## Question 512:

1

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The easiest way I think to approach this is to use insert some values and solve for the unknown sample size. Let's pick a standard deviation of 80, so we want to know what sample size we need to have a 95% margin of error of +/- 10.

1. For a 2-sided 95% confidence interval around the mean, we'd use the critical value from the normal distribution of 1.96, which we can find in a z-table or using the percentile to z-score calculator and selecting 2-sided area.
2. The margin of error is equal to the standard error of the mean (SEM) times the critical z value.
3. We have SEM*1.96 = 10
4. The SEM is made up of the standard deviation divided by the square root of the sample size. Which would be 80/SQRT(n)
5. Substituting that into our equation we have [ 80/SQRT(n) ]*1.96 = 10
6. Dividing both sides of the equation by 1.96  =  80/SQRT(n) = 5.10
7. Square both sides = 6400/n = 26.03
8. Solving for n = 26.03n = 6400
9. So n = 245.86

We'd round up to the nearest person and say a sample size of 246 would be needed to have a margin of error equal to 1/8th the standard deviation.

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