Data collection is fundamental for any statistical analysis in Science Laboratory Technology, where accurate and reliable data underpin laboratory experiments and quality control. This chapter focuses on the initial step of statistical methods: collecting raw data, which can be either ungrouped or grouped depending on the nature and volume of observations. Understanding how to organize and interpret these data types is essential for laboratory professionals handling chemical assays, biological measurements, or environmental samples in Kenya’s diverse scientific settings.

5.1 Collection of raw data

Data collection involves gathering measurements or observations from laboratory experiments or field studies. In scientific laboratories, data must be recorded systematically to ensure valid analysis and reproducibility. Raw data may be recorded as individual values (ungrouped) or summarized into classes or intervals (grouped), depending on the volume and variability of the data.

5.1.1 Ungrouped data

Ungrouped data consists of individual raw observations without any categorization or summarization. This data type is common in laboratory settings where each measurement is critical, such as recording pH values of water samples or concentration readings from spectrophotometry.

The simplest form of data, ungrouped data is analyzed using direct statistical formulas for measures like mean, median, mode, variance, and standard deviation without the need for frequency distribution tables.

$$ \text{Mean} \, \bar{x} = \frac{\Sigma x_i}{n} $$

where \(x_i\) are individual observations and \(n\) is the total number of observations.

Worked Examples

Example 1: A laboratory technician records the concentration of a chemical in five water samples (mg/L): 2.5, 3.0, 2.8, 3.2, 2.9. Find the mean concentration.

Given: \(x = \{2.5, 3.0, 2.8, 3.2, 2.9\}\), \(n=5\)

$$ \bar{x} = \frac{2.5 + 3.0 + 2.8 + 3.2 + 2.9}{5} $$$$ = \frac{14.4}{5} $$$$ = 2.88 \, \text{mg/L} $$

Answer: 2.88 mg/L

Example 2: Calculate the variance of the data set in Example 1.

Given: \(x = \{2.5, 3.0, 2.8, 3.2, 2.9\}\), \(n=5\), \(\bar{x} = 2.88\)

$$ s^2 = \frac{\Sigma (x_i - \bar{x})^2}{n-1} $$$$ = \frac{(2.5-2.88)^2 + (3.0-2.88)^2 + (2.8-2.88)^2 + (3.2-2.88)^2 + (2.9-2.88)^2}{4} $$$$ = \frac{(-0.38)^2 + (0.12)^2 + (-0.08)^2 + (0.32)^2 + (0.02)^2}{4} $$$$ = \frac{0.1444 + 0.0144 + 0.0064 + 0.1024 + 0.0004}{4} $$$$ = \frac{0.268}{4} $$$$ = 0.067 $$

Answer: 0.067 (mg/L)\(^2\)

Example 3: Find the standard deviation of the data in Example 1.

Given: \(s^2 = 0.067\)

$$ s = \sqrt{0.067} $$$$ = 0.259 $$

Answer: 0.259 mg/L

Example 4: A biochemist records enzyme activity rates (units/mL) as 15, 18, 16, 17, 19, 20. Calculate the median.

Given: \(x = \{15, 16, 17, 18, 19, 20\}\) (sorted data)

For even \(n=6\), median is average of middle two values:

$$ \text{Median} = \frac{17 + 18}{2} = \frac{35}{2} = 17.5 $$

Answer: 17.5 units/mL

Example 5: Determine the mode of the data set: 12, 15, 12, 16, 15, 12, 18.

Given: \(x = \{12, 15, 12, 16, 15, 12, 18\}\)

• Frequency: 12 appears 3 times, 15 appears 2 times, others less.

Answer: Mode = 12

5.1.2 Grouped data

Grouped data arises when raw data are organized into classes or intervals, a common practice when handling large data sets such as repeated measurements of temperature in incubation experiments or batch tests of water quality parameters. Grouping simplifies analysis by summarizing data frequencies in intervals.

The mean of grouped data is calculated using midpoints of class intervals weighted by their frequencies:

$$ \bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} $$

where \(f_i\) is the frequency of class \(i\) and \(x_i\) is the class midpoint.

Worked Examples

Example 1: A laboratory records the number of bacterial colonies in Petri dishes as follows:

Calculate the mean number of colonies.

