Module 5 - Hypethesis Testing + Video 15 Class Notes

Module 5: Hypothesis Testing Video 15 The difference between a sample and a population Population: From the sample, we can draw inference about the population. Sample: We draw a random sample that represents a population. We measure a sample statistic. We use this as an estimate of the true unknown population parameter. Population mean: μ Sample mean: x̄ Population variance: σ² Sample variance: s² Our sample needs to be representative and random: Representative: Similar in structure to the population it is drawn from. Random: Every member of the population has an equal chance of being chosen as part of the sample. X is a random variable. We call the probability distribution of a statistic a sampling distribution. Example: Measuring heights of 1st year students Let X be the height of a randomly drawn 1st year student and let X₁, X₂, ..., Xn be a randomly drawn sample from the normal distribution. True mean = μ, variance = σ². Remember: E(A + B) = E(A) + E(B) Var(A + B) = Var(A) + Var(B) for A and B drawn randomly and independently, i.e., drawn randomly from a normal distribution with mean μ and variance σ². X̄ = (1/n) * ΣXi E[X̄] = E[(1/n) * ΣXi] = (1/n) * ΣE[Xi] = (1/n) * nμ = μ Var(X̄) = Var[(1/n) * ΣXi] = (1/n²) * ΣVar(Xi) = (1/n²) * nσ² = σ²/n Large sample: stable estimate of mean, small variance (it has big n) Small sample: varying mean, big variance (imprecise) The Central Limit Theorem The average (or sum divided by n) of a large number of random variables always has a normal distribution. X̄ ~ N(μ, σ²/n) as long as n 30 Video 16: Confidence Intervals Point estimate: no information regarding the uncertainty of the estimate Interval: range of values that communicate how much precision or uncertainty is present in the estimate The sample mean is a random variable. Therefore, it has a Sampling Distribution X̄ ~ N(μ, σ²/n) Now, for a fair die tossed 30 times: X̄ ~ N(3.5, √(2.917) / √30) where X̄ is our observed sample mean from 30 tosses. z = (X̄ - μ) / (σ / √n) ~ N(0, 1) Pr(-1.96 z 1.96) = 0.95 We are 95% confident that the true mean value for a fair die (30 tosses) lies between: (X̄ - 1.96 * (σ / √n)) and (X̄ + 1.96 * (σ / √n)) Continuing example X̄ = 4.70 (4.70 - 1.96 * (√2.917 / √30)) and (4.70 + 1.96 * (√2.917 / √30)) (4.09, 5.31) We are 95% confident that the interval (4.09, 5.31) contains the true mean. It excludes the true mean of a fair die (3.5). There is reason to believe our friend is cheating. What affects the width of a confidence interval L = z(α/2) * (σ / √n) Therefore, it is apparent: If you increase n, the confidence interval gets narrower. If you increase z (confidence level), the interval will get wider. So... LARGE SAMPLES YIELD BETTER RESULTS THAN SMALL SAMPLES Confidence interval can be written as 100(1-α)% We need to ask the following questions: Within how many units of the true mean do we want our estimate to be? This is the L. How confident do you want to be of your estimate? That is your z. How variable is the population you are studying? This determines σ. L = z(α/2) * (σ / √n) n = (z(α/2) * σ / L)² If we don't know σ, we can construct an interval that we are 100(1-α)% confident contains μ, as follows: Pr(X̄ - z(α/2) * (s / √n) = μ = X̄ + z(α/2) * (s / √n)) = 1 - α Confidence interval method for p, σ² known If we have a random sample size n with sample mean x̄, then A% confidence intervals for μ are given by: (x̄ - z*σ/√n, x̄ + z*σ/√n) where the appropriate values of z* are given by: A% 90% 95% 98% 99% z* 1.64 1.96 2.33 2.58 Sample size n required to achieve desired accuracy L, when σ² is known To obtain a sample mean x̄ which is within L units of the population mean, with confidence level A%, the required sample size is: n = (z*σ / L)²