Estimate The Population Proportion Within ±0.05 At 95% Confi

Estimate the population proportion within ±0.05 at 95% confidence level

The estimate of the population proportion should be within plus or minus .05, with a 95% level of confidence. The best estimate of the population proportion is .15. How large a sample is required? (Use z Distribution Table.) (Round z-value to 2 decimal places and round up your answer to the next whole number.)

Sample Paper For Above instruction

To determine the necessary sample size for estimating a population proportion with a specified confidence level and margin of error, we utilize the formula involving the z-score, the estimated proportion, and the desired margin of error. The formula for the sample size (n) when estimating a proportion is:

\( n = \left( \frac{Z_{\alpha/2} \times \sqrt{p(1-p)}}{E} \right)^2 \)

Where:

  • \( Z_{\alpha/2} \) is the z-score corresponding to the desired confidence level (for 95%, \( Z_{0.025} \)),
  • \( p \) is the estimated population proportion,
  • \( E \) is the margin of error specified, in this case, 0.05.

Given \( p = 0.15 \), \( E = 0.05 \), and a 95% confidence level, the critical z-value from the standard normal distribution table is approximately 1.96. Plugging in the numbers:

\( n = \left( \frac{1.96 \times \sqrt{0.15 \times 0.85}}{0.05} \right)^2 \)

Calculate the square root component:

\( \sqrt{0.15 \times 0.85} = \sqrt{0.1275} \approx 0.357 \)

Then multiply by z-score:

\( 1.96 \times 0.357 \approx 0.699 \)

Find n by dividing and squaring:

\( n = \left( \frac{0.699}{0.05} \right)^2 = (13.98)^2 \approx 195.68 \)

Since sample size must be an integer and rounded up to the next whole number, the required sample size is approximately 196.

Therefore, at least 196 observations are needed to estimate the population proportion with a margin of error of ±0.05 at a 95% confidence level.

References

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