Question

Let the mean of the population be 38 instances of from 6" - 9" hatchings per nest, and let the standard deviation of the mean be 3. what sample mean would have a confidence level of 95% or a 2.5% margin of error?

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Answer 1

To find the sample mean that would have a confidence level of 95% or a 2.5% margin of error, we need to calculate the sample size. The necessary steps involve determining the critical z-value, calculating the margin of error, and solving the equation for the square root of the sample size. In this case, a sample size of 14400 would meet the desired criteria.

Step-by-step explanation:

To calculate the sample mean that would have a confidence level of 95% or a 2.5% margin of error, we need to find the critical z-value corresponding to the confidence level and then use it to calculate the margin of error.

1. Since the confidence level is 95%, we want to find the z-value that leaves 2.5% in each tail. The z-value for a 95% confidence level is approximately 1.96.
2. Next, we can calculate the margin of error using the formula:
   Margin of Error = z-value * (standard deviation / square root of sample size).
3. Substituting the given values, the margin of error is:
   Margin of Error = 1.96 * (3 / square root of sample size).
4. Since the margin of error is 2.5%, we can set up the equation:
   2.5% = 1.96 * (3 / square root of sample size) and solve for the sample size.
5. Simplifying the equation, we get:
   0.025 = 1.96 * (3 / square root of sample size)
6. Dividing both sides by 1.96 and multiplying by the square root of sample size, we get:
   0.025 * square root of sample size = 3
7. Finally, we can solve for the square root of the sample size:
   square root of sample size = 3 / 0.025 = 120
8. Squaring both sides of the equation, we find:
   sample size = 120^2 = 14400

Therefore, a sample size of 14400 would have a confidence level of 95% or a 2.5% margin of error.