Question

Let $X_1,X_2,X_3,X_4$ be a sample of size 4 from Bernoulli($\theta$), $0<\theta<1$. We test $H_0:\theta=\tfrac14$ vs $H_1:\theta>\tfrac14$, rejecting $H_0$ if $X_1+X_2+X_3+X_4>2$. If $\alpha$ is the Type-I error and $\gamma(\theta)$ the power function, then find $16\alpha+7\gamma\!\left(\tfrac12\right)$ (in integer).

Hint:

For small samples of Bernoulli data, the power is computed from Binomial probabilities under each parameter value.

Answer 1

Correct Answer: 3

Let $Y=X_1+X_2+X_3+X_4\sim\text{Binomial}(4,\theta)$. Reject $H_0$ when $Y>2$, i.e.\ $Y=3$ or $4$. Type-I error: \[ \alpha=P(Y>2\mid\theta=\tfrac14) ={4\choose3}\!\left(\tfrac14\right)^{3}\!\left(\tfrac34\right) +{4\choose4}\!\left(\tfrac14\right)^{4} =4\!\left(\tfrac{1}{64}\right)\!\left(\tfrac34\right)+\tfrac{1}{256} =\tfrac{13}{256}\approx0.0508. \] Power at $\theta=\tfrac12$: \[ \gamma\!\left(\tfrac12\right) =P(Y>2\mid\theta=\tfrac12) ={4\choose3}\!\left(\tfrac12\right)^{3}\!\left(\tfrac12\right) +{4\choose4}\!\left(\tfrac12\right)^{4} =0.3125. \] Hence \[ 16\alpha+7\gamma\!\left(\tfrac12\right) =16(0.0508)+7(0.3125) =0.8128+2.1875=3.0003\approx\boxed{3}. \]