By the end of this chapter, you will be able to:
- correctly classify different types of health statistics based on their specific functions
- identify reliable sources of health data by following standard operating procedures (SOPs) accurately
Mastering these skills helps you make informed decisions that improve community health outcomes effectively.
The application of health statistics is essential for community health professionals in Kenya to understand disease patterns, evaluate health programs, and make informed decisions. Health statistics enable the monitoring of population health indicators, prediction of future health trends, and assessment of interventions’ effectiveness. This chapter introduces the fundamental types of health statistics, descriptive, predictive, inferential, and correlational, and provides practical mathematical examples relevant to community health settings.
Health statistics are quantitative data used to describe, analyze, and predict health-related phenomena in populations. They support evidence-based decision-making in health policy, resource allocation, and program evaluation. Understanding different types of statistics allows community health workers to apply appropriate methods for data interpretation and health planning.
Descriptive statistics summarize and organize health data to provide clear insights into health status and disease distribution in a population. They include measures of central tendency (mean, median, mode), dispersion (range, variance, standard deviation), and frequency distributions.
Measures of central tendency describe the average or typical value in a dataset. The mean is the arithmetic average, the median is the middle value when data are ordered, and the mode is the most frequently occurring value.
$$\text{Mean } (\bar{x}) = \frac{\Sigma x_i}{n}$$
Example 1: A community health nurse records the number of new malaria cases per week in a village over 5 weeks: 12, 15, 10, 20, 13 cases. Calculate the mean number of cases.
Given: \(x_1=12\), \(x_2=15\), \(x_3=10\), \(x_4=20\), \(x_5=13\), \(n=5\)
$$\bar{x} = \frac{12 + 15 + 10 + 20 + 13}{5}$$
$$\bar{x} = \frac{70}{5}$$
$$\bar{x} = 14 \text{ cases}$$
Answer: 14 cases
Example 2: Find the median of the weekly malaria cases above.
Order data: 10, 12, 13, 15, 20
Median = middle value = 13 cases
Answer: 13 cases
Example 3: Find the mode of weekly malaria cases: 12, 15, 10, 20, 13, 15, 10, 15
Frequency: 15 (3 times), 10 (2 times), others once
Mode = 15 cases
Answer: 15 cases
Dispersion measures show how data values spread around the central tendency, indicating variability in health data.
$$\text{Variance } (s^2) = \frac{\Sigma (x_i - \bar{x})^2}{n-1}$$
$$\text{Standard deviation } (s) = \sqrt{s^2}$$
Example 1: Calculate the range of weekly malaria cases: 12, 15, 10, 20, 13.
Range = max, min = 20 - 10 = 10 cases
Answer: 10 cases
Example 2: Calculate variance and standard deviation for the same data.
Given: \(x_i = 12, 15, 10, 20, 13\), \(\bar{x} = 14\), \(n=5\)
Calculate squared deviations:
\((12 - 14)^2 = 4\)
\((15 - 14)^2 = 1\)
\((10 - 14)^2 = 16\)
\((20 - 14)^2 = 36\)
\((13 - 14)^2 = 1\)
Sum = 4 + 1 + 16 + 36 + 1 = 58
Variance:
$$s^2 = \frac{58}{5-1} = \frac{58}{4} = 14.5$$
Standard deviation:
$$s = \sqrt{14.5} = 3.81 \text{ cases}$$
Answer: Variance = 14.5 cases\(^2\), Standard deviation = 3.81 cases
Example 3: Interpret the standard deviation in the context of malaria cases.
The standard deviation of 3.81 cases indicates that the weekly malaria cases typically vary by about 4 cases from the mean of 14.
Frequency distributions organize health data into categories or intervals, showing how often each value occurs. Percentages express these frequencies relative to the total population, facilitating easy comparison.
Example 1: In a survey of 200 households, 50 reported diarrhoea cases last month. Calculate the percentage of affected households.
Given: cases = 50, total = 200
$$\text{Percentage} = \frac{50}{200} \times 100 = 25\%$$
Answer: 25%
Example 2: The ages of 100 patients are grouped as follows: 0-9 years (20), 10-19 years (30), 20-29 years (25), 30-39 years (15), 40+ years (10). Calculate the percentage for each group.
