Question 4.20: The %w/w Na2CO3 in soda ash can be determined by an acid–bas...

We begin by summarizing the mean and standard deviation for the data reported by each analyst. These values are

\bar{X} _A = 86.83%

s_A = 0.32

\bar{X} _B = 82.71%

s_B = 2.16

A two-tailed F-test of the following null and alternative hypotheses

H_0:     s_A²= s_B²          H_A:     s_A² ≠ s_B²

is used to determine whether a pooled standard deviation can be calculated. The test statistic is

F_{exp}=\frac{s_B^2}{s_A^2} =\frac{(2.16)^2}{(0.32)^2}=45.6

Since F_{exp} is larger than the critical value of 7.15 for F(0.05, 5, 5), the null hypothesis is rejected and the alternative hypothesis that the variances are significantly different is accepted. As a result, a pooled standard deviation cannot be calculated.
The mean values obtained by the two analysts are compared using a twotailed t-test. The null and alternative hypotheses are

H_0:\bar{X}_A=\bar{X}_B        H_A:\bar{X}_A≠\bar{X}_B

Since a pooled standard deviation could not be calculated, the test statistic, t_{exp}, is calculated using equation 4.19

t_{exp}=\frac{\left|\bar{X}_A -\bar{X}_B \right| }{s_{pool}\sqrt{(s_A^2/n_A)+(s_B^2/n_B)} } =\frac{\left|86.83-82.71\right| }{\sqrt{[(0.32)^2/6]+[(2.16)^2/6]} } =4.62

and the degrees of freedom are calculated using equation 4.22

\mathrm{v}=\frac{[(S_A^2/n_A)+(S_B^2/n_B)]^2}{[(S_A^2/n_A)^2/(n_A+1)]+[(S_B^2/n_B)^2/(n_B+1)]} -2      (4.22)

\mathrm{v}=\frac{[(0.32^2/6)+(2.16^2/6)]^2}{[(0.32^2/6)^2/(6+1)]+[(2.16^2/6)/(6+1)]} -2=5.3\simeq 5

The critical value for t(0.05, 5) is 2.57. Since the calculated value of t_exp is greater than t(0.05, 5) we reject the null hypothesis and accept the alternative hypothesis that the mean values for %w/w Na_2CO_3 reported by the two analysts are significantly different at the chosen significance level.