Data are sampled from a population. If the population distribution is Normal with mean 56 and standard deviation 21 calculate
a) how large the sample size n needs to be so that there will be at least a 90% chance that its mean is greater than 52.
b) for the sample size n found in a), what the chance is that the sample mean is between 50 and 51.
This is what is known as a "sample size calculation" and is used in clinical trials to determine how much data is needed to proove a practical improvement in patient outcome above the existing standard treatment.
The population distribution is N(56,21^2).
The distribution of the sample mean will be N(56,21^2/n)
where n is the size of the sample.
a) We want to choose the sample size so that at least 90% of the distribution of the sample mean is expected to be above 52.
The 10th percentile of a standard Normal distribution is (using lookup tables)
Converting this to the 10th percentile on our distribution of the sample mean we get
p(n) = -1.282 x 21/sqrt(n) + 56
We want this value to equal 52 at least so that at least 90% of the distribution is above 52.
Therefore we want n such that
-1.282 x 21/sqrt(n) + 56 >= 52
sqrt(n) >= (-1.282 x 21)/(-4) = 6.728
Therefore n >= 45.27
So n must be at least 46.
b) Given this value of n, the chance that the sample mean will be less than 50 is
`Phi((50-56)/(21/sqrt(46))) = Phi(-1.938) = 0.02632` x 100%
where `Phi` is the distribution function of the standard Normal distribution.
The chance that the sample mean will be less than 51 is
`Phi((51-56)/(21/sqrt(46))) = Phi(-1.615) = 0.05317` x 100%
Therefore the chance that the sample mean is between 50 and 51 is
5.317 - 2.632 = 2.69%
a) n needs to be at least 46
b) the chance is 2.69%