Calculate percentages:
0-9: \(\frac{20}{100} \times 100 = 20\%\)
10-19: \(30\%\)
20-29: \(25\%\)
30-39: \(15\%\)
40+: \(10\%\)
Answer: Percentages as above
Example 3: Construct a frequency distribution table for the above data.
| Age Group (years) | Frequency | Percentage |
|---|---|---|
| 0-9 | 20 | 20% |
| 10-19 | 30 | 30% |
| 20-29 | 25 | 25% |
| 30-39 | 15 | 15% |
| 40+ | 10 | 10% |
Predictive statistics use current and historical health data to forecast future health events or trends, guiding proactive interventions. They often involve regression models and time series analysis to estimate disease incidence or healthcare needs.
Simple linear regression models the relationship between a dependent health variable and one independent variable to predict future outcomes.
$$y = a + bx$$
Where:
\(y\) = predicted value (e.g., disease cases)
\(a\) = intercept
\(b\) = slope (rate of change)
\(x\) = independent variable (e.g., time)
Example 1: A clinic records monthly malaria cases for 6 months: 10, 12, 15, 20, 18, 22. Predict cases in month 7 using linear regression.
Given: \(x = 1,2,3,4,5,6\), \(y = 10,12,15,20,18,22\)
Calculate slope \(b\) and intercept \(a\).
Calculate means:
$$\bar{x} = \frac{1+2+3+4+5+6}{6} = 3.5$$
$$\bar{y} = \frac{10+12+15+20+18+22}{6} = 16.17$$
Calculate numerator and denominator for \(b\):
$$\Sigma (x_i - \bar{x})(y_i - \bar{y}) = (1-3.5)(10-16.17) + \dots + (6-3.5)(22-16.17)$$
Calculate each term:
\((1-3.5)(10-16.17) = (-2.5)(-6.17) = 15.43\)
\((2-3.5)(12-16.17) = (-1.5)(-4.17) = 6.25\)
\((3-3.5)(15-16.17) = (-0.5)(-1.17) = 0.59\)
\((4-3.5)(20-16.17) = (0.5)(3.83) = 1.92\)
\((5-3.5)(18-16.17) = (1.5)(1.83) = 2.75\)
\((6-3.5)(22-16.17) = (2.5)(5.83) = 14.58\)
Sum numerator = 15.43 + 6.25 + 0.59 + 1.92 + 2.75 + 14.58 = 41.52
Denominator:
$$\Sigma (x_i - \bar{x})^2 = (-2.5)^2 + (-1.5)^2 + (-0.5)^2 + 0.5^2 + 1.5^2 + 2.5^2$$
$$= 6.25 + 2.25 + 0.25 + 0.25 + 2.25 + 6.25 = 17.5$$
Calculate slope:
$$b = \frac{41.52}{17.5} = 2.37$$
Calculate intercept:
$$a = \bar{y} - b \bar{x} = 16.17 - 2.37 \times 3.5 = 16.17 - 8.30 = 7.87$$
Predict month 7:
$$y = 7.87 + 2.37 \times 7 = 7.87 + 16.59 = 24.46$$
Answer: Predicted malaria cases in month 7 = 24 cases
Example 2: Predict cases for month 8 using the same model.
$$y = 7.87 + 2.37 \times 8 = 7.87 + 18.96 = 26.83$$
Answer: 27 cases
Example 3: Interpret the slope \(b=2.37\) in the health context.
Each additional month increases malaria cases by approximately 2.37 cases.
Logistic regression predicts the probability of a binary health outcome (e.g., presence or absence of disease) based on predictor variables.
$$P = \frac{1}{1 + e^{-(a + bx)}}$$
Where:
\(P\) = probability of event
\(a\) = intercept, \(b\) = coefficient, \(x\) = predictor
Example 1: A health survey predicts the probability of hypertension based on age (years). Given \(a = -5\), \(b = 0.1\), calculate probability for a 50-year-old.
$$P = \frac{1}{1 + e^{-(-5 + 0.1 \times 50)}} = \frac{1}{1 + e^{-(-5 + 5)}} = \frac{1}{1 + e^{0}} = \frac{1}{1 + 1} = 0.5$$
Answer: Probability of hypertension = 50%
Example 2: Calculate probability for a 60-year-old.
$$P = \frac{1}{1 + e^{-(-5 + 0.1 \times 60)}} = \frac{1}{1 + e^{-(-5 + 6)}} = \frac{1}{1 + e^{-1}} = \frac{1}{1 + 0.3679} = 0.731$$
Answer: 73.1%
Example 3: Calculate probability for a 40-year-old.
$$P = \frac{1}{1 + e^{-(-5 + 0.1 \times 40)}} = \frac{1}{1 + e^{-(-5 + 4)}} = \frac{1}{1 + e^{1}} = \frac{1}{1 + 2.718} = 0.269$$
Answer: 26.9%
Time series analysis examines health data collected over time to identify trends and seasonal patterns for forecasting.
