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.
...idk what formula to use.
how do I find the margin of error for the given values of c,s,and n
c=0.95, s=2.1,n=100
n=159,949
p̂ = 30%
H0: p = p0
Ha: p ≠ p0 or p >p0 or p<p0
C = 0.99, z* = 2.58
m = z* √p̂(1-p̂)/n = 0.0031
p̂ +- m = (0.3869, 0.3931)
To find the margin of error for a 99% confidence interval, you can use the formula:
Margin of Error = Z * (√(p * (1-p) / n))
Where:
- Z is the Z-score that corresponds to the desired confidence level (in this case, 99%). You can find the Z-score value from a standard normal distribution table or use a calculator.
- p is the proportion or percentage of the sample that was reported to have participated in community service or volunteer work (39% or 0.39).
- n is the size of the sample (159,949).
Since the sample size is large (159,949 > 30), we can assume that the distribution of sample proportions approximately follows a normal distribution.
Using a Z-score table or calculator, you can find the Z-score for a 99% confidence level, which is approximately 2.576.
Substituting the values into the formula:
Margin of Error = 2.576 * (√(0.39 * (1-0.39) / 159,949)) ≈ 0.005
Therefore, the margin of error for a 99% confidence level is approximately 0.005 or 0.5%.