In an election poll, 56% of voters chose a certain candidate. find the margin of sampling error.
So far I have: ME=2sqrt0.56(1-0.56)/400 but I don't know how to finish the problem.
Any help would be great!
ME = 2 sqrt(.56(1-.56)/400)
= 2 sqrt(.56(.44)/400)
= 2 sqrt(.2464/400)
= 2 sqrt(0.000616)
= 2 x .0248
= 0.0496
or
4.96 percent margin of sampling error
in an election poll,56% of 400 voters choose a certain candidate, find the margin of sampling error.
To find the margin of sampling error, you need to use the formula you provided. Let's break it down step by step:
Margin of Sampling Error (ME) = 2 * sqrt(p * (1 - p) / n)
In your case, p represents the proportion of voters who chose a certain candidate, which is 56% or 0.56 in decimal form. n represents the sample size, which is not given in the question.
ME = 2 * sqrt(0.56 * (1 - 0.56) / n)
To complete the problem, you need to determine the sample size (n). The sample size is the number of voters included in the election poll. Once you know the sample size, plug it into the formula to calculate the margin of sampling error.
For example, let's assume the sample size is 400, as you've shown in your incomplete formula. Plug in this value:
ME = 2 * sqrt(0.56 * (1 - 0.56) / 400)
Now you can finish the calculation:
ME = 2 * sqrt(0.56 * 0.44 / 400)
ME = 2 * sqrt(0.2464 / 400)
ME = 2 * sqrt(0.000616)
ME ≈ 2 * 0.0248
ME ≈ 0.0496 or 4.96% (rounded to two decimal places)
Therefore, the margin of sampling error is approximately 4.96%, assuming a sample size of 400.