1. Problem statement: How many signals (sample size $n$) are needed so that a level 0.05 test of $H_0: \mu=10$ has at least 80% chance of rejection when the true mean is 11.2? 2. Assumptions: We assume a z-test with known population standard deviation $\sigma$ and a two-sided alternative by default. 3. Parameters: The significance level is $\alpha=0.05$ so $z_{1-\alpha/2}=z_{0.975}\approx 1.96$. 4. Power specification: Desired power is 0.80 so $1-\beta=0.80$ and $z_{1-\beta}=z_{0.8}\approx 0.8416$. 5. Effect size: The true difference is $\delta=11.2-10=1.2$. 6. Sample size formula: For a two-sided z-test the required sample size is $$n=\left(\frac{(z_{1-\alpha/2}+z_{1-\beta})\sigma}{\delta}\right)^2$$ 7. Substitute numerical z-values and simplify: Substitute $z_{1-\alpha/2}\approx 1.96$, $z_{1-\beta}\approx 0.8416$, and $\delta=1.2$ to obtain $$n=\left(\frac{(1.96+0.8416)\sigma}{1.2}\right)^2=\left(\frac{2.8016\sigma}{1.2}\right)^2=\left(2.334666\sigma\right)^2\approx 5.449\sigma^2$$ 8. Interpretation: The required sample size is $n\approx 5.449\sigma^2$, and you should take the ceiling (next integer) because $n$ must be an integer. 9. Examples: If $\sigma=1$ then $n\approx 6$. If $\sigma=2$ then $n\approx 22$. If $\sigma=3$ then $n\approx 50$. 10. Final answer: $$n=\left(\frac{(1.96+0.8416)\sigma}{1.2}\right)^2\approx 5.449\sigma^2$$ Plug in the known value of $\sigma$ and round up to get the required number of signals.