find the margin of error for the sample proportion, given a sample size of n=1400. Round to the nearest percent.
Thank you for using the Jiskha Homework Help Forum. Just entitling this HELP may get you no answer. Try the heading of MATH if that is what this is, and a math person is more likely to read it.
Math assistance needed.
To find the margin of error for the sample proportion, you need to use the formula:
Margin of Error = Z * sqrt((p * (1 - p)) / n),
where:
Z is the z-score corresponding to the desired level of confidence,
p is the estimated proportion,
n is the sample size.
Since you have not provided the estimated proportion or the desired level of confidence, I will assume a 95% confidence level, which is commonly used.
For a 95% confidence level, the z-score is approximately 1.96.
If you do not have an estimated proportion, you can assume a value of 0.5, which provides the maximum margin of error.
Plugging these values into the formula, we get:
Margin of Error = 1.96 * sqrt((0.5 * (1 - 0.5)) / 1400).
Calculating this expression, we find:
Margin of Error ≈ 1.96 * sqrt(0.25 / 1400) ≈ 1.96 * sqrt(0.00017857142) ≈ 1.96 * 0.0133630292 ≈ 0.0261444.
To round this to the nearest percent, we multiply by 100 and round to two decimal places:
Margin of Error ≈ 0.0261444 * 100 ≈ 2.61%.
Therefore, the margin of error for the sample proportion, rounded to the nearest percent, is approximately 2.61%.