EXPLOR. CHEM ANALYSIS (LL) - W/ ACHIEVE
EXPLOR. CHEM ANALYSIS (LL) - W/ ACHIEVE
5th Edition
ISBN: 9781319468859
Author: Harris
Publisher: MAC HIGHER
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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:

Fcalculated = s12s22 If Fcalculated>Ftable for n1-1 and n2-2, degrees of freedom, the standard deviations s1 and s2 are significantly different.

t test:

    tcalculated = |x1¯x2¯|spooledn1×n2n1+ n2spooleds12(n1-1)+ s22(n2-1)n1+ n2- 2

If tcalculated>ttable for 95 % confidence interval and n1+n2-2, degrees of freedom the difference is significant. Here, s1 and s2 are standard deviations, x1¯ and x2¯ are mean values.

Expert Solution & Answer
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Explanation of Solution

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

                         Enzymeactivity(5replications)Method 1139147160158135Mean, x1¯ = 147.8Method 2148159156164159Mean, x2¯ = 157.2

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

For method 1:

   s = i(xi - x¯)2(n1)x¯=averages = (139 - 147.8)2+(147 - 147.8)2+(160 - 147.8)2+(158 - 147.8)2+(135 -147.8)2(51)= 395.6(4)= 11.12

For method 2:

  s = (148 - 157.2)2+(159 - 157.2)2+(156 - 157.2)2+(164 - 157.2)2+(159 -157.2)2(51)= 395.6(4)= 5.9

The Fcalculated value for given two standard deviations is first determined.

    Fcalculated = s12s22= 11.1225.92Fcalculated= 3.55

Checking Table 4.3, using degrees of freedom value (4) for two given methods, shows the Ftable value as 6.39 hence, the difference is not significant since Fcalculated value is less than Ftable.

The tcalculated value for given two means are shown as follows,

    tcalculated = |x1¯x2¯|spooledn1×n2n1+ n2spooleds12(n1-1)+ s22(n2-1)n1+ n2- 2= 11.122(5-1)+ 5.92(5-1)5 + 5 - 2= 8.9|147.8¯157.2¯|8.95×55+5= 1.67

Observing the Table. 4.4, the ttable value is 2.306 which is greater than the calculated value hence the means are not significantly different.

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