Number of samples and Statistical Power I know that as size of the sample $N$ increases, the $Z$ value increases.
But as the $Z$ value increases, that means the $p$ value must decrease.
So what happens to $\alpha$ If we had set it to $.05$ and the $p$ value ends up being is lower than $.05$? How does that affect the $\beta$ region? And how does that affect the $1-\beta$ region?
I guess my real question is how does the $Z$ value end up affecting the $1-\beta$ region?
Thank you. 
 A: Suppose you have $n = 4$ observations from a normal population with unknown $\mu$ and $\sigma=5.$
Then test $H_0: \mu = 70$ vs. $H_a: \mu = 76$ so that the difference is $\Delta = 6.$ Test at level $\alpha = .05.$ Then you can solve for power $1 - \beta = 0.775.$
Results from Minitab 17 (other statistical software might have been used):
Power and Sample Size 

1-Sample Z Test

Testing mean = null (versus > null)
Calculating power for mean = null + difference
α = 0.05  Assumed standard deviation = 5

            Sample
Difference    Size     Power
         6       4  0.774919

The plot below shows power values (vertical axis) for various differences $\Delta.$

Now change the problem: Suppose you want to have $\alpha = \beta = 0.05,$ so that power is 95%. Keeping, $\sigma = 5$ and $\Delta = 6$ the same, you can find
the required sample size $n.$
Power and Sample Size 

1-Sample Z Test

Testing mean = null (versus > null)
Calculating power for mean = null + difference
α = 0.05  Assumed standard deviation = 5

            Sample  Target
Difference    Size   Power  Actual Power
         6       8    0.95      0.959877


For given $\sigma$ and $\Delta,$ the quantities $\alpha, \beta,$ and $n$ can
be considered to be in a three-way "contest." We would like $\alpha, \beta,$ and $n$ to be 'small'. If you make any one of the three smaller, then one or both
of the other two must get larger.
In a practical power computation, one has some idea of the value of the population $\sigma$
(perhaps based on earlier studies) and the size of the difference $\Delta$ that one would
like to detect. Then one can choose to test at level $\alpha = 0.05$ and
say what power is hoped for. Then find $n.$  If $n$ is too large to afford,
then perhaps settle for a larger $\beta$ (smaller power) --- or change the nature of the study, settling for
a larger detectable difference $\Delta.$
Addendum: Ott & Longnecker: "Statistical Methods and Data Analysis,"
gives the formula:
$$n = [(z_\alpha + z_\beta)\sigma/\Delta]^2,$$
which is easily derived using standardization. [The z-notation is defined as
$P(Z > z_p) = p,$ where $Z$ is standard normal.]
In the second Minitab output, this amounts to
$n = [(5/6)(1.645+1.645)]^2 \approx 7.52.$ Minitab rounds up to integer $n = 8,$
which gives 'actual power', somewhat above 'target power' 0.95. 
Notes: (1) In some editions of O&L the displayed formula for $n$
has misprints.
(2) If $\sigma$ is unknown, then the appropriate test is a t-test, not a z-test.
Finding the power of a one-sample t-test requires use of a non-central t-distribution.  
