For an 80% confidence interval, we would have a critical z value of
z_crit = norm.inv(0.9,0,1) = 1.281551566
The margin of error for the confidence interval will be z_crit x SEM
So 2.5 = 1.281551566 x sd/sqrt(n)
2.5 x sqrt(n) = 1.281551566 x sd
sqrt(n) = 1.281551566 x 5/2.5
n = 6.57
So, naively, we could say that the minimum required sample size is 7. In practice, we should take into account that we would be using the t-distribution, which would certainly increase the required sample size to 8-9, but since the degrees of freedom are dependent on n I'm not sure this is really what the example wants you to address (since it gets quite complex, mathematically).
Mar 12th, 2015
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