Procurement professionals in Kenya rely on accurate data to make effective purchasing decisions, manage supplier relationships, and ensure value for money. Understanding elementary statistics and data collection methods is essential for evaluating procurement performance, forecasting demand, and controlling costs. This chapter introduces fundamental concepts in data collection and elementary statistics tailored to procurement management contexts, providing practical calculation examples for data analysis.

4.1 Introduction to data collection

Data collection is the foundation of statistical analysis in procurement. It involves gathering relevant information about prices, supplier performance, delivery times, and market trends to support decision-making. Elementary statistics then summarise and interpret this data to identify patterns, averages, and variations critical to procurement planning and evaluation.

4.1.1 Measures of Central Tendency: Mean, Median, and Mode

Measures of central tendency summarise a data set by identifying a typical or central value. The three main measures are the mean, median, and mode.

• The mean is the arithmetic average of all data points, calculated by summing values and dividing by the number of observations.
• The median is the middle value when data is ordered, representing the 50th percentile.
• The mode is the most frequently occurring value in the data set.

The mean is sensitive to outliers, while the median better represents central tendency in skewed data. The mode is useful for categorical data analysis.

$$\text{Mean} = \frac{\Sigma x_i}{n}$$

$$\text{Median} = \text{middle value of ordered data}$$

$$\text{Mode} = \text{most frequent value}$$

Worked Examples

Example 1: A procurement officer records the delivery times (in days) of 5 suppliers: 7, 8, 6, 9, 7. Calculate the mean delivery time.

Given: Delivery times: 7, 8, 6, 9, 7 days; \( n = 5 \)

$$\text{Mean} = \frac{7 + 8 + 6 + 9 + 7}{5}$$

$$= \frac{37}{5}$$

$$= 7.4 \text{ days}$$

Answer: 7.4 days

Example 2: The prices (in Ksh) of a particular item from 7 suppliers are: 1500, 1450, 1500, 1600, 1550, 1500, 1490. Find the mode price.

Given: Prices: 1500, 1450, 1500, 1600, 1550, 1500, 1490

Mode = Most frequent value = 1500

Answer: Ksh 1500

Example 3: A set of bids (in Ksh) received are: 12000, 13000, 12500, 14000, 11000, 13500, 11500. Find the median bid.

Given: Bids sorted: 11000, 11500, 12000, 12500, 13000, 13500, 14000

Median = Middle value (4th value) = 12500

Answer: Ksh 12500

Example 4: For the delivery times: 5, 7, 8, 12, 20 days, find the mean and median.

Given: Delivery times: 5, 7, 8, 12, 20

Mean:

$$\frac{5 + 7 + 8 + 12 + 20}{5} = \frac{52}{5} = 10.4 \text{ days}$$

Median: Middle value = 8 days

Answer: Mean = 10.4 days, Median = 8 days

Example 5: Bids received for a contract are: 100000, 105000, 110000, 100000, 115000, 100000, 120000. Find mean, median, and mode.

Given: Bids: 100000, 105000, 110000, 100000, 115000, 100000, 120000

Mean:

$$\frac{100000 + 105000 + 110000 + 100000 + 115000 + 100000 + 120000}{7} = \frac{750000}{7} \approx 107142.86$$

Median: Sorted bids: 100000, 100000, 100000, 105000, 110000, 115000, 120000

Median = 105000

Mode = 100000 (occurs 3 times)

Answer: Mean = Ksh 107,142.86, Median = Ksh 105,000, Mode = Ksh 100,000

4.1.2 Measures of Dispersion: Range and Standard Deviation

Measures of dispersion describe the spread or variability in data, helping procurement managers understand consistency in supplier prices or delivery times.

• The range is the difference between the maximum and minimum values.
• The standard deviation quantifies how much values deviate from the mean on average.

Standard deviation is preferred over range because it considers all data points, not just extremes.

$$\text{Range} = x_{max} - x_{min}$$

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

Where \( s \) is sample standard deviation, \( \bar{x} \) is sample mean, and \( n \) is number of observations.

Worked Examples

Example 1: Supplier delivery times (days): 6, 8, 7, 10, 9. Find the range.

Given: Delivery times: 6, 8, 7, 10, 9

$$\text{Range} = 10 - 6 = 4 \text{ days}$$

Answer: 4 days

Example 2: Calculate the standard deviation of prices (Ksh): 1000, 1200, 1100, 1150, 1050.

Given: Prices: 1000, 1200, 1100, 1150, 1050; \( n=5 \)

Step 1: Calculate mean

$$\bar{x} = \frac{1000 + 1200 + 1100 + 1150 + 1050}{5} = \frac{5500}{5} = 1100$$

Step 2: Calculate squared deviations

$$(1000-1100)^2 = 10000$$

$$(1200-1100)^2 = 10000$$

$$(1100-1100)^2 = 0$$

$$(1150-1100)^2 = 2500$$

$$(1050-1100)^2 = 2500$$

Step 3: Sum squared deviations

$$10000 + 10000 + 0 + 2500 + 2500 = 25000$$

Step 4: Calculate variance

$$s^2 = \frac{25000}{5-1} = \frac{25000}{4} = 6250$$

Step 5: Calculate standard deviation

$$s = \sqrt{6250} \approx 79.06$$

Answer: Standard deviation = Ksh 79.06

Example 3: Delivery times (days): 5, 7, 8, 12, 20. Calculate range and standard deviation.

