Question 4.19: Tables 4.1 and 4.8 show results for two separate experiments...

To begin with, we must determine whether the variances for the two analyses are significantly different. This is done using an F-test as outlined in Example 4.18. Since no significant difference was found, a pooled standard deviation with 10 degrees of freedom is calculated

\begin{aligned} s_{\text {pool }} & =\sqrt{\frac{\left(n_{\mathrm{A}}-1\right) s_{\mathrm{A}}^{2}+\left(n_{\mathrm{B}}-1\right) s_{\mathrm{B}}^{2}}{n_{\mathrm{A}}+n_{\mathrm{B}}-2}} \\ & =\sqrt{\frac{(7-1)(0.00259)+(5-1)(0.00138)}{7+5-2}} \\ & =0.0459 \end{aligned}

where the subscript \mathrm{A} indicates the data in Table 4.1, and the subscript \mathrm{B} indicates the data in Table 4.8. The comparison of the means for the two analyses is based on the null hypothesis

H_{0}: \quad \bar{X}_{\mathrm{A}}=\bar{X}_{\mathrm{B}}

and a two-tailed alternative hypothesis

H_{\mathrm{A}}: \quad \bar{X}_{\mathrm{A}} \neq \bar{X}_{\mathrm{B}}

Since the standard deviations can be pooled, the test statistic is calculated using equation 4.20

t_{\text {exp }}=\frac{\left|\bar{X}_{\mathrm{A}}-\bar{X}_{\mathrm{B}}\right|}{s_{\text {pool }} \sqrt{\left(1 / n_{\mathrm{A}}+1 / n_{\mathrm{B}}\right)}}=\frac{|3.117-3.081|}{0.0459 \sqrt{(1 / 7+1 / 5)}}=1.34

The critical value for t(0.05,10), from Appendix 1 \mathrm{~B}, is 2.23 . Since t_{\exp } is less than t(0.05,10) the null hypothesis is retained, and there is no evidence that the two sets of pennies are significantly different at the chosen significance level.