % defective. Of 200 newspapers, 35 had defects. What is the 90% confidence interval estimate of the population proportion defective? Confidence Interval Solution* pˆ qˆ pˆ qˆ pˆ z / 2 p pˆ z / 2 n n .175 (.825) .175 (.825) .175 1.645 p .175 1.645 200 200 .1308 p .2192 Finding Sample Sizes Finding Sample Sizes for Estimating SE = Sampling Error (1) (2) X SE Z x x SE Z 2 x Z 2 ( Z 2 ) 2 (3) I don’t want to sample too much or too little! n ( SE ) 2 2 n Sample Size Example What sample size is needed to be 90% confident the mean is within 5? A pilot study suggested that the standard deviation is 45. ( Z 2 ) 2 n ( SE ) 2 2 1.645 45 2 5 2 2 219.2 220 Finding Sample Sizes for Estimating p SE = Sampling Error (1) (2) pˆ p SE Z pˆ pˆ SE Z 2 pˆ Z 2 pq n 2 (3) n ( Z 2 ) pq ( SE )2 If no estimate of p is available, use p = q = .5 Sample Size Example What sample size is needed to estimate p with 90% confidence and a width of .03? width .03 SE .015 2 2 ( Z 2 ) 2 n ( SE ) 2 2 1.645 .5 .5 2 .015 2 2 3006.69 3007 Thinking Challenge You work in Human Resources at Merrill Lynch. You plan to survey employees to find their average medical expenses. You want to be 95% confident that the sample mean is within ± $50. A pilot study showed that was about $400. What sample size do you use? Sample Size Solution* n ( Z 2 ) 2 2 ( SE ) 2 1.96 400 2 50 2 245.86 246 2 Finite Population Correction Factor Finite Population Correction Factor • Use when n, the sample size, is relatively large compared to N, the size of the population • If n/N > .05 use the finite population correction factor • Finite population correction factor: N n N Finite Population Correction Factor • Approximate 95% confidence interval for μ: s x 2 n N n s x2 N n N n N • Approximate 95% confidence interval for p: pˆ (1 pˆ ) N n pˆ (1 pˆ ) N n pˆ 2 p pˆ 2 n N n N Finite Population Correction Factor Example You want to estimate a population mean, μ, where x =115, s =18, N =700, and n = 60. Find an approximate 95% confidence interval for μ. Since n N 60 700 .086 is greater than .05 use the finite correction factor Finite Population Correction Factor Example You want to estimate a population mean, μ, where x =115, s =18, N =700, and n = 60. Find an approximate 95% confidence interval for μ. s x 2 n 18 115 2 60 N n s x 2 N n N n N 700 60 18 115 2 700 60 110.6 119.4 700 60 700 Conclusion 1. Stated What Is Estimated 2. Distinguished Point & Interval Estimates 3. Explained Interval Estimates 4. Computed Confidence Interval Estimates for Population Mean & Proportion 5. Computed Sample Size 6. Discussed Finite Population Correction Factor