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Chapter 8 Interval Estimation I. Basic Definitions II. Confidence Interval Estimation III. Sample Size and Precision I. Basic Definitions • Point estimate: a single value of sample statistic as an estimator of population parameter. • Sampling error = | Sample Statistic – Population Parameter | • Confidence interval: An interval with the confidence level. It is constructed on the basis of a sample: sample statistic + margin of error. • Confidence level: % of intervals formed by samples with sample size of n will contain the population parameter. (1-) • Margin of error: half-width of the confidence interval. II. Confidence Interval Estimation Outline: Procedure: 1. Sample point estimator ( X or p ) 2. Confidence level and Table Z or tn-1 3. Formulas compute UCL and LCL: point estimator margin of error Confidence Interval Estimation of Population Mean (σ Known) UCL x m.o.e. 1-α=.99 z=2.575 LCL x m.o.e. 1-α=.95 z=1.96 1-α margin of error z x z 1-α=.90 z=1.645 n 1 2 x UCL x LCL [--------------------- ---------------------] [--------------------- x ---------------------] [--------------------- x ---------------------] 1. Interval Estimation for Population Mean Example 1: p.306 #5 ( known case) In an effort to estimate the mean amount spent per customer for dinner at a major Atlanta restaurant. Data were collected for a sample of 49 customers over a three-week period. Assume a population standard deviation of $5. a. At the 95% confidence, what is the margin error? b. If the sample mean is $24.80, What is the 95% confidence interval for the population mean? HWK: p.306 #8 Example 1: p.306 #5 Answer: • n = 49 X $24.8 =$5 • Z: (1-)/2 = 0.95/2 = 0.475 Table 1: Z = 1.96 x 5 • 1. Z / 2 n (1.96 ) 49 1.4 x 2. UCL X Z / 2 24.8 1.4 26.2 n x LCL X Z / 2 24.8 1.4 23.4 n : [23.4, 26.2] Example 2: p.313 #16 ( unknown case) The mean flying time for pilots at Continental Airlines is 49 hours per month. This mean was based on a sample of 100 pilots and the sample standard deviation was 8.5 hours. a. At 95% confidence, what is the margin of error? b. What is the 95% confidence interval estimate of the population mean flying time? c. The mean flying time for pilots at United Airlines is 36 hours per month. Discuss difference between the flying times at two airlines. HWK: p.315 #16 Example 2: p.315 #16 Given: n = 100, X = 49, S = 8.5, 1- = .95 Think: What to estimate? Use Z or t? Answer: • Sample info (given): n = 100, X = 49, S = 8.5 • t: 1- =0.95, so /2=0.025, d.f.=n-1=99 Table 2: d.f.=100, /2=0.025 t=1.984 d.f.=80, /2=0.025 t=1.990 *Interpolation: t 1.984 (1.990 1.984) 100 99 1.9843 100 80 S 8.5 a. m.o.e.: m.o.e. t / 2 1.9843 1.69 n 100 b. UCL = 49+1.69 =50.69 LCL = 49 – 1.69 = 47.31 : [47.31, 50.69] c. 36 < LCL. The mean flying time is lower at United. Aside: Why t? t distribution: • flatter than Z - more spread out • n , t Z • Degree of freedom: a measure of precision. The d.f. more precise. 2. Interval Estimation for Population Proportion Example 3: p.323 #36 A survey asked 346 job seekers. The answer selected most (152 times) was “higher compensation.” a. What is the point estimate of the proportion of job seekers who would select “higher compensation” as the reason of changing jobs? b. What is the 95% confidence interval estimate of the population proportion? HWK: p.323 #35 Example 3: p.321 #36 Answer: x 152 a. Point estimate of p: p .4393 n 346 b. Confidence interval: • Z: (1-)/2=0.475, Table 1 Z=1.96 p (1 p ) (0.4393 )(1 0.4393 ) • Margin of error = Z n (1.96 ) 346 = 0.0523 • Confidence interval UCL p m.o.e. .4393 .0523 .4916 LCL p m.o.e. .4393 .0523 .3870 p: [.3870, .4916] III. Sample Size and Precision Quality of estimation: • Confidence level: 1 - • Precision: margin of error Confidence level: 1 - is guaranteed by procedure. P.303 Figure 8.3: sampling distribution for sample mean. Probability that X falls between Z n and Z n /2 x /2 x is 1 - . In general, any sample mean that is within this range will provide an interval that contains the population mean . Margin of error: Given n, then (1 - ) margin of error . Given 1 - , then n margin of error . Determine sample size to meet requirements for both confidence level and margin of error: 1. Determine sample size for estimation of (p.317) Z / 2 2 2 n 2 E E: desired margin of error [ ]: round up (“Be conservative”) 2. Determine sample size for estimation of p (p.321) Z / 2 p(1 p) 2 n E2 p = .5 (p.319 Table 8.5 “Be conservative”) E: desired margin of error [ ]: round up (“Be conservative”) Homework: p.318 #25, p.323 #39 Example 4: Bride’s magazine reported that the mean cost of a wedding is $19,000. Assume that the population standard deviation is $9,400. Use 95% confidence, a. What is the recommended sample size if the desired margin of error is $1,000? b. What is the recommended sample size if the desired margin of error is $500? Answer: a. 1 Z /2: .475 Z 1.96 /2 2 (1.96) 2 (9400) 2 n 2 339.44 340 1000 (1.96) 2 (9400) 2 b. n 2 1357.78 1358 500 Example 6: The League of American Theatres and Producers uses an ongoing audience tracking survey that provides up-to- date information about Broadway theater audiences. Every week, the League distributes a one-page survey on random theater seats at a rotation roster of Broadway shows. a. How large a sample should be taken if the desired margin of error on any proportion is 0.04? Use 95% confidence. Answer: 1 Z / 2 : .475, Table 1 Z / 2 1.96 2 (1.96) 2 .5(1 .5) n 2 600.25 601 (.04)