Question 1206941: Calculate the 95% margin of error in estimating a population mean 𝜇 for the following values. (Round your answer to three decimal places.)
n = 7,000, s2 = 64
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! i believe you want to find the margin of error associated with a 95% confidence interval.
you are given tht n = 7000 and s^2 = 64.
n is the sample size.
s^2 is the variance.
the standard deviation is equal to the square root of the variance = sqrt(64) = 8.
the standard error is equal to the standard deviation divided by the square root of the sample size = 8 / sqrt(7000) = .0956182887.
z-score for 95% confidence interval is z = plus or minus 1.96.
the margin of error is equal to (x-m) where x is the critical raw score and m is the mean).
on the high side of the confidence interval, the formula becomes 1.96 = (x-m)/s.
solve for (x-m) to get (x-m) = 1.96 * .0956182887 = .1874118459.
on the low side of the confidence interval, the formula becomes -1.96 = (x-m)/s.
solve for (x-m) to get (x-m) = -1.96 * .0956182887 = -.1874118459.
the margin of error is the same.
it is added to the mean to get the high side of the confidence interval critical raw score (x).
it is subtract from the mean to get the low side of the confidence interval critical raw score(x).
the mean is in the middle to the confidence interval.
in this type of problem, it can be anything as long as the same standard error is used.
for example, if the mean is .5 (randomly chosen), then:
the high side of the confidence interval will be 1.96 = (x - .5) / s which becomes 1.96 = (x - .5) / .0956182887.
solve for x to get b = 1.96 * .095.... + .5 = .6874118459
your margin of error is .6874118459 minus .5 = .1874118459 which is your desired margin of error.
same thing happens on the low side of the confidence interval where the z-score is -1.96.
the standard error of .0956182887 is what we had calculated earlier in this analysis.
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