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# Chapter 8 and Chapter 10.docx

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Department
Quantitative Methods
Course Code
QMS 202
Professor
Bob Hudyma

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Chapter 8 (pg 356) Example 8.5 Population Sample Statistic (Point Parameter Estimate) Mean Μ Proportio π p n Sample = 100 boxes probability = ? sample mean = below 365 g population mean (μ) = 368 σ = 15 ∴ area less than z = -2.00 → 0.0228 ∴ 2.28% of samples of 100 boxes have mean below 365 g for sampling distribution of mean Chapter 10: Confidence Interval Estimation (pg 402) central limit theorem & population distribution → % sample mean w/in pop mean deductive reasoning - conclu f/ s/t generally true for pop → apply to s/t specific (sample mean) inductive reasoning - specific make broader generalization - estim unknown pop (mean, proportion) - ensures is normally distrib when n is large Point Estimate - val of any statistic that estimates val of a parameter μ = 3.00 - rarely know if point estim correct b/c it's an estim of actual val = 2.95 sample mean estim pop Confidence interval estim - range of # around pt estim (μ) - prob of pop w/in interval sample mean = stat, pt estim pop mean = parameter Confidence Intervals - discrepancy of estim -- help estim what actual val of unknown pop mean - Point Estimate ± Margin of Error = Point Estimate ± (Critical Value)(Standard Error) o Standard Error - std dev of point estim o Critical Value - table val based on sampling distribution of pt estim and desired confidence level - - eval shape of sample data using histogram, stem and leaf, box plot, normal prob plot sample - pt estim of pop mean μ there is specified confidence μ is somewhere in range of # by interval - 95% confident level mean GPA (2.75 ≤ μ ≤ 2.85) = between - 5% chance mean GPA below/above 2.75/ 2.85 Sampling Error - sample stat vary f/ sample to sample - f/ selecting one sample f/ pop - size or error → variation in pop and sample size Level of confidence - α is proportion in the tails of distrib outside of confidence interval - level of confidence = (1 - α)(100%) 10.1: Confidence Interval Estim for Mean (σ KNOWN) Critical val - val ofα/2eeded for constructing a confidence interval for distrib - corresponds to area under curve equal to Z val CASIO - Determine Z Val: Z val = ?, confidence = 99% (prob = 0.99), mean = 0, std dev = 1 STAT → F5(DIST) → F1(NORM) → F2(InvN) → F2(Var) Inverse Normal Data: Variable Tail: CNTR (F3) Area: 0.99 σ: 1 μ: 0 Save Res: None EXE or F1(CALC) → X1 Inv = -2.575 and X2 Inv = 2.575 CASIO Confidence interval = 95%, = 10.998, σ = 0.02, n =100, μ = ? STAT → F4(INTR) → F1(Z) F1(1-S) Results: 1-Sample tInterval 1-Sample tInterval Data: F2(Var) Left = 10.9940801 C-Level: 0.95 Right = 11.0019199 σ: 0.02 = 10.998 = 10.998 n =100 n: 100 Save Res: None ∴ 95% confidence interval estim for pop mean = 10.994 to 11.001 in. find upper & lower limits around μ → μ ± (1.96) (σ/ √n) - sub as interval to estim unknown μ → ± (1.96) (σ/ √n) o μ = ± (1.96) (σ/ √n) = 362.3 ± (1.96)(15/√25) = 362.3 ± 5.88 = 368.18 or 356.42 ∴ 356.42 ≤ μ ≤ 365.18 ← estim Z-Table area in the tail is 0.025 → area in the body = 0.9750 Ex: - z = ±1.96 - sample = 520 μ = - pop μ = ? = and - σ = 100 = - n = 25 - ∴ I am 95% confident that the mean SAT ( = 520) score is b/ 480.80 and 559.20 - ∴ 480.80 ≤ μ ≤ 559.20 10.2: Confidence Interval Estim for Mean (σ UNKNOWN) Properties of the t Distribution - symmetrical, bell shaped, mean & med = 0 - more area in tail and less in center than std normal distb - b/c S used to estim unknown σ, val of t more variable than Z - Degree of freedom - n -1 o dir related to sample size n o increase in sample size degrees of freedom = S better estim σ and t distrib approaches std normal distib until little diff b/ t and Z distib - critical val - corresponds to area under curve equals t distrib o Cumulative Probabilities 0.75 0.90 0.95 0.975 0.99 0.995 Upper Tail Areas Degrees of 0.25 0.10 0.05 0.025 0.01 0.005 Freedom 1 1.000 3.077 6.3138 12.706 31.820 63.657 0 7 2 0.816 1.885 2.9200 4.3027 6.9646 9.9248 5 6 . 99 0.677 1.290 1.6604 1.9842 2.3646 2.6264 0 2 100 0.677 1.290 1.6602 1.9840 2.3642 2.6259 0 1 CASIO: Critical val f/ t Table for area of 0.025 in each tail with 99 degree freedom (df=99) STAT → F5(DIST) → F2(t) → F3(Invt) → F2(Var) Result: Inverse Normal XInv = ±1.988421 Data: Variable Area: 0.025 ∴ critical val for 99% confidence df: 99 Area = 0.95 interval with df =99 are -1.98 Save Res: None Area = 0.05/2 = 0.025 and 1.98 EXE t = 1.98421 The Concept of Degrees of Freedom 2 2 - Sample variation (S ) = ∑ (Xi- ) - n -
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