As part of a collaborative study of a new method for determining the amount of total

Question

As part of a collaborative study of a new method for determining the amount of total cholesterol in blood, two samples were sent to ten analysts with instructions to analyze each sample one time. The following results, in milligrams of total cholesterol per [latex]100 \mathrm{~mL}[/latex] of serum, were obtained

[latex]\begin{array}{ccc} ext{Analyst }& ext{Tablet }1 & ext{Tablet }2 \ \hline 1 & 245.0 & 229.4 \ 2 & 247.4 & 249.7 \ 3 & 246.0 & 240.4 \ 4 & 244.9 & 235.5 \ 5 & 255.7 & 261.7 \ 6 & 248.0 & 239.4 \ 7 & 249.2 & 255.5 \ 8 & 225.1 & 224.3 \ 9 & 255.0 & 246.3 \ 10 & 243.1 & 253.1 \end{array}[/latex]

Using this data estimate the values for [latex]\sigma_{ ext {rand }}[/latex] and [latex]\sigma_{ ext {sys }}[/latex] for the method assuming [latex]\alpha=0.05[/latex].

Accepted Answer

A two-sample plot of the data is shown in Figure 14.19, with the average value for sample 1 shown by the vertical line at 245.9, and the average value for sample 2 shown by the horizontal line at 243.5. To estimate [latex]\sigma_{ ext {rand }}[/latex] and [latex]\sigma_{ ext {sys }}[/latex], it first is necessary to calculate values for [latex]D_{i}[/latex] and [latex]T_{i}[/latex]; thus

[latex]\begin{array}{crc} extbf{Analyst }& \boldsymbol{D}_{\boldsymbol{i}} & \boldsymbol{T}_{\boldsymbol{i}} \ \hline 1 & 15.6 & 474.4 \ 2 & -2.3 & 497.1 \ 3 & 5.6 & 486.4 \ 4 & 9.4 & 480.4 \ 5 & -6.0 & 517.4 \ 6 & 8.6 & 487.4 \ 7 & -6.3 & 504.7 \ 8 & 0.8 & 449.4 \ 9 & 8.7 & 501.3 \ 10 & -10.0 & 496.2 \end{array}[/latex]

The standard deviations for the differences, [latex]s_{\mathrm{D}}[/latex], and the totals, [latex]s_{\mathrm{T}}[/latex], are calculated [latex]^{\star}[/latex] using equations 14.17 and 14.19 ,

[latex]s_{\mathrm{D}}\,=\,\sqrt{\frac{\sum(D_{i}-\overline{{{D}}})^{2}}{2(n-1)}}\,=\,s_{\mathrm{rand}}\approx\mathrm{\sigma_{rand}}[/latex] (14.17)

[latex]s_{T}\,=\,\sqrt{\frac{\sum(T_{i}-\overline{{{T}}})^{2}}{2(n-1)}}\,=\,s_{\mathrm{{tot}}}\approx\sigma_{\mathrm{{tot}}}[/latex] (14.19)

giving

[latex]s_{\mathrm{D}}=5.95 \quad s_{\mathrm{T}}=13.3[/latex]

To determine if the systematic errors of the analysts are significant, an F-test is performed using [latex]s_{\mathrm{T}}[/latex] and [latex]s_{\mathrm{D}}[/latex]

[latex]F=\frac{s_{\mathrm{T}}^{2}}{s_{\mathrm{D}}^{2}}=\frac{(13.3)^{2}}{(5.95)^{2}}=5.00[/latex]

Because the [latex]F[/latex]-ratio is larger than [latex]F(0.05,9,9)[/latex], which is 3.179 , we conclude that the systematic errors of the analysts are significant at the [latex]95 \%[/latex] confidence levels. The estimated precision for a single analyst is

[latex]\sigma_{ ext {rand }} \approx s_{\mathrm{D}}=5.95[/latex]

The estimated standard deviation due to systematic errors between analysts is calculated from equation 14.18.

[latex]\sigma_{\mathrm{tot}}^{2}=\sigma_{\mathrm{rand}}^{2}+2\sigma_{\mathrm{sys}}^{2}[/latex] (14.18)

[latex]\sigma_{ ext {sys }}=\sqrt{\frac{\sigma_{ ext {tot }}^{2}-\sigma_{ ext {rand }}^{2}}{2}} \approx \sqrt{\frac{s_{\mathrm{T}}^{2}-s_{\mathrm{D}}^{2}}{2}}=\sqrt{\frac{(13.3)^{2}-(5.95)^{2}}{2}}=8.41[/latex]

*Here is a short-cut that simplifies the calculation of [latex]s_D[/latex] and [latex]s_T[/latex]. Enter the values for [latex]D_i[/latex] into your calculator, and use its built-in functions to find the standard deviation. Divide this result by [latex]\sqrt{2}[/latex] to obtain [latex]s_D[/latex]. You can use the same approach to calculate [latex]s_T[/latex].

Question 14.7: As part of a collaborative study of a new method for determi...

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