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Data Interview Question

Estimating Confidence Range

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Solution & Explanation

When faced with the problem of estimating a confidence interval for the proportion of defective items in a batch, we can utilize the properties of the binomial distribution. Below is a step-by-step solution and explanation for calculating the confidence interval in this scenario:

Step 1: Understand the Problem

  • Given:
    • Total items (n): 100
    • Defective items: 25
  • Objective: Calculate the 95% confidence interval for the proportion of defective items.

Step 2: Identify the Proportion

  • Sample Proportion (p^\hat{p}): p^=Number of Defective ItemsTotal Items=25100=0.25\hat{p} = \frac{\text{Number of Defective Items}}{\text{Total Items}} = \frac{25}{100} = 0.25

Step 3: Calculate the Standard Error (SE)

The standard error for a proportion is calculated using the formula:

  • Standard Error (SE): SE=p^(1p^)n=0.25×(10.25)100=0.25×0.75100=0.0433SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.25 \times (1 - 0.25)}{100}} = \sqrt{\frac{0.25 \times 0.75}{100}} = 0.0433

Step 4: Determine the Z-Score for the Desired Confidence Level

  • 95% Confidence Level:
    • The Z-score for a 95% confidence level is approximately 1.96.

Step 5: Calculate the Margin of Error (ME)

  • Margin of Error (ME): ME=Z×SE=1.96×0.0433=0.0848ME = Z \times SE = 1.96 \times 0.0433 = 0.0848

Step 6: Calculate the Confidence Interval (CI)

  • Confidence Interval (CI): CI=p^±ME=0.25±0.0848CI = \hat{p} \pm ME = 0.25 \pm 0.0848
    • Lower Bound: 0.250.0848=0.16520.25 - 0.0848 = 0.1652
    • Upper Bound: 0.25+0.0848=0.33480.25 + 0.0848 = 0.3348

Final Result

  • 95% Confidence Interval for the Proportion of Defective Items: CI=[0.1652,0.3348]\text{CI} = [0.1652, 0.3348]

Explanation

  • Interpretation:
    • We are 95% confident that the true proportion of defective items in the entire population lies between 16.52% and 33.48%.
  • Assumptions:
    • The sample is randomly selected.
    • The sample size is large enough to assume a normal approximation of the binomial distribution (np^\hat{p} and n(1-p^\hat{p} are both greater than 5).
  • Importance:
    • Confidence intervals provide a range of plausible values for the population parameter, offering a more comprehensive insight than a single point estimate.