Chapter 4, Problem 4.9P

Interpretation Introduction

Interpretation:

The mean value obtained from method 1 has to be identified that whether it is significantly different from the mean value obtained from method 2 at 95% confidence level.

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.

Explanation of Solution

The mean values for method 1 and method 2 are first calculated as follows,

    $\begin{array}{l}Enzymeactivity\left(5replications\right)\\ \begin{array}{ccccccc}Method\text{ 1}& 139& 147& 160& 158& 135& Mean,\overline{x_1}\text{ = 147}\text{.8}\\ Method\text{ 2}& 148& 159& 156& 164& 159& Mean,\overline{x_2}\text{ = 157}\text{.2}\end{array}\end{array}$

The standard deviations for above two methods are identified as follows,

For method 1:

  $\begin{array}{c}\text{ s = }\sqrt{\frac{{\displaystyle \sum _i{\left(x_i\text{ - }\overline{x}\right)}^2}}{\left(n-1\right)}}\\ \\ \overline{x}=\text{average}\\ \\ \text{s = }\sqrt{\frac{{\left(\text{139 - }147.8\right)}^2+{\left(147\text{ - }147.8\right)}^2+{\left(160\text{ - }147.8\right)}^2+{\left(158\text{ - }147.8\right)}^2+{\left(135\text{ -}147.8\right)}^2}{\left(5-1\right)}}\\ \\ =\sqrt{\frac{395.6}{\left(4\right)}}\\ \\ =11.12\end{array}$

For method 2:

  $\begin{array}{c}\text{s = }\sqrt{\frac{{\left(\text{148 - }157.2\right)}^2+{\left(159\text{ - }157.2\right)}^2+{\left(156\text{ - }157.2\right)}^2+{\left(164\text{ - }157.2\right)}^2+{\left(159\text{ -}157.2\right)}^2}{\left(5-1\right)}}\\ \\ =\sqrt{\frac{395.6}{\left(4\right)}}\\ \\ =5.9\end{array}$

The ${\text{F}}_{\text{calculated}}$ value for given two standard deviations is first determined.

    $\begin{array}{c}{\text{F}}_{\text{calculated}}\text{ = }\frac{{\text{s}}_{\text{1}}^{\text{2}}}{{\text{s}}_{\text{2}}^{\text{2}}}\\ \\ =\frac{{11.12}^2}{{5.9}^2}\\ \\ {\text{F}}_{\text{calculated}}\text{= 3}\text{.55}\end{array}$

Checking Table 4.3, using degrees of freedom value (4) for two given methods, shows the ${\text{F}}_{\text{table}}$ value as 6.39 hence, the difference is not significant since ${\text{F}}_{\text{calculated}}$ value is less than ${\text{F}}_{\text{table}}$.

The ${\text{t}}_{\text{calculated}}$ value for given two means are shown as follows,

    $\begin{array}{c}{\text{t}}_{\text{calculated}}\text{ = }\frac{\left|\overline{x_1}-\overline{x_2}\right|}{{\text{s}}_{\text{pooled}}}\sqrt{\frac{{\text{n}}_{\text{1}}\times {\text{n}}_{\text{2}}}{{\text{n}}_{\text{1}}{\text{+ n}}_{\text{2}}}}\\ \\ {\text{s}}_{\text{pooled}}\text{= }\sqrt{\frac{{\text{s}}_{\text{1}}{}^{\text{2}}\left({\text{n}}_{\text{1}}\text{-1}\right){\text{+ s}}_{\text{2}}{}^{\text{2}}\left({\text{n}}_{\text{2}}\text{-1}\right)}{{\text{n}}_{\text{1}}{\text{+ n}}_{\text{2}}\text{- 2}}}\\ \\ =\sqrt{\frac{{11.12}^{\text{2}}\left(\text{5-1}\right)\text{+ 5}{\text{.9}}^{\text{2}}\left(\text{5-1}\right)}{\text{5 + 5 - 2}}}\\ \\ =\text{ 8}\text{.9}\\ \\ \text{= }\frac{\left|\overline{147.8}-\overline{157.2}\right|}{\text{8}\text{.9}}\sqrt{\frac{5\times 5}{\text{5+5}}}\\ \\ =\text{ 1}\text{.67}\end{array}$

Observing the Table. 4.4, the ${\text{t}}_{\text{table}}$ value is $\text{2}\text{.306}$ which is greater than the calculated value hence the means are not significantly different.