Sample size for a difference of means I have the following homework question which i do not know how to compute, any help is appreciated (i have know how to compute given 2 p's, but never a difference in means):
Samsung electronics has developed a new business process that they think can decrease customer complaints. Ms. Ko is required to conduct a randomized trial to determine its efficacy of the new business process. She has decided that she will sample transactions from the new business process (Group 1) and the old business process (Group 2) for analysis. The new business process will be deemed effective in reducing customer complaints if the difference between mean number of complaints is 28. From past experience you know that data will be normally distributed with standard deviation of 20. Ms. Ko has requested that you determine the appropriate sample sizes to reduce the likelihood of Type I and Type II errors. You are given the following values for:
Alpha (α) = .05
Beta (β) = .80

This a poorly worded question, but you have all of the information you need to solve this problem.
Just for a quick estimate, a difference of 28 in the means with a standard deviation of 20 has a z score of 1.4
From basic stats of normal distribution, the comparison of two mean is:
$Z = \frac{(\bar X - \mu)} { \frac{\sigma}{\sqrt(n)}}$
$(\bar X - \mu)$ is 28 - the number of reduction.
$\sigma$ is 20 given above as the standard deviation.
$Z$ is is obtained from the standard tables associated with $\alpha = 0.05$, you will need to decide on either one side or two sided Z value.
Now solve for $n$
The Type 2 calculation is performed the same way but with using the defined value for $\beta = 0.80$
Good luck.
