576. a) From a sample of size $n$, extracted from a normal general population with a known standard deviation $\sigma$, the sample mean $\bar{x}$ is found. At a significance level $\alpha$, it is required: 1) to find the critical region if the null hypothesis $H_{0}$: $a=a_{0}$ about the equality of the general mean $a$ to the hypothetical value $a_{0}$ is tested against the alternative hypothesis $H_{1}: a>a_{0}$; 2) to find the power function of the considered criterion, taking the hypothetical value of the general mean $a=a_{1}\left(a_{1}>a_{0}\right)$ as the argument; 3) to verify that increasing the sample size leads to an increase in the power of the criterion; 4) to verify that increasing the significance level leads to an increase in the power of the criterion.

$\mathrm{Solution.} \mathrm{1)} \mathrm{The} \mathrm{alternative} \mathrm{hypothesis} \mathrm{is} \mathrm{of} \mathrm{the} \mathrm{form} a>a_{0}$, so the critical region is right-sided. Using rule 2, we find the critical point $u_{\text {cr }}$ from the equality $\Phi\left(u_{\text {cr }}\right)=(1-2 \alpha) / 2$. Therefore, the right-sided critical region is defined by the inequality $U>u_{\text {cr }}$, or in more detail $\frac{\bar{x}-a_{0}}{\sigma \sqrt{n}}>u_{\text {cr }}$. Hence,

$$
\bar{x}>u_{\text {cr }}(\sigma / \sqrt{n})+a_{0} .
$$

At these values of the sample mean, the null hypothesis is rejected; in this sense, $\bar{x}=u_{\text {cr }}(\sigma / \sqrt{n})+a_{0}$ can be considered the critical value of the sample mean.

2) To calculate the power of the considered criterion, we first find its value under the condition that the alternative hypothesis is true (i.e., when $a=a_{1}$), setting $\bar{x}=u_{\text {cr }}(\sigma / \sqrt{n})+a_{0}:$

$$
U=\frac{\bar{x}-a_{1}}{\sigma / \sqrt{n}}=\frac{u_{\text {cr }}(\sigma / \sqrt{n})+a_{0}-a_{1}}{\sigma / \sqrt{n}}=u_{\text {cr }}-\frac{a_{1}-a_{0}}{\sigma / \sqrt{n}} .
$$

Thus,

$$
U=u_{\text {cr }}-\lambda, \text { where } \lambda=\left(a_{1}-a_{0}\right) \sqrt{n} / \sigma
$$

When $U>u_{\text {cr }}-\lambda$, the null hypothesis is rejected, so the power of the considered criterion at $a=a_{1}$ is

$$
\begin{aligned}
1-\beta & =P\left(U>u_{\text {cr }}-\lambda\right)=1-P\left(U<u_{\text {cr }}-\lambda\right)=1-\Phi\left(u_{\text {cr }}-\lambda\right) .
\end{aligned}
$$

3) To verify that increasing the sample size leads to an increase in the power of the criterion, we note that as $n$ increases, $\lambda$ increases, and thus $u_{\text {cr }}-\lambda$ decreases. Since $\Phi$ is an increasing function, $1-\Phi\left(u_{\text {cr }}-\lambda\right)$ increases, indicating an increase in the power of the criterion.

4) To verify that increasing the significance level leads to an increase in the power of the criterion, we note that as $\alpha$ increases, $u_{\text {cr }}$ decreases. Thus, $u_{\text {cr }}-\lambda$ decreases, and since $\Phi$ is an increasing function, $1-\Phi\left(u_{\text {cr }}-\lambda\right)$ increases, indicating an increase in the power of the criterion.

$\mathrm{2)} \mathrm{Given} \mathrm{the} \mathrm{values} \mathrm{of} \mathrm{the} \mathrm{parameters} \sigma=1, \alpha=0.05, a_{0}=3, n=16, \mathrm{and} \mathrm{the} \mathrm{alternative} \mathrm{hypothesis} \mathrm{is} \mathrm{of} \mathrm{the} \mathrm{form} a>a_{0}$. It is required: 1) to find the power of the right-sided criterion for testing the considered hypothesis for the hypothetical value of the general mean $a=a_{1}=3.2$; 2) to find the sample size $n_{1}$ for which the power of the criterion is 0.6.
Solution. 1) We use the formula

$$
1-\beta=0.5-\Phi\left(u_{\mathrm{kp}}-\lambda\right) .
$$

By rule 2, we find the critical point of the right-sided critical region $u_{\text {kp}}=1.65$.

We calculate $\lambda$, considering that, according to the condition, $a_{1}=3, a_{0}=2, n=16$, $\sigma=4:$

$$
\lambda=\left(a_{1}-a_{0}\right) \sqrt{ } \bar{n} / \sigma=(3-2) \sqrt{16} / 4=1 .
$$

Substituting $u_{\mathrm{kp}}=1.65$ and $\lambda=1$ into formula (*), we get

$$
1-\beta=0.5-\Phi(1.65-1)=0.5-\Phi(0.65)
$$

From the Laplace function table (see Appendix 2), we find $\Phi(0.65)=$ $=0.2422$. The required power $1-\beta=0.5-0.2422=0.2578$.

2) To find the "new" sample size $n_{1}$, for which the power of the criterion is 0.6, we find the "new" value of the parameter $\lambda$ (denote it as $\lambda_{1}$) from the relation $0.6=0.5-\Phi\left(1.65-\lambda_{1}\right)$. Hence,

$$
\Phi\left(\lambda_{1}-1.65\right)=0.1 \text {. }
$$

From the Laplace function table (see Appendix 2), we find $\lambda_{1}$ $-1.65=0.253$. Therefore, $\lambda_{1}=1.903$.

Considering that $\lambda_{1}=\left(a_{1}-a_{0}\right) \sqrt{n_{1}} / \sigma$, and, according to the condition, $a_{1}=3$, $a_{0}=2, \sigma=4$, we get $1.903=(3-2) \sqrt{n_{1}} / 4$. Hence, the required sample size $n_{1}=58$.

c) From a sample of size $n=9$, drawn from a normal general population with a known standard deviation $\sigma=4$, at a significance level of 0.05, the null hypothesis $H_{0}: a=a_{0}=15$ is tested for the equality of the general mean $a$ to the hypothetical value $a_{0}=15$ under the alternative hypothesis $a>15$. Required: 1) find the power of the right-sided criterion for the hypothetical value of the general mean $a=$ $=a_{1}=17 ; 2$ ) find the sample size $n_{1}$, for which the power of the criterion is 0.8 .