Given:

Classes and frequencies:

• \(f_1=5\), midpoint \(x_1 = \frac{0+10}{2} = 5\)
• \(f_2=8\), midpoint \(x_2 = \frac{11+20}{2} = 15.5\)
• \(f_3=12\), midpoint \(x_3 = \frac{21+30}{2} = 25.5\)
• \(f_4=7\), midpoint \(x_4 = \frac{31+40}{2} = 35.5\)
• \(f_5=3\), midpoint \(x_5 = \frac{41+50}{2} = 45.5\)

Calculate numerator:

$$ \Sigma f_i x_i = 5 \times 5 + 8 \times 15.5 + 12 \times 25.5 + 7 \times 35.5 + 3 \times 45.5 $$$$ = 25 + 124 + 306 + 248.5 + 136.5 = 840 $$

Calculate denominator:

$$ \Sigma f_i = 5 + 8 + 12 + 7 + 3 = 35 $$

Mean:

$$ \bar{x} = \frac{840}{35} = 24 $$

Answer: 24 colonies

Example 2: Using the data in Example 1, calculate the variance of the grouped data.

Given: \(f_i\), \(x_i\), \(\bar{x} = 24\), \(n=35\)

Variance formula for grouped data:

$$ s^2 = \frac{\Sigma f_i (x_i - \bar{x})^2}{n-1} $$

Calculate each term:

• \((5-24)^2 = 361\), contribution: \(5 \times 361 = 1805\)
• \((15.5-24)^2 = 72.25\), contribution: \(8 \times 72.25 = 578\)
• \((25.5-24)^2 = 2.25\), contribution: \(12 \times 2.25 = 27\)
• \((35.5-24)^2 = 132.25\), contribution: \(7 \times 132.25 = 925.75\)
• \((45.5-24)^2 = 462.25\), contribution: \(3 \times 462.25 = 1386.75\)

Sum of contributions:

$$ 1805 + 578 + 27 + 925.75 + 1386.75 = 4722.5 $$

Calculate variance:

$$ s^2 = \frac{4722.5}{35 - 1} = \frac{4722.5}{34} = 138.9 $$

Answer: 138.9 colonies\(^2\)

Example 3: Calculate the standard deviation from Example 2.

Given: \(s^2 = 138.9\)

$$ s = \sqrt{138.9} = 11.79 $$

Answer: 11.79 colonies

Example 4: A soil laboratory measures particle sizes with frequencies:

Calculate the mean particle size.

Midpoints:

• \(x_1 = 25\)
• \(x_2 = 75\)
• \(x_3 = 125\)
• \(x_4 = 175\)
• \(x_5 = 225\)

Calculate numerator:

$$ 10 \times 25 + 20 \times 75 + 30 \times 125 + 25 \times 175 + 15 \times 225 $$$$ = 250 + 1500 + 3750 + 4375 + 3375 = 13250 $$

Denominator:

$$ 10 + 20 + 30 + 25 + 15 = 100 $$

Mean:

$$ \bar{x} = \frac{13250}{100} = 132.5 \, \mu m $$

Answer: 132.5 µm

Example 5: Calculate the variance for the data in Example 4.

Calculate squared deviations and contributions:

• \((25 - 132.5)^2 = 11556.25\), contribution: \(10 \times 11556.25 = 115562.5\)
• \((75 - 132.5)^2 = 3306.25\), contribution: \(20 \times 3306.25 = 66125\)
• \((125 - 132.5)^2 = 56.25\), contribution: \(30 \times 56.25 = 1687.5\)
• \((175 - 132.5)^2 = 1806.25\), contribution: \(25 \times 1806.25 = 45156.25\)
• \((225 - 132.5)^2 = 8556.25\), contribution: \(15 \times 8556.25 = 128343.75\)

Sum:

$$ 115562.5 + 66125 + 1687.5 + 45156.25 + 128343.75 = 357875 $$

Variance:

$$ s^2 = \frac{357875}{100 - 1} = \frac{357875}{99} = 3615.9 $$

Answer: 3615.9 µm\(^2\)

Chapter Summary

This chapter covers the application of statistical methods, beginning with the collection of raw data, distinguishing between ungrouped and grouped data sets. It then explores various methods of data presentation, including pictograms, histograms, pie charts, bar charts, and frequency polygons, which help visualize data effectively. The processing of raw data is addressed to prepare it for meaningful analysis. Measures of central tendency are explained, focusing on the calculation and interpretation of the mean, mode, and median. These concepts provide essential tools for summarizing data sets and understanding their distribution. Overall, the chapter equips students with foundational skills to collect, organize, display, and analyze statistical data systematically.