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Leader: Course: Instructor: Date: Stat 226 SI: 3/5/13 Supplemental Instruction Iowa State University Carly Stat 226 (Various) 3/5/13 Exam 2 Material Review o Chapter 13: “Samples and Surveys” Accuracy vs. Precision The meaning of 𝑋̅ and how it is different from 𝑥̅ and μ Sampling Strategies SRS Voluntary Convenience Stratified Cluster Bias in sampling Shape, center, spread in histograms o Chapter 14: “Sampling Dist. of the Sample Mean and the Central Limit Theorem” Shape, center, spread for the dist. of 𝑋̅ 3 Cases X~N(μ, σ2) 2 𝜎 o 𝑋̅~N(𝜇, ( 𝑛) ) √ X is symmetrically distributed and bell-shaped (but not normal) 2 𝜎 o 𝑋̅~approx. N(𝜇, ( 𝑛) ) √ X is not normal, nor symmetric (skewed, multimodal, etc.) 2 𝜎 o 𝑋̅~approx. N(𝜇, ( 𝑛) ) if n is sufficiently large (n ≥ 30) √ o Use of the CLT Standard Deviation of 𝑋̅ “Standard Error” Generic Problems for Sampling Distribution of 𝑋̅ 1.) Variable X is normally distributed with a mean of 50 and standard deviation of 6. a.) What can we say about the shape of the sampling distribution of 𝑋̅? 𝑋̅ is normally distributed because X is normally distributed. b.) Do we need to use the CLT to comment on the shape of the dist. of 𝑋̅? No, we do not need to use the CLT because X follows a normal dist. What are the mean and standard error of the sampling distribution of 𝑋̅? 𝜇𝑋̅ = 𝜇 = 50 𝜎 6 SE(𝑋̅) = 𝜎𝑋̅ = 𝑛 = 𝑛 (A value for “n” was not given.) √ √ 2.) Variable X is not normally distributed, but it is symmetric and bell-shaped. It has a mean of 50 and a standard deviation of 6. c.) 1060 Hixson-Lied Student Success Center 515-294-6624 [email protected] http://www.si.iastate.edu a.) What can we say about the sampling distribution of 𝑋̅? 𝑋̅ is approximately normally distributed. b.) Does the standard error of 𝑋̅ change if we go from a normal sampling distribution of 𝑋̅to one that is only approximately normal? If so, how? 𝜎 No, the standard error is still equal to 𝜎𝑋̅ = 𝑛. √ c.) Is the mean for a distribution of X different from the mean of a sampling distribution of 𝑋̅? If so, when? Regardless of the shape of the distribution of X, the sample mean will follow a normal distribution with a mean 𝜇𝑋̅ = 𝜇, according to the CLT. A farmer in Iowa owns an apple orchard. He claims that the number of apples per tree on his apple orchard is normally distributed with a mean of 100 apples and a standard deviation of 20 apples. Assume a sample size of 300. What is the shape of the distribution of X? X is normally distributed. What is the shape of the sampling distribution of 𝑋̅? 𝑋̅ is normally distributed because X is normally distributed. 𝑋̅ X Mean Standard Deviation Mean μ = 100 σ = 20 𝜇𝑋̅ = 𝜇 = 100 Standard Error 𝜎𝑋̅ = 𝜎 √𝑛 = 20 √300 = 1.155 Can we make conclusions about the probability distribution of X? Yes, because X is normally distributed. What is the probability of “obtaining” a single tree with more than 115 apples? P(X > 115) = 1 – P(X < 115) 𝑥− 𝜇 115− 100 = 1– P( 𝜎 < ) 20 = 1 – P(Z < 0.75) = 1 – 0.7734 = 0.2266 = 22.66% What is the probability of obtaining a sample mean greater than or equal to 103 apples? P(𝑋̅ ≥ 103) = 1 – P(𝑋̅ ≤ 103) = 1 – P( 𝑥̅ − 𝜇 𝜎 √𝑛 ≤ 103− 100 = 1 – P(Z ≤ 2.60) = 1 – 0.9953 = 0.0047 = 0.47% 20 √300 )