Lecture 10: Depicting Sampling Distributions of a Sample Proportion

Lecture 10: Depicting Sampling Distributions of a Sample Proportion Chapter 5: Probability and Sampling Distributions 2/10/12

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Sample Proportion •  “1” is assigned to population members having a specified characteristic and “0” is assigned to those who don’t. The parameter of interest in this situation is p (or called π), the proportion of the population that has the characteristic of interest. •  Denote pˆ as the proportion of members having such characteristic (or say, successes), also called the sample proportion, in a random sample of size n. 2/10/12

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Sampling Distribution of Sample Proportion •  If X ~ B(n, p), the sample proportion is defined as

X count of successes in sample pˆ = = . n size of sample •  Mean & variance of a sample proportion:

µ pˆ = p,

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σ pˆ = p(1 − p) / n .

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Example: Clinton's vote •  43% of the population voted for Clinton in 1992. •  Suppose we survey a sample of size 2300 and see if they voted for Clinton or not in 1992. •  We are interested in the sampling distribution of the sample proportion pˆ , for samples of size 2300. •  What's the mean and variance of pˆ ? 2/10/12

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Count & Proportion of “Success” •  A BoilerMaker basketball player is a 95% freethrow shooter. •  Suppose he will shoot 5 free-throws during each practice. •  X: number of free-throws he makes during practice. •  pˆ : proportion of made-free-throws during practice. •  P( pˆ =0.6) = ? •  P(X=3) = ? 2/10/12

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Normal Approximation for Counts and Proportions

•  Let X ~ B(n, p) and pˆ = X / n is the sample proportion. •  If n is large*, then X is approx. N (np , np (1-p) ) pˆ is approx. N ( p,

p (1-p) / n ).

•  *Rule of Thumb: np ≥ 5, n(1 - p) ≥ 5.

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Switches Inspection •  A quality engineer selects an SRS of 100 switches from a large shipment for detailed inspection. •  Unknown to the engineer, 10% of the switches in the shipment fail to meet the specifications. •  What is the probability that at most 9 switches fail the standard test in the sample? •  Use Normal Approximation here. Try to use continuity correction as well. (Similar to ex. 43 in Hw4) 2/10/12

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Switches Inspection •  Express the probability in terms of X: P(X ≤ 9) •  The normal approximation to the probability of no more than 9 bad switches is the area to the left of X = 9 under the normal curve, which has

µ X = np = (100)(.1) = 10, σ X = np(1 − p) = 100(.1)(.9) = 3. •  In this case np = 10, and n(1-p) = 90, both satisfying the condition for “rule of thumb”. •  Also, don’t forget the CONTINUITY CORRECTION when applying normal approximation to the Binomial distribution. 2/10/12

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Continuity Correction

•  The normal approximation is more accurate if we consider X = 9 to extend from 8.5 to 9.5 •  Example (Cont.):

X − 10 9.5 − 10 P ( X ≤ 9) = P ( X ≤ 9.5) = P ( ≤ ) 3 3 ≈ P( Z ≤ −.17) = .4325.

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Continuity Correction

P(X ≤ 8) replaced by P(X < 8.5) P(X ≥ 14) replaced by P(X > 13.5) P(X < 8) = P(X5, n(1-p) = 1000(1-.3) > 5. Therefore we can use normal approximation to sample proportion here: ⎛ ⎞ ˆ p − p . 32 − . 30 ⎟ = P( Z > 1.38) = .0838. P( pˆ > .32) = P⎜ > ⎜ p(1 − p) n .01449 ⎟ ⎝ ⎠ 2/10/12

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Summary •  Sampling Distribution •  Sampling distribution for a sample mean –  Mean and Variance/S.D. for a sample mean –  Central Limit Theorem –  Normal calculation

•  Sampling distribution for a sample proportion –  Continuity correction for Binomial Distribution –  Rule of thumb, when n is large 2/10/12

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After Class… •  Review Sec 5.4 through 5.6 . •  Start Hw 4 – due by next Monday, Feb 14th •  Review all the notes up to today, go over the practice test. Come to Office Hours for Hw#1-#4 grading issues. •  About Exam 1. –  Start editing your own cheat-sheet –  Practice Test is posted. Try to do it in an hour

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