Computing Type II Error for a One-Sided Normal Test. 
For a random sample $X_1, X_2, \ldots, X_{49}$ taken from a population having standard deviation $4$, the sample mean is found to be $20.3$.
Test the claim that the mean of the population is not less than $21$ at $1\%$ level of significance. Find the probability of Type II error if the population mean is $19.1$.


So to test for $H_0: \mu\geq 21$ versus alternate hypothesis $H_1: \mu < 21$ we calculate the test statistic $Z = \dfrac{20.3-21}{\frac{4}{\sqrt{49}}}=-1.225$ and since $-z_{0.01} = -2.326<-1.225$, we cannot reject the null hypothesis at 1% level of significance.
I am confused how to do the second part. I know that the Type 2 error will be $P(\text{accept} \ H_0 \mid \mu = 19.1)$. How do I do this?
 A: Intuitively, for the test you have $H_0: \mu \ge 21$ and $H_a: \mu < 21.$
From data you have $\bar X = 20.3,$ which is smaller then $\mu_0 = 21.$
However, the critical value for a test at level 1% is $c = 19.67.$
Because $\bar X > c,$ you find that $\bar X$ is not significantly smaller
than $\mu_0.$
Computation using R: Under $H_0$ we have $\bar X \sim \mathsf{Norm}(21, 4/7);\,P(\bar X \le 19.671) = .01.$
qnorm(.01, 21, 4/7)     # 'qnorm' is normal quantile function (inverse CDF)
## 19.67066             # 1% critical value
pnorm(19.671, 21, 4/7)  # 'pnorm' is normal CDF
## 0.01001595           # verified

Now you wonder, whether a specific alternative value $\mu_a = 19.1 < 21$ might have yielded a value of $\bar X$ small enough to lead to rejection.
The Answer from @spaceisdarkgreen (+1) has done the power computation by
standardizing, so that probabilities can be read from printed normal tables.
If we leave the problem on the original measurement scale, the following
figure illustrates the situation. The blue curve (at right)  is the hypothetical
normal distribution of $\bar X \sim \mathsf{Norm}(\mu_0 = 21, \sigma = 2/7).$
The 1% significance level is the area under this curve to the left of the
vertical line. 
The orange curve is the alternative normal distribution of
$\bar X \sim \mathsf{Norm}(\mu_a =19.1, \sigma = 2/7).$ The area to the
left of the vertical line under this curve represents the power against
alternative $H_a: \mu = \mu_a,$ which is $0.840.$ [The power is $1 - P(\text{Type II Error}).$]

Computation: Under $H_a: \mu_a = 19.1,$ we have $\bar X \sim \mathsf{Norm}(19.1, 4/7).$
pnorm(19.671, 19.1, 4/7)
## 0.8411632               # power against alternative 19.1
1 - pnorm(19.671, 19.1, 4/7)
## 0.1588368               # Type II error probability

Note: Some statistical calculators can be used to find the same normal probabilities I have found using R statistical software.

Addendum: Some textbooks reduce the computations shown by @spaceisdarkgreen
to the following formula for Type II error of a one-sided test at level $\alpha$ against an alternative $\mu_a:$
$$\beta(\mu_a) = P\left(Z \le z_\alpha - \frac{|\mu_0-\mu_a|}{\sigma/\sqrt{n}}  \right).$$
In your case this is $P(Z \le 2.326 - 3.325 = -0.999) = \Phi(-0.999) = 0.1589.$
Ref.: The displayed formula is copied from Sect 5.4 of Ott & Longnecker: Intro. to Statistical Methods and Data Analysis.
A: For the $1\%$ significance level, you have calculated you should reject if $$ \frac{\sqrt{49}(21-\bar X)}{4} > 2.326,$$ so you need to calculate the probability that $$ \frac{\sqrt{49}(21-\bar X)}{4} < 2.326$$ when $X_i\sim N(19.1,16).$ 
Under this assumption, you know that $$ \frac{\sqrt{49}(19.1-\bar X)}{4} \sim N(0,1)$$ so you just need to rearrange: $$ P\left(\frac{\sqrt{49}(21-\bar X)}{4} < 2.326\right)=P\left(\frac{\sqrt{49}(19.1-\bar X)}{4}+\frac{\sqrt{49}(1.9)}{4} < 2.326\right)\\=P\left(\frac{\sqrt{49}(19.1-\bar X)}{4}<-0.999\right) =\Phi(-0.999) = 15.9\%$$
Another way of saying this is that at effect size $\mu = 19.1$ and sample size $n=49,$ the test has $1-15.9\% = 84.1\%$ power.
