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Bicarbonate in replicate samples of horse blood was measured four times by each of two methods with the following results:

Method 1:31.40, 31.24, 31.18, 31.43 mM

Method 2:30.70, 29.49, 30.01, 30.15mM

(a) Find the mean, standard deviation, and standard uncertainty (= standard deviation of the mean) for each analysis.

(b) Are the standard deviations significantly different at the 95%confidence level?

Short Answer

Expert verified

a)The mean value for method 1 is x1¯=31.31mM.

The standard deviation value for method 1 is s1=0.12mM.

The standard uncertainty value for method 1 isu1=0.06.

The mean value for method 2 isx2¯=30.08mM.

The standard deviation value for method 2 is s2=0.49mM.

The standard uncertainty value for method 2 isu2=0.24.

b)The standard deviations are significantly different.

Step by step solution

01

Definition of standard deviation and mean.

  • Arithmetic mean:By dividing the amount of measured values by the number of measurements n, the arithmetic meanx¯ is determined. The average is often referred to as the "mean."

Mean:x¯=∑ixin

  • Standard deviation: The standard deviation is a measurement of how tightly the data cluster around the mean.

The standard deviation (s) is given by the formula:

s=∑ixi-x¯2n-1

Standard deviation of the mean of sets of n values is:

σn=θn

02

Find the meanstandard deviation, standard uncertainty.

(a)

Let x1¯, s1, and u1be the mean, standard deviation, and standard uncertainty of method 1.

Method 's mean, standard deviation, and standard uncertainty arex2¯,s2andu2respectively.

For Method 1 :

Mean:

localid="1663562989158" x1¯=∑ixin=31.40+31.24+31.18+31.434=31.31=31.312mMroundedoff

Therefore, the mean value for method 1 isx1¯=31.31mM

Standard deviation:

s1=∑i(xi-x2n-1=31.40-31.3122+.....31.43-31.31224-1=0.121=0.12mMroundedoff

Therefore, the standard deviation value for method 1 is s1=0.12mM.

Standard uncertainty:

role="math" localid="1663563250038" u1=s1n=0.121mM4=0.061=0.06roundedoff

Therefore, the standard uncertainty value for method 1 is u1=0.06

For Method2:

Mean:

x2¯=∑ixin=30.70+29.49+30.01+30.154=30.088=30.08mMroundedoff

Therefore, the mean value for method 2 is x2¯=30.08mM.

Standard deviation:

s2=∑i(xi-x)2n-1=30.70-30.0882+.....30.15-30.08824-1=0.497=0.49mMroundedoff

Therefore, the standard deviation value for method 2 is s2=0.49mM.

Standard uncertainty:

u2=nn=0.497mM4=0.249=0.24roundedoff

Therefore, the standard uncertainty value for method 2 is u2=0.24.

03

Find if the standard deviations significantly different at the $95 \%$ confidence level.

(b)

It must be determined whether the standard deviation differs considerably at the 95% confidence level.

Comparison of Standard Deviation with F Test:

We determine if the standard deviations of two sets of measurements are "statistically different" when comparing mean values. The Ftest is used to do this.

The Ftest with quotient F is given as:

Fcalculated=k12x22

Where, s1ands2 are standard deviations for the set of measurements using original instrument and substitute instrument.

The difference betweenFcalculated>Ftable is substantial.

IfFcalculated>Ftable, then there is a considerable difference..

Degrees of Freedom:

The degrees of freedom for aset of measurements will ben-1 .

The F-test is used to compare standard deviations, which is computed as follows:

Fcalculated=s12s22=0.49720.1212=16.8=16roundedoff

The Ftableis found as Ftable=9.28for three degrees of freedom in both standard deviations.

Fcomputed16>Ftable=9.28after comparing the two values.

As a result, the standard deviations varied dramatically.

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