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Unformatted text preview: IENG 213: Probability and Statistics for Engineers Instructor: Steven E. Guffey, PhD, CIH 20022007 2 9.2 Statistical Inference Statistical inference: methods employed to make generalizations about a population. Two major areas: Estimation: estimate mean, variance, or other statistic (Chapter 9). Tests of hypotheses: determine probability that a statement is true (Chapter 10). 3 9.3 Classical Methods of Estimation Point estimates of : x X ~ (median) Point estimates of the proportion, p p^ = x/n 4 Interval Estimation The point estimate is unlikely to be a perfect estimator. Range of values all have some likelihood. Therefore have confidence interval . x lower < < x upper Confidence (i.e., probability) that mean within the range: P(x lower < < x upper ) = 1  5 Interpretation of Interval Estimates Ideally, narrow confidence interval with high confidence (i.e., small ) Example: For sample mean = 540 Confidence interval: 520 < < 560 P( 520 < < 560 ) = 1 = 1 0 .0 5 = 0 .9 5 Confidence (i.e., probability) that mean within the range: P(x lower < < x upper ) = 1  6 Looking at Zvalues for a particular P(Z /2 < Z < Z 1 /2 ) = 1 Everyone leaves off abs(). It is understood: Note: Z 1 /2 = Z /2 Since Z /2 is negative: Z /2 =  abs(Z / 2 ) P ( abs [Z /2 ] < Z < + abs [Z /2 ] ) = 1 P (Z /2 < Z < Z /2 ) = 1 7 Confidence interval for a particular / x X X Z n  = = Definition of Zvalues: Substituting: /2 /2 ( ) 1 / X P Z Z n  < < =  Algebra: P (Z /2 < Z < Z /2 ) = 1 /2 /2 1 Z Z P X X n n  < < + =  8 Interpretation of Interval Estimates /2 /2 1 Z Z P X X n n  < < + =  decreases (more certainty) interval width decreases (more precision) 9 Interpretation of Interval Estimates Example: if = 0.05, then: P(x lower < < x upper ) = 1 0.05 = 95% Z upper = 1.96 Z lower = 1.96 /2 = 0 .0 2 5 /2 = 0 .0 2 5 P(Z /2 < < Z /2 ) = 1  2 * P(Z < Z /2 ) 10 Confidence Interval of mean with and Known cont. n Z X n Z X 2 / 2 / + < < Works well for n > 30 even if underlying population distribution not completely normal 11 Example From a sample of 36, the average diameter of a pushrod was 5 cm. Its known (i.e., pop) standard deviation is 1 cm. Find the 95% confidence interval for the true mean. Solution: / 2 / 2 1 Z Z P X X n n  < < + =  1.96(1) 1.96(1) 5 5 1 0.05 95% 36 36 P  < < + = = 0 .0 5 /2 = 0 .0 2 5 = 1 .9 6 P(4.67 < < 5.33) = 95% 12 Example From a sample of 36, the average diameter of a pushrod was 5 cm. Its known standard deviation is 1 cm. Find the 99% confidence interval for the true mean....
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This note was uploaded on 04/03/2009 for the course IENG 213 taught by Professor Staff during the Spring '08 term at WVU.
 Spring '08
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