Calculate the standard deviation for an exp data set


Discussion:

Calculate the standard deviation for an exp. data set.

The standard deviation of a data set is a measure of the 'spread' of a distribution of data or other numerical values; it is a measure of variability within a particular data set. To calculate standard deviation of a sample, generally the variance of the sample is first calculated. The variance is the average squared deviation of each number from its mean, and is given by the following equation: where m is the mean and N is the number of values within the data set of interest. For example, for the data set: 1, 2, and 3, the mean is 2 and the variance is given by the following:

The standard deviation of a sample is simply the square root of the variance, and is probably the most commonly used measure of data 'spread.'

For this CLC exercise;

a) Calculate the standard deviation for the experimentally obtained values of Keq, DH, and the values calculated for DG, and DS. Then post the results of these individual experiments, not the mean values, to the CLC group, and calculate standard deviation for those same values, using the entire data set generated by your CLC group.

b) How do your values compare to the values of the entire data set generated by your CLC group?

c) How 'tight' is the spread of the data?

d) Compare the values of both your personal data set and those of the CLC groups data set with the following accepted values as reported by Pickering (Pickering 1987) shown in the table below:

Variable

Value

DH/kJ mol-1

17.6 ­­­± 7.2

DG/kJ mol-1

-5.77 ­­­± 0.63

DS/J mol-1 K-1

77.0 ­­­± 28

i)  How do your value compare to these experimentally obtained values?

ii) What are some possible sources of error in your measurement?

iii) Where there any extreme outliers in the data set you analyzed that might affect the pooled values?

Solution Preview :

Prepared by a verified Expert
Chemistry: Calculate the standard deviation for an exp data set
Reference No:- TGS01917530

Now Priced at $20 (50% Discount)

Recommended (98%)

Rated (4.3/5)