Determine the sample size required to estimate the mean score on a standardized test within 44 points of the true mean with 9999% confidence. Assume that sequals=1313 based on earlier studies.
To determine the sample size required, we can use the formula for sample size estimation for a mean:
n = (Z * σ / E)^2
Where:
n = Sample size
Z = Z-score corresponding to the desired confidence level
σ = Standard deviation of the population
E = Maximum error or margin of error
In this case, the desired confidence level is 99.99% or 0.9999, which corresponds to a Z-score of 3.890. The maximum error or margin of error, E, is given as 44 points.
Substituting these values into the formula:
n = (3.890 * 1313 / 44)^2
n = (3.890 * 29972 / 44)^2
n = (116396.8 / 44)^2
n = 2645.38^2
n = 6999688.68
Therefore, the sample size required to estimate the mean score on a standardized test within 44 points of the true mean with 99.99% confidence is approximately 6,999,688.
To determine the sample size required to estimate the mean score on a standardized test within 44 points of the true mean with 99.99% confidence, you can use the formula for sample size calculation:
n = (Z * σ / E)^2
Where:
- n is the sample size.
- Z is the Z-score corresponding to the desired confidence level. For 99.99% confidence, the Z-score is approximately 3.891.
- σ is the standard deviation of the population, given as 1313.
- E is the maximum error or the margin of error. In this case, it is 44.
Plugging in these values into the formula:
n = (3.891 * 1313 / 44)^2
n ≈ (5096.283 / 44)^2
n ≈ 116.095^2
n ≈ 13498.521
Therefore, the sample size required to estimate the mean score on a standardized test within 44 points of the true mean with 99.99% confidence is approximately 13,499.