in a sample of 159,949 first year college students, the national survey of student engagement reported that 39% participated in community service or volunteer work.find the margin of error for 99% confidence.

what formula do I use?
I know the 99% interval for z is 2.576
n is 159949..what would I do with 39%?

To find the margin of error for a 99% confidence level, you can use the formula:

Margin of Error = Z * (sqrt(p * (1-p) / n))

Where:
- Z is the z-value associated with the desired confidence level (in this case, for 99% confidence, Z = 2.576).
- p is the proportion of the sample that participated in community service or volunteer work (39% in this case, which is equivalent to 0.39).
- n is the sample size (159,949 in this case).

Now, substitute the values into the formula:

Margin of Error = 2.576 * sqrt(0.39 * (1 - 0.39) / 159,949)

Simplifying further:

Margin of Error = 2.576 * sqrt(0.39 * 0.61 / 159,949)

Margin of Error ≈ 0.012

So, the margin of error for a 99% confidence level is approximately 0.012.

To find the margin of error for 99% confidence, we can use the formula:

Margin of Error = z * (sqrt(p * (1-p) / n))

where:

z: the z-value for the desired confidence level (99% in this case)
p: the proportion of the sample that participated in community service (39% or 0.39)
n: the sample size (159,949 in this case)

So, plugging in the values:

Margin of Error = 2.576 * (sqrt(0.39 * (1-0.39) / 159,949))

First, we need to square the proportion of students who participated in community service and multiply it by the proportion who did not participate. This gives us the value inside the square root. Then, we divide it by the sample size and take the square root of the entire expression. Finally, we multiply it by the z-value to get the margin of error.