The question asks for the size of the test.
- The size of the test is the probability of making a Type I error ( $\alpha$ ).
- A Type I error is rejecting $H_0$ when $H_0$ is true.
- The critical region (rejection region) is given as $C=\{\bar{X} \leq 4.75\}$.
- Therefore, $\alpha=P$ (Reject $H_0 \mid H_0$ is true).
- $\alpha=P(\bar{X} \leq 4.75 \mid \mu \geq 5)$.
To calculate the size, we use the "worst-case" scenario, which is the value of $\mu$ at the boundary of $H_0$.
- We must calculate: $\alpha=P(\bar{X} \leq 4.75 \mid \mu=5)$.
The sample $X_i$ is from a uniform distribution $U[0, \theta]$.
- Population Mean $(\mu): E[X]=\frac{0+\theta}{2}=\frac{\theta}{2}$
- Population Variance $\left(\sigma^2\right): \operatorname{Var}(X)=\frac{(\theta-0)^2}{12}=\frac{\theta^2}{12}$
We are calculating the size under the assumption $\mu=5$.
- If $\mu=5$, then $\frac{\theta}{2}=5 \Longrightarrow \theta=10$.
- Using $\theta=10$, the population variance is: $\sigma^2=\frac{10^2}{12}=\frac{100}{12}=\frac{25}{3}$.
The CLT states that for a large sample ( $n=300$ ), the sample mean $\bar{X}$ is approximately normally distributed: $\bar{X} \approx N\left(\mu_{\bar{X}}, \sigma_{\bar{X}}^2\right)$
- Mean of $\bar{X}\left(\mu_{\dot{X}}\right)$ : This is the population mean, which we assume to be $\mu=5$.
- Variance of $\bar{X}\left(\sigma_{\bar{X}}^2\right)$ : This is the population variance divided by n. $\sigma_{\bar{X}}^2=\frac{\sigma^2}{n}=\frac{25 / 3}{300}= \frac{25}{900}=\frac{1}{36}$
- Standard Deviation of $\bar{X}\left(\sigma_{\bar{X}}\right): \sigma_{\bar{X}}=\sqrt{\frac{1}{36}}=\frac{1}{6}$
So, under $H_0$, the distribution of the sample mean is $\bar{X} \approx N(\mu=5, \sigma=1 / 6)$.
