581. Two instruments measured six parts in the same order and the following measurement results were obtained (in hundredths of a millimeter):

\[
\begin{aligned}
& x_{1}=2, x_{2}=3, x_{3}=5, x_{4}=6, x_{5}=8, x_{6}=10 ; \\
& y_{1}=10, y_{2}=3, y_{3}=6, y_{4}=1, y_{5}=7, y_{6}=4
\end{aligned}
\]

At a significance level of 0.05, determine whether the measurement results differ significantly or not, assuming they are normally distributed.
Solution. Let's find the differences $d_{i}=x_{i}-y_{i}$; subtracting the numbers of the second row from the numbers of the first, we get:

$$
d_{1}=-8, d_{2}=0, d_{3}=-1, d_{4}=5, d_{5}=1, d_{6}=6
$$

Let's find the sample mean, considering that $\sum d_{i}=3$: $\bar{d}=$ $=3 / 6=0.5$.

Let's find the "corrected" standard deviation $s_{d}$, considering that $\sum d_{i}^{2}=127$ and $\sum d_{i}=3$:

$$
s_{d}=\sqrt{\frac{\sum d_{i}^{2}-\left[\sum d_{i}\right]^{2} / n}{n-1}}=\sqrt{\frac{127-9 / 6}{6-1}}=\sqrt{25.1}
$$

Let's find the observed value of the criterion:

$$
T_{\mathrm{na} \cdot .1}=\bar{d} \cdot \sqrt{n} / s_{d}=0.5 \cdot \sqrt{\overline{6 /}} \sqrt{25.1}=0.24
$$

From the table of critical points of the Student's distribution (see Appendix 6), at the significance level 0.05, placed in the top row of the table, and the number of degrees of freedom $k=n-1=6-1=5$, we find the critical point $t_{\text {two-tailed crit }}(0.05 ; 5)=2.57$.

Since $T_{\text {obs. }}<t_{\text {two-tailed crit }}$ - there is no reason to reject the null hypothesis. In other words, the mean measurement results differ insignificantly.