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Understanding Distribution of Sample Proportion in Statistics

Learn about sampling distribution of sample proportion, shape, center, and spread. Compute probabilities using sample proportion in statistics.

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Understanding Distribution of Sample Proportion in Statistics

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  1. Lesson 8 - 2 Distribution of the Sample Proportion

  2. Objectives • Describe the sampling distribution of a sample proportion • Compute probabilities of a sample proportion

  3. Vocabulary • Sample proportion – p-hat is x / n ; where x is the number of individuals in the sample with the specified characteristic (x can be thought of as the number of successes in n trials of a binomial experiment). The sample proportion is a statistic that estimates the population portion, p.

  4. Conclusions regarding the distribution of the sample proportion • Shape: as the size of the sample, n, increases, the shape of the distribution of the sample proportion becomes approximately normal • Center: the mean of the distribution of the sample proportion equals the population proportion, p. • Spread: standard deviation of the distribution of the sample proportion decreases as the sample size, n, increases

  5. Sampling Distribution of p-hat • For a simple random sample of size n such that n ≤ 0.05N (sample size is ≤ 5% of the population size) • The shape of the sampling distribution of p-hat is approximately normal provided np(1 – p) ≥ 10 • The mean of the sampling distribution of p-hat is μ p-hat = p • The standard deviation of the sampling distribution of p-hat is σ = √(p(1 – p)/n)

  6. Summary of Distribution of p x, number of individuals with specified characteristic p-hat or p = --- n, number of sample size • For a simple random sample of size n such that n≤ 0.05N(that is less than 5% of the population size) • Shape of sampling distribution of p is approximately normal, provided np(1 – p) ≥ 10 • Mean of the sampling distribution of p is μp = p • Standard Deviation of the sampling distribution of p is p(1 – p) σp = ------------ n

  7. a Example 1 Assume that 80% of the people taking aerobics classes are female and a simple random sample of n = 100 students is taken What is the probability that at most 75% of the sample students are female? P(p < 75%) μp = 0.80 n = 100 σp = (0.8)(0.2)/100 = 0.04 p - μp Z = ------------- σx -0.05 = ----------------- 0.04 0.75 – 0.8 = ----------------- 0.04 = -1.25 normalcdf(-E99,-1.25) = 0.1056 normalcdf(-E99,0.75,0.8,0.04) = 0.1056

  8. a Example 2 Assume that 80% of the people taking aerobics classes are female and a simple random sample of n = 100 students is taken If the sample had exactly 90 female students, would that be unusual? P(p > 90%) μp = 0.80 n = 100 σp = (0.8)(0.2)/100 = 0.04 p - μp Z = ------------- σx 0.1 = ----------------- 0.04 0.90 – 0.8 = ----------------- 0.04 = 2.5 normalcdf(2.5,E99) = 0.0062 less than 5% so it is unusual normalcdf(0.9,e99,0.8,0.04) = 0.0062

  9. a Example 3 According to the National Center for Health Statistics, 15% of all Americans have hearing trouble. In a random sample of 120 Americans, what is the probability at least 18% have hearing trouble? P(p > 18%) μp = 0.15 n = 120 σp = (0.15)(0.85)/120 = 0.0326 p - μp Z = ------------- σx 0.03 = ----------------- 0.0326 0.18 – 0.15 = ----------------- 0.0326 = 0.92 normalcdf(0.92,E99) = 0.1788 normalcdf(0.18,E99,0.15,0.0326) = 0.1787

  10. a Example 4 According to the National Center for Health Statistics, 15% of all Americans have hearing trouble. Would it be unusual if the sample above had exactly 10 having hearing trouble? P(x < 10) μp = 0.15 p = 10/120 = 0.083 n = 120 σp = (0.15)(0.85)/120 = 0.0326 p - μp Z = ------------- σx -0.067 = ----------------- 0.0326 0.083 – 0.15 = ----------------- 0.0326 = -2.06 normalcdf(-E99,-2.06) = 0.0197 which is < 5% so unusual normalcdf(-E99,0.083,0.15,0.0326) = 0.01993

  11. Summary and Homework • Summary: The sample proportion, like the sample mean, is a random variable • If the sample size n is sufficiently large and the population proportion p isn’t close to either 0 or 1, then this distribution is approximately normal • The mean of the sampling distribution is equal to the population proportion p • The standard deviation of the sampling distribution is equal to p(1-p)/n • Homework • pg 439 – 441; 1, 2, 9, 13, 17, 19

  12. Homework 1: p = 0.44 2: sample proportion = x / n 9: apx normal μ = 0.103 σ≈ 0.010 13: a. apx normal μ = 0.103 σ≈ 0.015 b. P(x ≥ 390) = 0.00383normalcdf(0.39,E99,0.35,0.015) c. P(x ≤ 320) = 0.02275normalcdf(-E99,0.32,0.35,0.015) 17: a. apx normal μ = 0.26 σ≈ 0.0196 b. P(p-hat < 0.24) = 0.15377normalcdf(-E99,0.24,0.26,0.0196) c. P(x ≥ 150) = 0.02063, Yes normalcdf(0.3,E99,0.26,0.0196) 19: a. P(p-hat ≤ 0.40) = 0.04324 normalcdf(-E99,0.4,0.43,0.0175) b. P(p-hat ≥ 0.45) = 0.1265, No normalcdf(0.45,E99,0.43,0.0175)

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