Chapter 4, Problem 4.12P

Interpretation Introduction

Interpretation:

Does the two given methods gives statistically equivalent results has to be identified.

Concept-Introduction:

F Test:

    ${\text{F}}_{\text{calculated}}\text{ = }\frac{{\text{s}}_{\text{1}}^{\text{2}}}{{\text{s}}_{\text{2}}^{\text{2}}}$ If ${\text{F}}_{\text{calculated}}>{\text{F}}_{\text{table}}$ for ${\text{n}}_{\text{1}}{\text{-1 and n}}_{\text{2}}\text{-2}$, degrees of freedom, the standard deviations ${\text{s}}_{\text{1}}{\text{ and s}}_{\text{2}}$ are significantly different.

t test:

    $\begin{array}{c}{\text{t}}_{\text{calculated}}\text{ = }\frac{\left|\overline{x_1}-\overline{x_2}\right|}{{\text{s}}_{\text{pooled}}}\sqrt{\frac{{\text{n}}_{\text{1}}\times {\text{n}}_{\text{2}}}{{\text{n}}_{\text{1}}{\text{+ n}}_{\text{2}}}}\\ \\ {\text{s}}_{\text{pooled}}\text{= }\sqrt{\frac{{\text{s}}_{\text{1}}{}^{\text{2}}\left({\text{n}}_{\text{1}}\text{-1}\right){\text{+ s}}_{\text{2}}{}^{\text{2}}\left({\text{n}}_{\text{2}}\text{-1}\right)}{{\text{n}}_{\text{1}}{\text{+ n}}_{\text{2}}\text{- 2}}}\end{array}$

If $t_{\text{calculated}}>t_{\text{table}}$ for $\text{95 \%}$ confidence interval and ${\text{n}}_{\text{1}}{\text{+n}}_{\text{2}}\text{-2}$, degrees of freedom the difference is significant. Here, ${\text{s}}_{\text{1}}{\text{ and s}}_{\text{2}}$ are standard deviations, $\overline{{\text{x}}_{\text{1}}}\text{ and }\overline{{\text{x}}_{\text{2}}}$ are mean values.