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Chapter 7: Sampling Distribution of

7.6 Overview

  • Focus on understanding sampling distribution of proportions (p).

  • Learn about random variables with specific mean (𝜇) and standard deviation (𝜎).

Learning Goals

  • Understand the concept of random variable with mean p and standard deviation p.

  • Gain insight into the Sampling Distribution of p and its normal distribution properties.

  • Learn to calculate probabilities related to sampling distributions.

7.6 Example 1: Frequent Flyer Customers

  • An airline is assessing customer interest in new routes targeting frequent flyer customers.

  • A random sample of 10,000 frequent flyers is surveyed for their usage plans with an incentive of 1500 miles.

Important Elements

  • Population: All frequent flyer customers.

  • Parameter: Proportion of all frequent flyers planning to use the new hubs (denoted as p).

  • Sample: The selected 10,000 frequent flyers.

  • Sample Statistics (Point Estimator): Proportion of surveyed customers intending to use new hubs; denoted as p.

Sample Proportion Calculation

  • Represented as:

    • p = x/n

  • Where:

    • x = number of frequent flyer customers in the sample planning to use new hubs.

    • n = sample size.

Concept of Sampling Distribution

  • If multiple random samples are taken, various values of p will be produced.

  • The sampling distribution of p consists of all possible values of p.

  • Key characteristics:

    • Mean (μ) of sampling distribution: μp = p

    • Standard deviation (σ): σp = √[p(1−p)/n]

Conditions for Normal Distribution

  • The sampling distribution of p can be approximated by a normal distribution if:

    • np ≥ 5

    • n(1−p) ≥ 5

Importance of Sampling Distributions

  • Allows probability statements about how close sample proportions (p) are to population parameters (p).

Computing Probabilities Example #2

  • Scenario with Doerman Distributors:

    • Belief: 30% of orders are from first-time customers.

    • Random sample: 100 orders taken.

    • Questions:

      • Probability that sample proportion exceeds 0.35.

      • Probability that it falls between 0.20 and 0.40.

Solution Calculation

  • Random variable p = proportion from sample.

  • Define:

    • n = 100

    • p = 0.3

    • σ = √[(0.3)(0.7)/100] = 0.046.

  • Normality check:

    • 0.3 * 100 > 5, 0.7 * 100 > 5 (condition satisfied).

Probability Analysis

  • (A): P(p > 0.35) calculated via Z transformation:

    • Z = (0.35 - 0.3)/0.046 = 1.087.

    • Result: P(Z > 1.087) = 0.14.

  • (B): P(0.2 < p < 0.4) involves:

    • Z values: Z for p = 0.2 is -2.17 and for p = 0.4 is 2.17.

    • Result: P(-2.17 < Z < 2.17) = 0.97.

Example #3: Age and Startups

  • Report: 55% of entrepreneurs start by age 29.

  • Random sample of 200 entrepreneurs.

  • Probability calculation for proportion within ±0.05 of the population proportion (0.55).

Solution Steps

  • Define:

    • n = 200, mean p = 0.55, σ = √[(0.55)(0.45)/200] = 0.035.

    • Normality conditions checked (successes and failures > 5).

Probability Transformation

  • Required: P(0.50 < p < 0.60).

    • Calculate Z scores for boundaries and determine distribution probabilities.

    • Result: P(Z between -1.42 and 1.42) = 0.84.

Sample Size Impacts

  • Analyzing n = 400 affects the standard deviation (0.025) and probabilities closer to the mean (µ).

  • Larger sample sizes yield results closer to the population parameter, p.

Practice Exercises

  • Complete exercises from pages 326 to 328 of the textbook.

Chapter 7 Addendum: Sampling Distribution of

Learning Goals (Section 7.5)

  • Explore random variable distribution, Central Limit Theorem (CLT), and probability calculations.

Analysis of Battery Life

  • Random variable X denotes the battery lifetime with mean = 400 hours, σ = 40 hours.

Sampling Process

  • For n samples, calculate mean and repeat for further samples to assess distribution behavior.

Mean and Standard Deviation

  • Characteristics of the sampling distribution of X:

    • Mean: μX = μ.

    • Standard Deviation: σX = σ/√n.

Central Limit Theorem (CLT)

  • Asserts distributions approach normality given sufficiently large n (>30).

Normal Distribution Cases

  • Case distinctions: Analysis varies based on normality of the original data distribution.

Probability Statements

  • Utilize sampling distributions to gauge proximity of sample mean (X) to the population mean (𝜇).

Probabilities Computation Steps

  • Refer back to Chapter 6 for probability computing procedures based on sampling distributions.

Example #2: Professors' Work Week

  • Study focused on weekly work hours with normal distribution: μ = 52, σ = 6 for 3 professors. Result calculated for working hours exceeding 60.

Example #3: Car Insurance Costs

  • Insurance cost example included to exhibit probability calculations using sample size changes.

Key Takeaway

  • Larger sample sizes minimize the deviation of sample results from the population mean.

Summary on Sampling Distributions

  • Essential understanding of sampling distributions to justify inferences in statistics.