e-Statistics
Introduction
  • Advanced User Guide
    Assignment & Quiz
  • User Account
  • Guide to Worksheet 1-2
  • Guide to Worksheet 3-4
  • Guide to Worksheet 5-6
  • Guide to Worksheet 7-8
    Data in Statistical Studies
  • Overview
  • Data from SMDA
  • Data and Story Library
    Exploratory Graphics
  • Overview
  • Histogram
  • Stem and Leaf Plot
  • Boxplot
  • Scatter Plot
  • QQ plot
  • Bar Graph
    Descriptive Statistics
  • Overview
  • Summary Statistics
  • Grouped Data
  • Frequency Table
    Discrete Distributions
  • Frequency Function
  • Binomial Distribution
  • Normal Approximation
    Normal Distribution
  • Normal Density Function
  • Critical Point
  • Z-score
  • Central Limit Theorem
    Inference on Mean
  • t-Distribution
  • Confidence Interval
  • Hypothesis Test
  • Power of Test
  • One-sided CI
    Comparison of Two Groups
  • Overview
  • t-Test
  • Nonparametric Test
  • Test for Paired Data
    Inference on Variances
  • Chi-square Distribution
  • Testing a Standard Deviation
  • F-Distribution
  • F-Test
    Explore Statistics in Drug Safety

    • Normal Approximation

    Suppose that $ X$ is a binomial random variable with $ n =$ and $ p =$ . If the size $ n$ is adequately large, then the distribution of $ X$ can be approximated by the normal distribution with parameter $ (np, np(1-p))$ . That is, the normal distribution $ N(np,np(1-p))$ approximates the binomial distribution $ B(n,p)$ . A general rule for ``adequately large'' $ n$ is to satisfy $ np \ge 5$ and $ n(1-p) \ge 5$ .

    Calculation of probability with continuity correction. Let $ Y$ be a normal random variable with parameter $ (np, np(1-p))$ . Then the distribution of the binomial random variable $ X$ can be approximated by that of $ Y$ . Having taken into account the fact that $ Y$ is a continuous random variable, the approximation of discrete probability becomes

    $\displaystyle P(i \le X \le j) \approx P(i-0.5 \le Y \le j+0.5) = \Phi\left(\fr... ...np}{\sqrt{np(1-p)}}\right) - \Phi\left(\frac{i-0.5-np}{\sqrt{np(1-p)}}\right). $