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

Studypool's Notebank makes it easy to buy and sell old notes, study guides, reviews, etc.