APPENDIX

A. Raw dataTABLE 1. Weight of Samples

Sample No.

Weight, g

1 3.5961 D A T A

S E T

2

D.

S E T 1

2 3.50773 3.58414 3.50205 3.59946 3.50427 3.57828 3.58879 3.5939

10 3.5620

TABLE 2. Grubb’s TestData Set Suspect Values Qtab Qexp Conclusion

1H: 3.5994 1.8871 1.0343 RetainL: 3.5020

2H: 3.5994L: 3.5020 2.2900 1.4640 retain

TABLE 3. Reported ValuesData Set X s RSD R RR CL

1 3.5489 0.048810 13.754 0.097400 27.4453.60013.4977

2 3.5616 0.040731 11.346 0.097400 27.3473.59073.5325

B. Working equations and sample calculations1. Divide the samples into Data Sets 1 and 2.Data set 1 consists of samples 1 to 6, while data set 2 consists of samples 1 to 10.

2. Perform the Grubb’s test (at 95% confidence level) for the suspected outlier of each data set.In data set 1, taking the absolute values of the differences of the highest and the lowest values from the mean

|3.5994−3.5489|=0.0505|3.5020−3.5489|=0.0469

The experimental value, g, is then calculated using the questionable measurement, the highest value (3.5994). (We use the mean computed in #3).

g=

maxi=1…n

|X i−X|s

=|3.5994−3.5489|

0.048810= 0.05050.048810

=1.0343

The suspected outlier is then compared with the tabulated critical g value.

N 3 4 5 6 7 8 9 10g95% 1.1543 1.4812 1.7150 1.8871 2.0200 2.1266 2.2150 2.2900g99% 1.1547 1.4962 1.7637 1.9728 2.1391 2.2744 2.3868 2.4821

TABLE 4. Critical Values for the Grubbs Test (taken from the Analytical Chemistry Laboratory Manual)

Gtab=1.8871Since the experimental value is less than the tabulated one, the suspected outlier should be retained.

For data set 2, taking the absolute values of the differences of the highest and the lowest values from the mean

|3.5994−3.5616|=0.0378|3.5020−3.5616|=0.0596

The experimental value, g, is then calculated using the questionable measurement, the lowest value (3.5020). (Again, the mean to be used is the one computed in #3)

g=

maxi=1…n

|X i−X|s

=|3.5020−3.5616|

0.040731= 0.05960.040731

=1.4640

The suspected outlier is then compared with the tabulated critical g value.Gtab=2.2900

Because the experimental value is less than the tabulated one, the suspected outlier should be retained.

3. Calculate the following statistics for each data set:a. MeanSince the suspected outliers are retained in #2 (Grubbs Test), the sample means remain the same. The formula for the mean is

X=∑i=1

n

X i

nFor data set 1, the mean is 3.5489.

X=∑i=1

n

X i

n=

(3.5961+3.5077+3.5841+3.5020+3.5994+3.5042)6

=3.5489

For data set 2, the mean is 3.5616.

X=∑i=1

n

X i

n=

(3.5961+3.5077+3.5841+3.5020+3.5994¿+3.5042+3.5782+3.5887+3.5939+3.5620)10

=3.5616

b. Standard deviationBecause the conclusion is to retain the suspected outliers, the standard deviations also remain the same. In getting the standard deviation, the formula

s=√∑i=1n

( X i−X )2

n−1shall be used.For data set 1, the standard deviation is 0.04881.

s=√∑i=1n

(X i−X)2

n−1=√ (3.5961−3.5489 )2+(3.5077−3.5489 )2

+(3.5841−3.5489 )2+(3.5020−3.5489 )2

+(3.5994−3.5489)2+(3.5042−3.5489)2

5=0.04881

While in data set 2, the standard deviation is 0.040731.

s=√∑i=1n

(X i−X)2

n−1=√

(3.5961−3.5616 )2+(3.5077−3.5616 )2

+ (3.5841−3.5616 )2+(3.5020−3.5616 )2

+ (3.5994−3.5616 )2+ (3.5042−3.5616 )2

+ (3.5782−3.5616 )2+(3.5887−3.5616 )2

+(3.5939−3.5616)2+(3.5620−3.5616 )2

9=0.040731

c. Relative standard deviation (in ppt)Computing for the relative standard deviation using the formula

RSD= sX×1000 ppt

the value for data set 1 is 13.754

RSD= sX×1000 ppt=0.048810

3.5489×1000 ppt=13.754

and 11.436 for data set 2

RSD= sX×1000 ppt=0.040731

3.5616×1000 ppt=11.436

d. RangeThe formula for getting the range is

R=Xh ig hest−X lowestData set 1 has a range of 0.097400.

R=Xh ig hest−X lowest=3.5994−3.5020=0.097400Data set 2, also, has a range of 0.097400.

