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Paul Cartledge

7.1 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 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 ˆp= p. p (1− p pq Rule 2:σ ˆp = . (a.k.a. standard error, see above) n n Ex7.1) Suppose the population proportion is 0.5. a) What is the standard deviation of p for 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) p(1 p ) 0.5(1 0.5) n = Æ n = 2 = 2 =16 σ pˆ σ p 0.125 ˆ is Rule 3: When n is large and p is not too near 0 or 1, the sampling distribution of p approximately normal. The farther from p = 0.5, the larger n must be for accurate normal approximation of p . Conservatively speaking, if np and n(1 – p) are both sufficiently large (e.g. ≥ 15), then it is safe to use a normal approximation. Using all 3 rules, the distribution of p − p Z = p(1− p) n is well approximated by the standard normal distribution. Ex7.2) 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) µ p 0.870 σ pˆ= = = 0.0268 n 158 ˆ ? 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 op should be well approximated by a normal curve. c) Find P( p> 0.94). Note: not a z-score! ⎛ ⎞ ⎜ p − p 0.94 −.87 ⎟ P( p > 0.94) ÆP ⎜ > ⎟ = P(Z > 2.64) ⎜ p(1− p) 0.0268 ⎟ ⎜ ⎟ ⎝ n ⎠ = 1 – P(Z < 2.64) = 1 – 0.9959 = 0.0041 7.2 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)? Ex7.3) An epically gigantic jar contains a large number of balls, each labeled 1, 2, or 3, with the same proportion for each value. Let X be the label on a randomly selected ball. Find µ and σ. µ = ∑ xiP(X = x i = 1(1/3) + 2(1/3) + 3(1/3) = 2 2 2 2 2 2 σ = ∑ (x − µ
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