Question

If the difference between the two mean values were half as great as Rayleigh found, but the standard deviation were unchanged, would the difference still be significant?

Short answer

The two given masses of nitrogen are found to be different significantly

Step-by-step solution

Formula Used

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

Given Information

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

From the air:

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

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" $\overline{\left(x_2\right)}=2.29947\mathrm{g}.$

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.

Check whether the given two masses are different

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.

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

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

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

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

$=10.8\hspace{0.33em}(\mathrm{Rounded}\mathrm{to}\mathrm{the}\mathrm{correct}\mathrm{significant}\mathrm{figure}).$

Calculate the Degree of Freedom

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.