R=Xh ig hest−X lo west=3.5994−3.5020=0.097400

e. Relative range (in ppt)The formula for computing the relative range in ppt is

RR= RX×1000 ppt

Data set 1’s relative range is 27.445.

RR= RX×1000 ppt=0.097400

3.5489×1000 ppt=27.445

Data set 2’s relative range is 27.347.

RR= RX×1000 ppt=0.097400

3.5616×1000 ppt=27.347

f. Confidence limits (95% confidence level)For the confidence limit, the formula to be used is

Confidence limit=X ± ts√n

where t is dependent on the confidence level and degrees of freedom.

TABLE 5. Values of t for various levels of probability (taken from the Analytical Chemistry Laboratory Manual)

n-1 1 2 3 4 5 6 7 8 9 10T90% 6.31 2.92 2.35 2.13 2.02 1.94 1.90 1.86 1.83 1.81T95% 12.7 4.30 3.18 2.78 2.57 2.45 2.36 2.31 2.26 2.23

T99% 63.7 9.92 5.84 4.60 4.03 3.71 3.50 3.36 3.25 3.17

In data set 1, the upper confidence limit is 3.6001,

Confidence limit=X ±ts

√n=3.5489+

(2.57 )(0.048810)√6

=3.6001

while the lower confidence limit is 3.4977.

Confidence limit=X ±ts

√n=3.5489−

(2.57 )(0.048810)√6

=3.4977

In data set 2, the upper confidence limit is 3.5907,

Confidence limit=X ±ts

√n=3.5616+

(2.26 )(0.040731)10

=3.5907

while the lower confidence limit is 3.5325.

Confidence limit=X ±ts

√n=3.5616−

(2.26 )(0.040731)10

=3.5325

APPLICATION OF STATISTICAL CONCEPTS IN THE DETERMINATION OF WEIGHT VARIATION IN SAMPLES

M. FRANCISCONATIONAL INSTITUTE OF MOLECULAR BIOLOGY AND BIOTECHNOLOGY, COLLEGE OF SCIENCEUNIVERSITY OF THE PHILIPPINES, DILIMAN, QUEZON CITY 1101, PHILIPPINESDATE SUBMITTED: 3 FEBRUARY 2015DATE PERFORMED: 29 JANUARY 2015

GUIDES FOR DISCUSSION

1. Give the significance of the Grubbs Test.

In a data set of samples, presence of outliers is possible. An outlier is an observation that deviates from the other observations in the set of samples. This adjusts the mean to be closer to the outlier, since the mean is sensitive if there are outliers. It is, therefore, important to identify potential outliers in the sample. An outlier can show error in the data. If it is proven that it is faulty, the outlier should be removed or corrected if possible. However, if it is not erroneous, the value should be retained. The Grubbs’ test is one of the formal outlier tests used when testing for a single outlier in a sample with normal distribution. The highest and the lowest values are taken and tested. The largest absolute value difference from the mean is then taken. The suspected outlier is then compared with the tabulated Grubbs value. If the obtained value is less than or equal to the tabulated one, the value is retained. If not, it is rejected and, therefore, omitted. Once done, it is safe to say that the sample is representative of the population.

2. Give the significance of the mean and standard deviation.

The mean is the most common and well-known form of average. It measures the center of a set of data, and it can be used with both discrete and continuous data. Essentially, it is a model of the data set and shows the most common value. One property of the mean is that it takes into account every value included in the data set. It is the easiest to calculate; thus, it is most frequently used. However, the data that this measure presents is still not enough to describe the set of data.

The standard deviation shows the spread of the data around the mean. It also shows whether the data is extreme or not, and it normalizes the data, giving us an idea of how far apart our data is from the average (mean). It is used to summarize continuous data and is only appropriate when the data set is not skewed or present with outliers.

Both help determine if the findings are valid.

3. Give the significance of the confidence interval.

Confidence intervals are a measure of how good an estimate of the population is, since it is important to know how closely the mean of the sample is representative of the mean for the population, which can be known from the confidence interval. It shows limitations of the estimates. Also, it produces a lower and upper limit for the mean. The interval shows an indication of how great the uncertainty is in the estimate of the true mean. The estimate is more precise if the interval is narrower.

In taking the sample (representative of the population), it is important and necessary to know how close the measure of the sample is to that of the mean.

4. How do the statistics calculated from data set 1 differ from those obtained from data set 2?

The statistics calculated from data set 1 differ from those in data set 2 by only a few decimals apart, except for the relative standard deviation. They do not differ greatly. Also, since the highest and the lowest value for both sets are the same, the range is similar.

REFERENCES

Alfassi, Z. Statistical Treatment of Analytical Data. 2005. London: Blackwell. 70-71.Thode, H. Testing for Normality. 2002. CRC Press. 123-127.Walpole, R. Introduction to Statistics. 1982. New York: Macmillan. 23-26.