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Data Interview Question

Minimizing Margin of Error

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Solution & Explanation

To solve the problem of reducing the margin of error from 3 to 0.3, we need to understand the relationship between the margin of error and the sample size. The margin of error (ME) in a statistical sample is often calculated using the formula:

Margin of Error=Z×σn\text{Margin of Error} = Z \times \frac{\sigma}{\sqrt{n}}

Where:

  • ZZ is the Z-score, which corresponds to the desired confidence level.
  • σ\sigma is the population standard deviation.
  • nn is the sample size.

Given:

  • Initial margin of error ME1=3ME_1 = 3
  • Desired margin of error ME2=0.3ME_2 = 0.3

Objective:

  • Find the additional number of samples, kk, required to achieve the desired margin of error.

Steps to Solution:

  1. Initial Equation Setup:

    • From the initial condition:
      3=Z×σn3 = Z \times \frac{\sigma}{\sqrt{n}}
  2. Desired Condition Setup:

    • For the desired margin of error:
      0.3=Z×σn+k0.3 = Z \times \frac{\sigma}{\sqrt{n+k}}
  3. Express nn in terms of ZZ and σ\sigma:

    • Rearranging the initial equation, we have:
      n=(Z×σ3)2n = \left( \frac{Z \times \sigma}{3} \right)^2
  4. Express n+kn+k in terms of ZZ and σ\sigma:

    • Rearranging the desired equation, we have:
      n+k=(Z×σ0.3)2n+k = \left( \frac{Z \times \sigma}{0.3} \right)^2
  5. Find the relationship between nn and n+kn+k:

    • From the two equations:
      (Z×σ0.3)2=10.09×(Z×σ3)2\left( \frac{Z \times \sigma}{0.3} \right)^2 = \frac{1}{0.09} \times \left( \frac{Z \times \sigma}{3} \right)^2
    • Simplifying, we find:
      n+k=10.09×n=100nn+k = \frac{1}{0.09} \times n = 100n
  6. Calculate kk:

    • Substitute n+k=100nn+k = 100n into the equation:
      k=100nn=99nk = 100n - n = 99n

Therefore, to reduce the margin of error from 3 to 0.3, the sample size needs to be increased by a factor of 100, meaning that 99 additional nn samples are required.

Conclusion

The additional number of samples required to achieve a margin of error of 0.3, given an initial margin of error of 3, is 99 times the original sample size. This significant increase in sample size reflects the inverse square root relationship between sample size and margin of error.