Central limit Theorem

Central Limit Theorem Two assumptions 1. The sampled values must be independent 2. The sample size, n, must be large enough • The mean of a random sample has a sampling distribution whose shape can be approximated by a Normal model. • The larger the sample, the better the approximation will be. • This is regardless of the shape of the distribution of the population being sampled from or the shape of the distribution of the sample. Central limit Theorem Sample Distribution Models for Means and Proportions Distribution of sample proportions • Population has a fixed proportion ( p ) • To find population proportion a sample is taken and a sample proportion is calculated ( p ) If samples are repeatedly taken with the same sample size • The mean of the sample distribution would be the population proportion, μ p̂ = p • The standard deviation would be σ p̂ = pq n Conditions to check for the assumptions 1. Success/Failure: The expected number of successes and failure is both greater than 10 np ≥ 10 and nq ≥ 10 2. 10% Condition: Each sample is less than 10% of the population 3. Randomization: The sample was obtained through random sample techniques or we can at least assume that the sample is representative. All conditions have been met to use the Normal model for the distribution of sample proportions. Example: Skittles • If samples were repeatedly taken with the same sample size then from the CLT, the distribution would be approximately Normal ⎛ pq ⎞ pˆ ~ N ⎜ p, ⎟ n ⎠ ⎝ Conditions: 1. 10% condition: 58 skittles is less than 10% of all skittles produced. 2. Success/Failure: np = 58i0.20 = 11.6 ≥ 10 nq = 58i0.80 = 46.4 ≥ 10 There are at least 10 successes and failures 3. Randomization: Though not from a random sample we can assume the bag is representative of the population. All conditions have been met to use the Normal model for the distribution of sample proportions. • According to the manufacturer of the candy Skittles, 20% of the candy produced is the color red. What is the probability that given a large bag of skittles with 58 candies that we get at least 17 red? • Mean: μ = p = 0.20 • Standard Deviation: pq 0.20i0.80 σ= = = 0.0525 n 58 • So the model for p̂ becomes N(0.20,0.0525) • Sample proportion: pˆ = 17 = 0.293 58 • Then to find the probability that we get a sample proportion of 0.293 or higher: P ( pˆ ≥ 0.293) = 0.0383 p. 433 #12,16 12. Smoking p = 0.264 n = 50 1. Random: stated as a random sample 2. 10% condition: 50 adults is less than 10% of all adults 3. Success/Failure: np = 50(.264) = 13.2 ≥ 10 nq = 50(.736) = 36.8 ≥ 10 There are at least 10 successes and failures. All conditions have been met to use the Normal model for the distribution of sample proportions. ⎛ (0.264)(0.736) ⎞ pˆ ~ N ⎜ 0.264, = N ( 0.264,0.0623 ) ⎜ ⎟⎟ 50 ⎝ ⎠ 16. Contacts p = 0.30 n = 100 1. Random: stated as a random sample 2. 10% condition: 100 college students are less than 10% of all students at a university 3. Success/Failure: np = 100(0.3) = 30 ≥ 10 nq = 100(0.7) = 70 ≥ 10 There are at least 10 successes and failures. All conditions have been met to use the Normal model for the distribution of sample proportions. ⎛ (0.3)(0.7) ⎞ pˆ ~ N ⎜ 0.3, ⎟⎟ = N ( 0.3,0.0458 ) ⎜ 100 ⎝ ⎠ P ( pˆ > 1/ 3) = 0.2334 Distribution of Sample Means • A sample of size n is taken from a population with mean, µ, and standard deviation, σ. Then the distribution of the samples means: • The mean is the mean of the population: μx = μ • The standard deviation is the population standard deviation divided by the square root of sample size, n. σx = σ n Assumptions • There are two assumptions that must be met for this to be true 1. The sampled values must be independent 2. The sample size, n, must be large enough If met then from the CLT the distribution of the sample means, x , will be approximately Normal. ⎛ σ ⎞ x ∼ N ⎜ μ, ⎟ n⎠ ⎝ Example: SAT Scores • Suppose that male seniors have a mean score of 1200 with a standard deviation of 130. • If we take a random sample of 100 male seniors what would be the distribution of sample means look like? • What would be the probability that we would get an average of less than 1150? • Conditions: 1. Randomization: The sample was obtained through random sample techniques or we can at least assume that the sample is representative. 2. Independence(10% Condition) 3. Large enough sample: • If the population distribution is thought to be unimodal and symmetric a small sample size will do. • If it is more skewed, larger samples are needed. n > 30 All conditions have been met to use the Normal model for the distribution of sample means μ = 1200 σ = 130 n = 100 Conditions 1. Randomization: Stated as a random sample 2. 10% Condition: 100 male seniors is less than 10% of all male seniors 3. Large Enough Condition: n > 30, A sample of 100 would be a large enough sample All conditions have been met to use the Normal Model for the distribution of sample means. • Parameters: μ x = μ = 1200 σx = σ 130 = = 13 n 100 • So we have a Normal model N(1200,13). • What would be the probability that we would get an average of less than 1150? P ( x ≤ 1150) = 6 × 10−5 34. Home Values μ = $140,000 σ = $60,000 n = 100 Conditions: 1. Random: stated as a random sample. 2. 10% Condition: 100 homes is less than 10% of all homes in a small city 3. Large Enough: 100 > 30, the sample size is ok. All conditions have been met to use the Normal model for the distribution of sample means. 60,000 ⎞ ⎛ x ~ N ⎜ 140,000, ⎟ = N (140,000, 6,000 ) 100 ⎝ ⎠ 38. Rainfall μ = 35.4 " σ = 4.2" x − 35.4 a) N(35.4, 4.2) b) z = −0.842 = 4.2 P(x > 40) = 0.1368 x = 31.86" c) n = 4 Conditions: 1. Random: not a random sample. Assume it is a representative sample 2. 10% Condition: 4 years is less than 10% of all years in Ithica 3. Large Enough: 4 < 30, The population is Normally distributed so the sample size is ok. All conditions have been met to use the Normal model for the distribution of sample means. 4.2 ⎞ ⎛ x ~ N ⎜ 35.4, ⎟ = N ( 35.4,2.1) 4⎠ ⎝ d ) P ( x < 30) = 0.0051