Normal approximation to binomial can be applied when np 5 and nq 5
p = 0.15
If np = 5
n x 0.15 = 5
n = 33.33
q = 1 - p = 0.85
n x 0.85 = 5
n = 5.88
Take the larger of n values and round up
Sample size required = 34
When p = 0.85
np = 5
n = 5.88
q = 1 - 0.15
nq = 5
q = 33.33
Sample size required = 34
Because the probability values 0.15 and 0.85 are complement of each other, when one is p, the other one is q. Since both np and nq must be greater than 5, the same n works for both of the previous problems.
Get Answers For Free
Most questions answered within 1 hours.