Question 362:

Asked on September 1, 2008

Tags: Sample Size , Confidence Intervals , Small Sample , Margin of Error

1

Answer:

No answer provided yet.

We don't know the standard deviation, but can get a crude estimate by dividing the range by 6, since one of the properties of the normal curve is that 99% of values fall within +/- 3 standard deviations of the mean. That estimate of the standard deviation is then (200-100)/6 = 16.67 seconds.

We need a sample size which will generate a margin of error of +/- 5 seconds. The margin of error is calculated by multiplying the standard error of the mean by the critical value of the t or z-distribution. As the sample size increase the values of t and z converge. For this example, I'll just use the z or normal deviates. For a two-sided 95% confidence interval, the normal deviate or z-score is 1.96.

We can then just setup and equation and solve for the unknown:

1. The margin of error is the standard error of the mean SEM * the critical z value of 1.96, so working backwards from a margin of 5, the SEM = 5/1.96 = 2.55.
2. The SEM is 2.55 which is made up of the SD/ SQRT(n). Solving for the unknown sample size n we have the equation 16.67/sqrt(n) = 2.55.
3. Squaring both sides = 277.7/n  = 6.51 
4. Isolating n gets us 6.51n = 277.7 or n = 277.7/6.51 = 42.65
5. Rounding up to the nearest kernel gets us 43.

So a sample size of 43, given our estimate of the standard deviation will generate a +/- 5 second margin of error.