Suppose x has a normal distribution with sigma = 1.2. Find the minimal sample size required so that for a 95% confidence interval, the maximal margin of error E = 0.5.
To find the minimal sample size required for a 95% confidence interval with a maximal margin of error E = 0.5, we can use the formula:
n = (Z * sigma / E)^2
Where:
- n is the sample size
- Z is the Z-score corresponding to the desired confidence level. For a 95% confidence level, Z = 1.96 (based on the standard normal distribution)
- sigma is the standard deviation of the population
- E is the maximal margin of error
Plugging in the values given in the problem:
n = (1.96 * 1.2 / 0.5)^2
Calculating this:
n = (2.352 / 0.5)^2
n = 4.704^2
n ≈ 22.11
Therefore, the minimal sample size required is approximately 22.11. Since you cannot have a fraction of a sample, you should round up to the nearest whole number.
Therefore, the minimal sample size required is 23.
To find the minimal sample size required for a given confidence interval and margin of error, we can use the formula:
n = (Z * σ / E)^2
Where:
- n is the sample size
- Z is the Z-score corresponding to the desired confidence level
- σ is the standard deviation of the population
- E is the maximal margin of error
In this case, we want a 95% confidence interval, which means the Z-score will correspond to a confidence level of 0.95. The Z-score for a 95% confidence level is approximately 1.96.
Plugging in the values:
n = (1.96 * 1.2 / 0.5)^2
n = (2.352 / 0.5)^2
n = 4.704^2
n ≈ 22
Therefore, the minimal sample size required to achieve a 95% confidence interval with a maximal margin of error of 0.5, assuming a normal distribution with a standard deviation of 1.2, is approximately 22.
n = (z times sigma divided by E)^2
n = (1.96)(1.2)/.05 = 15.68 now square 245.8 or 246