# M312_5 - 1 Probability and Statistics II M-312 Lecture 5 5...

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Unformatted text preview: 1 Probability and Statistics II M-312 Lecture 5 5. Inferences Based on Two Samples 5.1. z Tests and Confidence Intervals for a Difference Between Two Population Means Proposition 5.1. The expected value of Y X- is µ 1- µ 2 , so Y X- is an unbiased estimator for µ 1- µ 2 . The standard deviation of Y X- is n m Y X 2 2 2 1 σ σ σ + =- TEST PROCEDURES FOR NORMAL POPULATIONS WITH KNOWN VARIANCES Because the population distributions are normal, both X and Y have normal distributions. This implies that Y X- is normally distributed, with expected value µ 1- µ 2 and standard deviation Y X- σ given in the foregoing proposition. Standardizing Y X- gives the standard normal variable ( 29 n m Y X Z 2 2 2 1 2 1 σ σ μ μ +--- = (5.1) We then arrive at the following testing scheme. Basic Assumptions 1. m X X X , , , 2 1 is a random sample from a population with mean µ 1 and variance 2 1 σ . 2. n Y Y Y , , , 2 1 is a random sample from a population with mean µ 2 and variance 2 2 σ . 3. The X and Y samples are independent from one another. 2 Example 5.1. Analysis of a random sample consisting of m = 20 specimens of cold-rolled steel to determine yield strengths resulted in a sample average strength of 8 . 29 = x ksi. A second random sample of n = 25 two-sided galvanized steel specimens gave a sample average strength of 7 . 34 = y ksi. Assuming that the two yield-strength distributions are normal with σ 1 = 4.0 and σ 2 = 5.0, does the data indicate that the corresponding true average yield strengths µ 1 and µ 2 are different? Let’s carry out a test at significance level α = .01. 1. The parameter of interest is µ 1- µ 2 , the difference between the true average strengths for the two types of steel. 2. The null hypothesis is H : µ 1- µ 2 = 0. 3. The alternative hypothesis is H a : µ 1- µ 2 ≠ 0; if H a is true then µ 1 and µ 2 are different. 4. With Δ = 0, the test statistic value is n m y x z 2 2 2 1 σ σ +- = 5. The inequality in H a implies that the test is two-tailed. For α = .01, α/2 = .005 and z α/2 = z 0.005 = 2.58. H will be rejected if z ≥ 2.58, or if z ≤ -2.58. 6. Substituting , 7 . 34 , 25 , . 16 , 8 . 29 , 20 2 1 = = = = = y n x m σ and . 25 2 2 = σ into the formula for z yields Null Hypothesis: H : µ 1- µ 2 = Δ Test statistic value: n m y x z 2 2 2 1 σ σ + ∆-- = Alternative Hypothesis Rejection Region for Level α Test H a : µ 1- µ 2 > Δ z ≥ z α (upper-tailed test) H a : µ 1- µ 2 < Δ z ≤ - z α (lower-tailed test) H a : µ 1- µ 2 ≠ Δ either z ≥ z α/2 or z ≤ - z α/2 (two- tailed test) Because these are z tests, a P-value is computed as it was for the z tests in Lecture 4 (e.g., P-value = 1 – Φ( z ) for an upper tailed test). 3 66 . 3 25 . 25 20 . 16 7 . 34 8 . 29- = +- = z That is, the observed value of y x- is more that three standard deviations below what would be expected were H true....
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## This note was uploaded on 05/11/2011 for the course M 312 taught by Professor Sheng during the Spring '11 term at ITT Tech Pittsburgh.

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M312_5 - 1 Probability and Statistics II M-312 Lecture 5 5...

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