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Paul Cartledge

Ch. 18 - Sampling Distributions Expanded def’n: A parameter is: - a numerical value describing some aspect of a pop’n - usually regarded as constant - usually unknown A statistic is: - a numerical value describing some aspect of a sample - regarded as random before sample is selected - observed after sample is selected The observed value depends on the particular sample selected from the population; typically, it varies from sample to sample. This variability is called sampling variability. The distribution of all the values of a statistic is called its sampling distribution. Def’n: p = proportion of ppl with a specific characteristic in a random sample of size n p = population proportion of ppl with a specific characteristic The estimate of the standard deviation of a sampling distribution is called a standard error. General Properties of the Sampling Distribution of p ˆ: Let pnd p be as above. Also, µ apd σ arepthe mean and standard deviation for the distribution of p. Then the following rules hold: Rule 1: µ p = p. (Textusesk µ(p) ) p(1− p pq Rule 2:σ p= = . (standardor Æ σ )p n n Ex18.1) Suppose the population proportion is 0.5. a) What is the standard deviation of pfor a sample size of 4? p(1 − p) 0.5(1 0.5) σ ˆp= = = 0.25 n 4 b) How large must n (sample size) be so that the sample proportion has a standard deviation of at most 0.125? p(1− p) n = Æ n = p (1p ) = 0.5(1 0.5) =16 σ p σ p 0.125 2 Rule 3: When n is large and p is not too near 0 or 1, the sampling distribution of pis approximately normal. The farther from p = 0.5, the larger n must be for accurate normal approximation of pˆ. Thus, if np and n(1 – p) are both sufficiently large (≥ 15), then it is safe to use a normal approximation. Further assumptions: the sample should always be random and, if sampling without replacement, the sample should be less than 10% of the population. Using all 3 rules, the distribution of papproximately normal. pp− Z N )1,~▯( pp1− ) n Ex18.2) Suppose that the true proportion of people who have heard of Sidney Crosby is 0.87 and that a new sample consists of 158 people. a) Find the mean and standard deviation of p. p(1− p) 0.87(1− 0.87) µp= 0.870 σ p = = = 0.0268 n 158 b) What can you say about the distribution of p? np = 158(0.87) = 137.46 n(1 – p) = 158(1 – 0.87) = 20.54 Since both values are > 15, the distribution ofshould be well approximated by a normal curve. c) What is the probability of getting a sample proportion greater than 0.94? ⎛ ⎞ ⎜ ⎟ P( p > 0.94) Æ P ⎜ p − p > 0.94 −.87 ⎟ = P(Z > 2.62) ⎜ p(1− p) 0.0268 ⎟ ⎜ ⎟ ⎝ n ⎠ = 1 – P(Z < 2.62) = 1 – 0.9956 = 0.0044 Sampling Distribution of Mean How does the sampling distribution of the sample mean compare with the distribution of a single observation (which comes from a population)? Ex18.3) An epically gigantic jar contains a large number of balls, each labeled 1, 2, or 3, with the same proportion for each value. Let Y be the label on a randomly selected ball. Find µand σ .
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