Hypothesis Testing a small sample for the binomial parameter p The following is a question from a homework set that I truly do not understand how to even begin.
The following is a Minitab printout of the binomial pdf $p_{x}(k) {9 \choose k}(0.6)^k(0.4)^{9-k}$, $k=0,1,...,9$. Supposed $H_{0}:p=0.6$ is to be tested against $H_{1}:p>0.6$ and we wish the level of significance to be exactly 0.05. Combine two different critical regions into a single randomized decision rule for which $\alpha=0.05$. 
MTB>pdf;
SUBC> binomial $9$ $0.6$.
Probability Density Function
Binomial with $n=9$ and $p=0.6$
The data:
x; P(X=x)
0; 0.000262
1; 0.003539
2; 0.021234
3; 0.074318  
4; 0.167215
5; 0.250823
6; 0.250823
7; 0.161243
8; 0.060466
9; 0.010078
 A: If the null hypothesis $p=0.6$ holds, then the probability that our random variable is $\ge 8$ is approximately $0.070544$ (we added the last two terms). So if we reject when the number of successes is $\ge 8$, the probability of Type $1$ error is too big.   
On the same hypothesis, the probability that our random variable is $9$ is approximately $0.010078$. Rejecting only for number of successes $9$ leads to a Type $1$ error that is "too small."  Now one might think small probability of error is good. But if we make the probability of Type $1$ error small, we almost always increase the probability of Type $2$ error (not rejecting the null hypothesis when it is false).  
Let's employ a mixed strategy, using a "coin flip." So with probability $p$ we reject $H_0$ if the number of successes is $8$. With probability $1-p$, we only reject $H_0$ if there are $9$ successes. It remains to find $p$ so that the probability of Type $1$ error is $0.05$. 
If we use the mixed strategy, then the probability of Type $1$ error is approximately
$$(p)(0.060466)+(1-p)(0.010078).$$
Set the above expression equal to $0.05$, and solve the resulting linear equation for $p$.  
Remark: The numerical computations were a good beginning, though we ended up using only the last two numbers in your table. It would have been more efficient to calculate the probability of $9$, $8$, and so on until the sum was $\ge 0.05$. In our case there is no "and so on."
