558. Two methods were used to measure the same physical quantity. The following results were obtained:

a) in the first case $x_{1}=9.6 ; \quad x_{2}=10.0 ; x_{3}=9.8$; $x_{4}=10.2 ; x_{5}=10.6$

b) in the second case $y_{1}=10.4 ; \quad y_{2}=9.7 ; \quad y_{3}=10.0$; $y_{4}=10.3$.

Can we consider that both methods provide the same measurement accuracy if we accept the significance level $\alpha=0.1$? It is assumed that the measurement results are normally distributed and the samples are independent.
Solution. We will judge the accuracy of the methods by the magnitudes of the variances. Thus, the null hypothesis is $H_{0}: D(X) = D(Y)$. As a competing hypothesis, we will take $H_{1}: D(X) \neq D(Y)$.

Let's find the sample variances. For simplification of calculations, we will transition to conditional variants:

$$
u_{i}=10 x_{i}-100, \quad v_{i}=10 y_{i}-100
$$

As a result, we get the conditional variants

$$
\begin{array}{rrrrrr}
u_{i} & -4 & 0 & -2 & 2 & 6 \\
v_{i} & 4 & -3 & 0 & 3 &
\end{array}
$$

Let's find the corrected sample variances:

$$
\begin{gathered}
s_{u}^{2}=\frac{\sum u_{i}^{2}-\left|\sum u_{i}\right|^{2} / n_{1}}{n_{1}-1}=\frac{(16+4+4+36)-2^{2} / 5}{5-1}=14.8 \\
s_{v}^{2}=\frac{\sum v_{i}^{2}-\left[\sum v_{i}\right]^{2} / n_{2}}{n_{2}-1}=\frac{(16+9+9)-4^{2} / 4}{4-1}=10
\end{gathered}
$$

Let's compare the variances. Find the ratio of the larger corrected variance to the smaller one (each of the variances increased by a factor of $10^{2}$, but their ratio did not change):

$$
F_{\text {obs}}=s_{X}^{2} / s_{Y}^{2}=s_{u}^{2} / s_{v}^{2}=14.8 / 10=1.48
$$

According to the condition, the competing hypothesis is $D(X) \neq D(Y)$, so the critical region is two-sided and in accordance with rule 2, when finding the critical point, we should take a significance level that is half of the given one.

From the table in Appendix 7, at the significance level $\alpha / 2=0.1 / 2=0.05$ and degrees of freedom $k_{1}=n_{1}-1=5-1=4$ and $k_{2}=n_{2}-1=4-1=3$, we find the critical point $F_{\text {crit}}(0.05 ; 4 ; 3)=9.12$.

Since $F_{\text {obs}} < F_{\text {crit}}$ - there is no reason to reject the null hypothesis about the equality of the population variances. In other words, the corrected variances differ insignificantly, and therefore both methods provide the same measurement accuracy.