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# 1/29/2016 Class Notes MAT 258

SUNY Oswego

GPA 3.6

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This 3 page Class Notes was uploaded by Andrew on Friday January 29, 2016. The Class Notes belongs to MAT 258 at State University of New York at Oswego taught by Gregory Schneider in Winter 2016. Since its upload, it has received 26 views. For similar materials see Intro To Statistics B in Mathematics (M) at State University of New York at Oswego.

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Date Created: 01/29/16

MAT 258 Gregory Schneider 1/29/2016 Class Notes Recap of 1/27 notes: o Sampling distributions associated with values of a statistic over all samples of a given size n - Sampling Distribution of sample mean x o If X ~ NM, ơ )… Then x ~ N (M, ơ/ √n ) ) o (CLT): If n ≥ 30 and x has mean M, standard deviation √n 2 isơ with any distribution, then x ≈ N (M,ơ/ ) ) In other words: M = M, and ơ ơ/ √n x x = o Fundamentals: evaluating Normal Probs - Interpret desired prob - P (x ≤ a), p(x ≥ a), P(a ≤ x ≤ b) - P (x < a) = P(x ≤ a) - Rewrite Prob for evaluation - Look up Table V in text Gives areas (probabilities) to the left P (x ≤ a) P (x ≥ a) = 1 – P (x ≤ a) P (a ≤ x ≤ b) = P(x ≤ b) – P (x ≤ a) - Calculate Z – Scores - Z = (x-m)/ ơ - P (x ≤ a) = P (Z ≤ (a-M)/ ơ) - Use Table V - (Round to 2 decimal places) o Sampling Distribution of Sample portion p - Definition: Sample proportion p in a sample size n where x individuals possess a desired characteristic p = x/n Example: n=28, x=10 p = 10/28 = 0.3571 If we take lots of samples, from a population with proportion p, what kinds of sample proportions will we get? Aside: The Binomial Distribution: (Sections 6.1, 6.2) The # of successes X in n independent trials where each trial has same probability of p success The example taken in class of blue eyed people within a sample size of 28 is equal to 28 independent trials Q: Are trials independent? Solution: No, not really Large #’s make it OK (Note: for trials to be independent, you need approximately n ≤ 0.05N) mean)lation 1) If np(1-p) ≥ 10 OR sometimes np ≥ 5 AND n(1-P) ≥ 5 … then Bin (n, p) ≈ N (np, ( √np(1−p)¿ 2) √np(1−p) 2 √ 2 2) If x ≈ N (np, ( ) ), then it can show x/n ≈(p, ( (p(1p))/n)) Then: if a sample of size N is taken from a population of size N, having proportion p with a specified characteristic, where n satisfies N ≤ 0.05N Np (1-p) ≥ 10 Then p ≈ N(p, ( √ (p(1p))/n) ) ie, Mp = p and ơp = √ (p(1p))/n) Assuming p = 0.166, what is the probability of selecting a sample size of 28 with a proportion p ≥ 0.3571 N ≤ 0.05N? …yes, 28 ≤ 0.05 (318 million) Np (1-p) ≥ 10? 28(0.166)(0.834) = 3.876 (note: results may be unreliable) Another Example: 10% of Americans are afraid to fly. What is the probability that fewer than 88 in a sample of 1100 are afraid to fly? p = 88/1100 = 0.08 p( p < 0.08) N ≤ 0.05? ✓ 1100 ≤ 5% of U.S. population Np(1-p) = 1100 (0.1)(0.9) = 99 ≥ 10 ✓ o p ≈ N (p, ( √ (p(1p))/) o P(p < 0.08) = p(p ≤ 0.08) 0.1 0.4 /1100¿ = p( z ≤ (0.08-0.1)/ √ ( ) ) o = P ( z ≤ -2.21) = 0.0136 ≈ 1.4%

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