Math Problem Statement

A simple random sample of size n = 1380 is obtained from a population whose size is N = 1,100,000 and whose population proportion with a specified characteristic is p = 0.54. Describe the sampling distribution of p̂. Round the standard deviation of the sampling distribution to three decimal places.

Solution

The question you provided asks for the standard deviation of the sampling distribution of the sample proportion (p^\hat{p}), given the following information:

  • Population proportion: p=0.54p = 0.54
  • Sample size: n=1380n = 1380
  • Population size: N=1,100,000N = 1,100,000

The standard deviation of the sampling distribution for p^\hat{p} can be calculated using the formula for the standard error of the sample proportion:

σp^=p(1p)n×NnN1\sigma_{\hat{p}} = \sqrt{\frac{p(1 - p)}{n} \times \frac{N - n}{N - 1}}

Where:

  • pp is the population proportion.
  • nn is the sample size.
  • NN is the population size.

Let's calculate this step-by-step:

  1. Calculate p(1p)p(1 - p): p(1p)=0.54×(10.54)=0.54×0.46=0.2484p(1 - p) = 0.54 \times (1 - 0.54) = 0.54 \times 0.46 = 0.2484

  2. Divide by the sample size nn: p(1p)n=0.248413800.000180\frac{p(1 - p)}{n} = \frac{0.2484}{1380} \approx 0.000180

  3. Calculate the finite population correction factor: NnN1=1,100,00013801,100,00010.9987\frac{N - n}{N - 1} = \frac{1,100,000 - 1380}{1,100,000 - 1} \approx 0.9987

  4. Multiply the results together: σp^=0.000180×0.99870.00017970.0134\sigma_{\hat{p}} = \sqrt{0.000180 \times 0.9987} \approx \sqrt{0.0001797} \approx 0.0134

So the standard deviation of the sampling distribution is approximately 0.013, rounded to three decimal places.

Now, let's analyze the choices:

  • A. Approximately normal; μp^=0.54\mu_{\hat{p}} = 0.54, σp^=0.150\sigma_{\hat{p}} = 0.150
  • B. Exactly normal; μp^=0.54\mu_{\hat{p}} = 0.54, σp^=0.013\sigma_{\hat{p}} = 0.013
  • C. Approximately normal; μp^=0.54\mu_{\hat{p}} = 0.54, σp^=0.013\sigma_{\hat{p}} = 0.013
  • D. Exactly normal; μp^=0.54\mu_{\hat{p}} = 0.54, σp^=0.150\sigma_{\hat{p}} = 0.150

Given that the sample size is large and n/Nn/N is small, the sampling distribution of p^\hat{p} is approximately normal. Therefore, the correct answer is C: "Approximately normal; μp^=0.54\mu_{\hat{p}} = 0.54, σp^=0.013\sigma_{\hat{p}} = 0.013."

Would you like more details, or do you have any questions?

Here are 5 related questions to explore further:

  1. What is the Central Limit Theorem, and why does it apply in this scenario?
  2. How does the finite population correction factor impact the standard deviation?
  3. What conditions must be met for the sampling distribution of p^\hat{p} to be approximately normal?
  4. How would the answer change if the population size were infinite?
  5. What is the difference between "exactly normal" and "approximately normal"?

Tip: Always check if a finite population correction factor is needed when sampling from a finite population!

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Math Problem Analysis

Mathematical Concepts

Statistics
Sampling Distributions
Normal Distribution

Formulas

σ_p̂ = sqrt[ (p(1 - p) / n) × ((N - n) / (N - 1)) ]

Theorems

Central Limit Theorem

Suitable Grade Level

College/University Level