You conduct a survey and find that 422 students out of 813 would vote for Hillary Clinton, if they could vote (same numbers as other problem). What is the margi

Question

You conduct a survey and find that 422 students out of 813 would vote for Hillary Clinton, if they could vote (same numbers as other problem). What is the margin of error for your survey?

Answer 1

Margin of Error = Z_critical* standard error, where Z_critical is Z value needed for confidence interval, usually 95%, and standard error is sqrt[p(1-p)/n] That value of Z is 1.96. p is defined as number that voted for Hillary divided by total 422/813 = 0.519

 

Margin of Error = 1.96*sqrt[0.519(1-0.519)/813] = 0.03435 or 3.4535%

Answer 2

 

Hi Kim,

The margin of error for a binomial variable can be computed using this formula:

MoE = Zα * √p*(1-p)/N

Zα is the z-score for the level of confidence that we want. Since the question doesn't say, I will assume 95% (95% is pretty typical). The z-score for 95% is 1.96 (this can be found in a z-score table)

p is the proportion of students who say they would vote for Hilary. In your question, this equals 422/813 = 0.519

N is the total sample size In your case, this is 813

If we plug everything in, we get:

1.96 * √0.519*(1 - 0.519)/813 = 1.96 * 0.018 = 0.035

This means that the margin of error is .035 (or 3.5%). Again, this assumes that we want 95% confidence. If not, just substitute the z-score for the level of confidence you really want.

Hope that helps.