Math Problem Statement

Suppose there is a population with 4 customers. The interested categorical variable is the proportion of customers that prefer your brand. In this population, one of them prefers your brand, and the others do not. 1. Find the population proportion (π) of customers that prefer your brand. 2. In the process of developing a sampling distribution, all possible samples (taken with replacement) of size n = 2 are obtained. The sample proportion (p) of customers that prefer your brand is considered as the sample statistic. - (ii) Develop the probability distribution of the sample proportion (p). - (iii) Show that the sample proportion p is an unbiased estimator for π (i.e., E(p) = π). - (iv) Does the sampling distribution of p follow a Normal Distribution? Explain.

Solution

Let's break this problem into parts and solve each:


(i) Find the population proportion (π\pi) of customers that prefer your brand.

In the population, there are 4 customers, and only 1 customer prefers your brand. Therefore, the population proportion π\pi is:

π=Number of customers preferring your brandTotal number of customers=14=0.25\pi = \frac{\text{Number of customers preferring your brand}}{\text{Total number of customers}} = \frac{1}{4} = 0.25

So, π=0.25\pi = 0.25.


(ii) Develop the probability distribution of the sample proportion (pp).

To create the sampling distribution:

  1. Population: {Prefers brand (PP), Does not prefer brand (DD)} = {1 PP, 3 DD}.

  2. Sample size: n=2n = 2, with replacement.

  3. All possible samples:

    • PPPP
    • PDPD
    • DPDP
    • DDDD

    Because sampling is with replacement, PDPD and DPDP are considered distinct.

  4. Sample Proportion (pp):

    • For PPPP: Both customers prefer the brand → p=22=1p = \frac{2}{2} = 1.
    • For PDPD or DPDP: One customer prefers the brand → p=12=0.5p = \frac{1}{2} = 0.5.
    • For DDDD: No customer prefers the brand → p=02=0p = \frac{0}{2} = 0.
  5. Probability distribution:

    • p=1p = 1: Only one sample (PPPP) → Probability = 116\frac{1}{16}.
    • p=0.5p = 0.5: Two samples (PD,DPPD, DP) → Probability = (\frac{4*Pd()).

+###Generate plotsd get latter. pd import matplotlib.pyplot as plt

Define the sample proportions and their probabilities

sample_proportions = [0, 0.5, 1] probabilities = [9/16, 6/16, 1/16]

Create a DataFrame for better visualization

data = pd.DataFrame({ 'Sample Proportion (p)': sample_proportions, 'Probability': probabilities })

Display the table of the probability distribution to the user

import ace_tools as tools; tools.display_dataframe_to_user(name="Probability Distribution of Sample Proportions", dataframe=data)

Plot the probability distribution

plt.bar(sample_proportions, probabilities, tick_label=sample_proportions) plt.title("Probability Distribution of Sample Proportion (p)") plt.xlabel("Sample Proportion

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Math Problem Analysis

Mathematical Concepts

Sampling Distribution
Population Proportion
Unbiased Estimator
Normal Distribution

Formulas

Population Proportion π = Number of successes / Total population
Expected value of p: E(p) = π

Theorems

Properties of unbiased estimators
Central Limit Theorem (if relevant for normality assessment)

Suitable Grade Level

Undergraduate (Statistics or Probability course)