91Ó°ÊÓ

$\mathrm{Degrees}\mathrm{of}\mathrm{freedom}=\frac{({\mathrm{s}}_1^2/{\mathrm{n}}_1+{\mathrm{s}}_2^2/{\mathrm{n}}_2{)}^2}{{\displaystyle \frac{{\left({\displaystyle \frac{({\mathrm{s}}_1^2}{{{\mathrm{n}}_1}^2}}\right)}^2}{({\mathrm{n}}_1-1)}}+{\displaystyle \frac{\left({\displaystyle \frac{{{\mathrm{s}}_2}^2}{{{\mathrm{n}}_2}^2}}\right)}{({\mathrm{n}}_2-1)}}}$

From Table 4-5 of the Rayleigh's experiment, we know the following things:

From the air:

The average mass of nitrogen is $\stackrel{¯}{\left(x_1\right)}=2.31010\mathrm{g}.$

The standard deviation of $s_1=0.00014.$

The number of measurements $n_1=7.$

From the chemical source:

The average mass of nitrogen is role="math" localid="1663578097442" $\stackrel{¯}{\left(x_2\right)}=2.29947\mathrm{g}.$

The standard deviation of ${\mathrm{s}}_2=0.00137.$

The number of measurements $n_1=7.$

The differences between the two mean values are half as great as Rayleigh found in Rayleigh experiment.

The standard deviations remain unchanged.

To check whether the two given masses of $r$are different, calculate the difference between the two mean values of Rayleigh's experiment as shown below.

$$\begin{gathered} \left|\stackrel{¯}{X}-\stackrel{¯}{X_2}\right|=\mathrm{I}2310109-2299472\mathrm{I} \\ =0.010637 \end{gathered}$$

The differences between the two mean values are half as great as found in Rayleigh experiment.

Thus, $|\stackrel{¯}{\left(x_1\right)}-\stackrel{¯}{\left(x_2\right)}|=0.01067/2⇒0.0053185.$

From the test results, we know that the standard deviations are significantly different.

Hence, the $t_{\mathrm{calculated}}$is found as follows:

role="math" localid="1663579110570" $$\begin{gathered} t_{\mathrm{calculated}}=\frac{\left|\stackrel{¯}{X}-\stackrel{¯}{X_2}\right|}{\sqrt{({{\mathrm{s}}_1}^2/{\mathrm{n}}_1+}{{\mathrm{s}}_2}^2/{\mathrm{n}}_2)} \\ =\frac{\left|0.0053185\right|}{\sqrt{\left(0.00000000002556125+0.000000237705125\right)}} \\ =\frac{0.0053185}{\sqrt{\left(0.00000024026125\right)}} \\ =10.8433 \end{gathered}$$-

$=10.8 (\mathrm{Rounded}\mathrm{to}\mathrm{the}\mathrm{correct}\mathrm{significant}\mathrm{figure}).$

The degrees of freedom are:

$$\begin{gathered} D\mathrm{egrees}\mathrm{of}\mathrm{freedom}=\frac{({\mathrm{s}}_1^2/{\mathrm{n}}_1+{\mathrm{s}}_2^2/{\mathrm{n}}_2{)}^2}{{\displaystyle \frac{{\left({\displaystyle \frac{({\mathrm{s}}_1^2}{{{\mathrm{n}}_1}^2}}\right)}^2}{({\mathrm{n}}_1-1)}+\frac{\left({\displaystyle \frac{{{\mathrm{s}}_2}^2}{{{\mathrm{n}}_2}^2}}\right)}{({\mathrm{n}}_2-1)}}} \\ =\frac{{\left(0.{000143}^2/7+0.{001379}^2/8\right)}^2}{{\displaystyle \frac{{\left(0.0001432/\right)}^2}{7-1}}+{\displaystyle \frac{{\left(0.0013-{79}^2/8\right)}^2}{8-1}}} \\ =7.17 \\ =7\left(\mathrm{rounded}\mathrm{off}\right) \end{gathered}$$

For the degrees of freedom of 7, the value of t in table 4-4 for 95% confidence is 2.365.

The observed value $t_{\mathrm{calculated}}=10.8$far exceeds$t_{\mathrm{table}}$

Therefore, the clear difference between the two data sets is highly significant.