Given: Delivery times: 5, 7, 8, 12, 20; \( n=5 \)

Range:

$$20 - 5 = 15 \text{ days}$$

Mean:

$$\frac{5 + 7 + 8 + 12 + 20}{5} = \frac{52}{5} = 10.4$$

Squared deviations:

$$(5-10.4)^2 = 29.16$$

$$(7-10.4)^2 = 11.56$$

$$(8-10.4)^2 = 5.76$$

$$(12-10.4)^2 = 2.56$$

$$(20-10.4)^2 = 92.16$$

Sum:

$$29.16 + 11.56 + 5.76 + 2.56 + 92.16 = 141.2$$

Variance:

$$\frac{141.2}{4} = 35.3$$

Standard deviation:

$$\sqrt{35.3} \approx 5.94$$

Answer: Range = 15 days, Standard deviation = 5.94 days

Example 4: Bids (Ksh): 100000, 105000, 110000, 100000, 115000, 100000, 120000. Calculate range and standard deviation.

Given: Bids: 100000, 105000, 110000, 100000, 115000, 100000, 120000; \( n=7 \)

Range:

$$120000 - 100000 = 20000$$

Mean:

$$\frac{100000 + 105000 + 110000 + 100000 + 115000 + 100000 + 120000}{7} = 107142.86$$

Squared deviations:

$$(100000 - 107142.86)^2 = 51020408.16$$

$$(105000 - 107142.86)^2 = 4493877.55$$

$$(110000 - 107142.86)^2 = 8163265.31$$

$$(100000 - 107142.86)^2 = 51020408.16$$

$$(115000 - 107142.86)^2 = 61632653.06$$

$$(100000 - 107142.86)^2 = 51020408.16$$

$$(120000 - 107142.86)^2 = 165306122.45$$

Sum:

$$51020408.16 + 4493877.55 + 8163265.31 + 51020408.16 + 61632653.06 + 51020408.16 + 165306122.45 = 397346042.85$$

Variance:

$$\frac{397346042.85}{6} = 66224340.48$$

Standard deviation:

$$\sqrt{66224340.48} \approx 8137.17$$

Answer: Range = Ksh 20,000, Standard deviation = Ksh 8,137.17

Example 5: Prices (Ksh): 1500, 1450, 1500, 1600, 1550, 1500, 1490. Calculate range and standard deviation.

Given: Prices: 1500, 1450, 1500, 1600, 1550, 1500, 1490; \( n=7 \)

Range:

$$1600 - 1450 = 150$$

Mean:

$$\frac{1500 + 1450 + 1500 + 1600 + 1550 + 1500 + 1490}{7} = \frac{10590}{7} = 1512.86$$

Squared deviations:

$$(1500 - 1512.86)^2 = 163.27$$

$$(1450 - 1512.86)^2 = 3955.10$$

$$(1500 - 1512.86)^2 = 163.27$$

$$(1600 - 1512.86)^2 = 7654.69$$

$$(1550 - 1512.86)^2 = 1377.55$$

$$(1500 - 1512.86)^2 = 163.27$$

$$(1490 - 1512.86)^2 = 522.45$$

Sum:

$$163.27 + 3955.10 + 163.27 + 7654.69 + 1377.55 + 163.27 + 522.45 = 13899.60$$

Variance:

$$\frac{13899.60}{6} = 2316.60$$

Standard deviation:

$$\sqrt{2316.60} \approx 48.13$$

Answer: Range = Ksh 150, Standard deviation = Ksh 48.13

Chapter Summary

This chapter provides a comprehensive introduction to elementary statistics, beginning with the fundamentals of data collection and its significance in statistical analysis. It explains the two main methods of data collection: primary data gathered directly from sources and secondary data obtained from existing records. Sampling techniques are explored, distinguishing between probability sampling, which involves random selection, and non-probability sampling, where selection is based on non-random criteria. The chapter then details various methods of data presentation, emphasizing the use of tables and diagrams such as frequency distribution tables, bar charts, pie charts, histograms, and frequency polygons to effectively organize and visualize data. Different types of graphs are also covered, including basic time series graphs, z-charts, Lorenz curves, and semi-log graphs, each serving unique purposes in data interpretation. Finally, the chapter discusses cumulative frequency curves, commonly known as ogives, which are valuable for understanding data distribution and percentiles. Together, these topics establish a solid foundation for collecting, sampling, presenting, and interpreting statistical data.