The %w/w Na2CO3 in soda ash can be determined by an acid–base titration. The results obtained by two analysts are shown here. Determine whether the difference in their mean values is significant at α = 0.05.

Question

The %w/w [latex]Na_2CO_3[/latex] in soda ash can be determined by an acid–base titration. The results obtained by two analysts are shown here. Determine whether the difference in their mean values is significant at α = 0.05.

Analyst A	Analyst B
86.82	81.01
87.04	86.15
86.93	81.73
87.01	83.19
86.20	80.27
87.00	83.94

Accepted Answer

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

[latex]\bar{X} _A[/latex] = 86.83%

[latex]s_A[/latex] = 0.32

[latex]\bar{X} _B[/latex] = 82.71%

[latex]s_B[/latex] = 2.16

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

[latex]H_0: s_A²= s_B² H_A: s_A² ≠ s_B²[/latex]

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

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

Since [latex]F_{exp}[/latex] 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

[latex]H_0:\bar{X}_A=\bar{X}_B H_A:\bar{X}_A≠\bar{X}_B[/latex]

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

[latex]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[/latex]

and the degrees of freedom are calculated using equation 4.22

[latex]\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[/latex] (4.22)

[latex]\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[/latex]

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

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

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