Please explain how to find the margin of error?

Find the margin of error for a survey that has a given sample size. Round your answers to the nearest tenth of a percent.

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margin of error is the z value times the standard deviation divided by the square root of the sample size. Lol sorry that's in words instead of written formula. But the z value is taken from the amount that is in the tail, so if there is .025 in the left tail, you go to the z table (negative values) and find .025 in the area part. I believe it is -1.96...either that or -2.33....I don't remember, but anyways you take that number and that is your z. But disregard the negative sign because ultimately you are taking the average and adding/subtracting the margin of error.

If your sample size is less than 30 you need to use the t-table. For that, you need to know the df, which is n-1 or the sample size minus one; then take the amount in the tail, for example, .025, and see where they cross. That is your t-value which you then multiply by the same thing you multiplied z in the first time...standard deviation divided by the square root of the sample size.

Hope this helped, Good luck :)

Is this for proportion or mean?

For proportion:

E = z* sqrt(p^ q^ /n)

For mean with sigma:

E = z* sigma/sqrt(n)

For mean without sigma:

E = t* s/sqrt(n)