Solution
Power of a Ten-Flip Test
Show the problem again
A coin is flipped 10 times. To test $H_0:p=1/2$ against $H_1:p>1/2$, reject $H_0$ when 8 or more heads are observed. If the coin's true probability of heads is $p=3/4$, what is the power of this test? Give a decimal rounded to three places, or the exact fraction.
Worked solution
Power $=\sum_{k=8}^{10}\binom{10}k(3/4)^k(1/4)^{10-k}=\dfrac{45\cdot3^8+10\cdot3^9+3^{10}}{4^{10}}=\dfrac{295245+196830+59049}{1048576}=\dfrac{551124}{1048576}=\dfrac{137781}{262144}\approx0.5256$, which rounds to 0.526. So the type II error is about 0.474: even for a coin this biased the test misses it nearly half the time, while its size under $H_0$ is $56/1024\approx0.055$.
Source: Original problem written for AxiomIQ.