A trainee in a medical lab will be released to work on her own when her results agree with those of an experienced worker at the 95% confidence level. Results for a blood urea nitrogen analysis are shown below.

Trainee: $\overline{\mathbf{x}}\mathbf{=}\mathbf{14}\mathbf{.}\mathbf{57}\mathbf{mg}\mathbf{/}\mathbf{dL}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{s}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{53}\mathbf{mg}\mathbf{/}\mathbf{dL}\mathbf{}\mathbf{}\mathbf{}\mathbf{n}\mathbf{=}\mathbf{6}$ samples

Experienced worker: $\overline{\mathbf{x}}\mathbf{=}\mathbf{13}\mathbf{.}\mathbf{95}\mathbf{}\mathbf{mg}\mathbf{/}\mathbf{dL}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{s}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{42}\mathbf{mg}\mathbf{/}\mathbf{dL}\mathbf{}\mathbf{}\mathbf{}\mathbf{n}\mathbf{=}\mathbf{5}$samples

(a) What does the abbreviation dL stand for?

(b) Should the trainee be released to work alone?

(a)

The abbreviation dL stands for deciliter, a unit of volume equal to one litre. $1\times {10}^{-1}$ liter or 0.1 liter.

(b)

To determine whether the trainee should be released to work alone or not, her lab results must be compared to those of an experienced worker using statistical analyses-F test and t-test, with the trainee's lab results equal to the experimental values and the experienced worker's lab results equal to the accepted/reference values.

F- test

$$\begin{gathered} {\mathrm{F}}_{\mathrm{Calc}}=\frac{{\mathrm{s}}_1^2}{{\mathrm{s}}_2^2} \\ =\frac{{\left(0.53\right)}^2}{{\left(0.42\right)}^2} \\ =1.59 \end{gathered}$$

${\mathrm{F}}_{\mathrm{table}}=6.26$ (degrees of freedom for based on Table 4-3: $s_1=5$and $s_2=4$)

There is no significant difference between the variances and standard deviations of the two lab results since ${\mathrm{F}}_{\mathrm{calc}}<{\mathrm{F}}_{\mathrm{table}}$

t- test

$\mathrm{F}$the outcome of a test ${\mathrm{F}}_{\mathrm{calc}}<{\mathrm{F}}_{\mathrm{table}}$

Case 2 an of the test will be performed.

$$\begin{gathered} {\mathrm{S}}_{\mathrm{pooled}}=\sqrt{\frac{{\mathrm{s}}_1^2\left({\mathrm{n}}_1-1\right)+{\mathrm{s}}_2^2\left({\mathrm{n}}_2-1\right)}{{\mathrm{n}}_1+{\mathrm{n}}_2-{\mathrm{n}}_{\mathrm{t}}}} \\ =\sqrt{\frac{{\left(0.53\right)}^2\left(6-1\right)+{\left(0.42\right)}^2\left(5-1\right)}{6+5-2}} \\ =0.4842061085 \\ {\mathrm{t}}_{\mathrm{calc}}=\frac{\left|\overline{{\mathrm{x}}_1}-\overline{{\mathrm{x}}_2}\right|}{{\mathrm{S}}_{\mathrm{pooled}}}\sqrt{\frac{{\mathrm{n}}_1{\mathrm{n}}_2}{{\mathrm{n}}_1+{\mathrm{n}}_2}} \\ =\frac{\left|14.57-13.95\right|}{0.4842061085}\sqrt{\frac{\left(6\right)\left(5\right)}{6+5}} \\ =2.11 \end{gathered}$$

${\mathrm{t}}_{\mathrm{table}}=2.262$(4 degrees of freedom based on Table 4-4

$-{\mathrm{n}}_1+{\mathrm{n}}_2-{\mathrm{n}}_{\mathrm{t}}=6+5-2=9;$ 95% level of assurance)

Since ${\mathrm{t}}_{\mathrm{calc}}<{\mathrm{t}}_{\mathrm{table}}$then there isn't a substantial difference between the two lab findings' means.