Hypothesis Testing: Probability $\beta$ of Type II Error in Proportions $\newcommand{\szdp}[1]{\!\left(#1\right)}$
Background Question: A manufacturer claimed that at least $20\%$ of the public
preferred his product. A sample of $100$ persons is taken to check his claim.
With $\alpha=0.05,$ how small would the sample percentage have to be before
the claim could legitimately be refuted? (Notice that this would involve a
one-tailed test of the hypothesis.) Note: This is Exercise 10.18 in Mathematical Statistics with Applications, 5th Ed., by Wackerly, Mendenhall, and Sheaffer.
Answer to Background Question: We would have $H_0: p=0.2, H_a: p<0.2.$ At the $\alpha=0.05$ level, the
rejection region would be $z<-1.645.$ This would require:
\begin{align*}
z&<-1.645\\
\frac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}&<-1.645\\
\hat{p}-p_0&<-1.645\sqrt{\dfrac{p_0(1-p_0)}{n}}\\
\hat{p}&<p_0-1.645\sqrt{\dfrac{p_0(1-p_0)}{n}}\\
\hat{p}&<0.2-1.645\sqrt{\dfrac{0.2(0.8)}{100}}\\
\hat{p}&<0.1342.
\end{align*}
Problem Statement: Refer to the Background Question. Calculate the value of $\beta$ for the alternative $p_a=0.15.$ Note: This is Exercise 10.27 in Mathematical Statistics with Applications, 5th. Ed., by Wackerly, Mendenhall, and Sheaffer.
My Work So Far: In the background question, we had $p_0=0.2,\; n=100.$ We found that the rejection
region was $z<-2.575,$ corresponding to $\hat{p}<0.097.$ We have that
\begin{align*}
\beta
&=P\szdp{\frac{\hat{p}-p_a}{\sqrt{p_0(1-p_0)/n}}\ge 
         \frac{p_0-p_a}{\sqrt{p_0(1-p_0)/n}}}\\
&=P(z\ge 1.25)\\
&=0.1056.
\end{align*}
My Question: The book does not have an example of calculating $\beta$ for a proportion, so I'm trying to go by analogy, and it's not working. The book's answer is $0.67.$ Where am I going wrong?

I think I was plugging numbers in wrong, even before I corrected the $0.097$ to $0.1342.$ It should be
\begin{align*}
\beta
&=P\!\left(\frac{\hat{p}-p_a}{\sqrt{p_a(1-p_a)/n}}\ge 
         \frac{0.1342-0.15}{\sqrt{0.15(0.85)/100}}\right)\\
&=P(z\ge -0.448)\\
&=0.67
\end{align*}
