588. A batch of products is accepted if the probability that a product will be defective does not exceed 0.02. Among 480 randomly selected products, 12 were found to be defective. Can the batch be accepted?
Solution. The null hypothesis $H_{0}$ is $p=p_{0}=0.02$. Let's find the relative frequency of defects:

$$
m / n=12 / 480=0.025
$$

We will take the alternative hypothesis $H_{1}: p>0.02$ and the significance level $\alpha=0.05$.

Let's find the observed value of the criterion:

$$
U_{\text {obs}}=\frac{\left(m / n-p_{0}\right) \cdot \sqrt{n}}{\sqrt{p_{0} q_{0}}}=\frac{(0.025-0.02) \cdot \sqrt{480}}{\sqrt{0.02 \cdot 0.98}}=0.78
$$

Given that the alternative hypothesis is $p>p_{0}$, the critical region is right-tailed. Let's find the critical point $u_{\text {cr }}$ of the right-tailed critical region from the equation (rule 2)

$$
\Phi\left(u_{\text {cr}}\right)=(1-2 \cdot 0.05) / 2=0.45
$$

From the Laplace function table (see Appendix 2), we find

$$
u_{\text {cr}}=1.645
$$

Since $U_{\text {obs}}<u_{\text {cr}}$-there is no reason to reject the hypothesis that the probability of defects in the batch does not exceed 0.02. Therefore, the batch can be accepted.