What is the minimal sample size needed for a 99% confidence interval to have a maximal margin of error of 0.06 if there is no preliminary estimate for p?

To calculate the minimal sample size needed for a 99% confidence interval with a maximal margin of error of 0.06 (assuming no preliminary estimate for p), you will need to use the formula:

n = (Z^2 * 0.25) / E^2

Where:
- n is the required sample size
- Z is the Z-score corresponding to the desired confidence level (99% in this case)
- E is the desired margin of error (0.06 in this case)
- 0.25 is the maximum variance value when there is no preliminary estimate for p

We need to find the Z-score for a 99% confidence level, which corresponds to a two-tailed test. In standard normal distribution, this Z-score is approximately 2.576 (you can look up this value in a Z-table or use a calculator).

Plugging in the values, the formula becomes:

n = (2.576^2 * 0.25) / 0.06^2

Simplifying the equation:

n = (6.650176 * 0.25) / 0.0036
n ≈ 110.28

Therefore, the minimal sample size needed for a 99% confidence interval to have a maximal margin of error of 0.06, without a preliminary estimate for p, is approximately 111.

To determine the minimal sample size needed for a 99% confidence interval with a maximal margin of error of 0.06, we can use the formula for sample size calculation without a preliminary estimate for p.

The formula for sample size calculation without a preliminary estimate is given by:

n = (Z^2 * (1/4)) / (E^2)

Where:
- n represents the required sample size
- Z is the Z-score corresponding to the specified confidence level (99% confidence level corresponds to a Z-score of approximately 2.576)
- E is the desired margin of error (0.06 in this case)

Substituting the given values into the formula:

n = (2.576^2 * (1/4)) / (0.06^2)
n = 6.656 / 0.0036
n ≈ 1852.22

Rounding up to the nearest whole number, the minimal sample size needed is approximately 1853.

Therefore, a sample size of at least 1853 would be required to have a 99% confidence interval with a maximal margin of error of 0.06 when there is no preliminary estimate for p.

301

Formula to find sample size:

n = [(z-value)^2 * p * q]/E^2
... where n = sample size, z-value is found using a z-table for 99% confidence (which is 2.58), p = .5 (when no value is stated in the problem), q = 1 - p, ^2 means squared, * means to multiply, and E = 0.06

Plug values into the formula and calculate n.

I hope this will help get you started.