Example 1: Monthly clinic visits for respiratory infections over 6 months: 120, 130, 125, 135, 140, 150. Calculate the average monthly increase.
Given: data points \(y_i\)
Calculate differences:
130 - 120 = 10
125 - 130 = -5
135 - 125 = 10
140 - 135 = 5
150 - 140 = 10
Average increase:
$$\frac{10 + (-5) + 10 + 5 + 10}{5} = \frac{30}{5} = 6 \text{ visits/month}$$
Answer: Average monthly increase is 6 visits
Example 2: Predict visits in month 7.
Last month visits = 150
Predicted visits = 150 + 6 = 156 visits
Answer: 156 visits
Predictive models require evaluation using metrics such as mean squared error (MSE) or accuracy to assess their performance in health data forecasting.
Example 1: Calculate MSE for predicted malaria cases vs actual: predicted (20, 22, 25), actual (18, 24, 27).
Calculate errors:
(20-18)^2 = 4
(22-24)^2 = 4
(25-27)^2 = 4
MSE:
$$\frac{4 + 4 + 4}{3} = \frac{12}{3} = 4$$
Answer: MSE = 4 cases\(^2\)
Example 2: Calculate root mean squared error (RMSE).
$$\sqrt{4} = 2 \text{ cases}$$
Answer: RMSE = 2 cases
Inferential statistics allow community health professionals to draw conclusions about a population based on sample data, using hypothesis testing and confidence intervals.
Hypothesis testing evaluates assumptions about population parameters using sample data, commonly via t-tests or chi-square tests.
Example 1: A sample of 30 children has an average weight of 15 kg with standard deviation 2 kg. Test if the mean weight differs from the national average of 14 kg at 5% significance.
Given: \(n=30\), \(\bar{x}=15\), \(s=2\), \(\mu_0=14\), \(\alpha=0.05\)
Calculate t-statistic:
$$t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} = \frac{15 - 14}{2/\sqrt{30}} = \frac{1}{2/5.477} = \frac{1}{0.365} = 2.74$$
Degrees of freedom = 29
Critical t-value (two-tailed) ≈ 2.045
Since 2.74 > 2.045, reject null hypothesis.
Answer: Mean weight significantly differs from 14 kg
Example 2: Test if the proportion of vaccinated children differs from 80% in a sample of 100 with 85 vaccinated.
Calculate sample proportion:
$$p = \frac{85}{100} = 0.85$$
Standard error:
$$SE = \sqrt{\frac{0.8 \times 0.2}{100}} = \sqrt{0.0016} = 0.04$$
Calculate z-score:
$$z = \frac{0.85 - 0.8}{0.04} = 1.25$$
Critical z-value at 5% significance = 1.96
Since 1.25 < 1.96, fail to reject null hypothesis.
Answer: No significant difference from 80%
Confidence intervals estimate the range within which a population parameter lies with a given level of confidence.
$$CI = \bar{x} \pm z \times \frac{s}{\sqrt{n}}$$
Example 1: Calculate a 95% confidence interval for mean blood pressure of 40 patients with mean 120 mmHg, standard deviation 15 mmHg.
Given: \(n=40\), \(\bar{x}=120\), \(s=15\), \(z=1.96\)
Calculate margin of error:
$$ME = 1.96 \times \frac{15}{\sqrt{40}} = 1.96 \times 2.37 = 4.64$$
Confidence interval:
$$120 \pm 4.64 = (115.36, 124.64) \text{ mmHg}$$
Answer: 95% CI = 115.36 to 124.64 mmHg
Example 2: Calculate 90% confidence interval for average height of 25 children, mean 130 cm, standard deviation 10 cm.
\(z = 1.645\)
$$ME = 1.645 \times \frac{10}{\sqrt{25}} = 1.645 \times 2 = 3.29$$
CI:
$$130 \pm 3.29 = (126.71, 133.29) \text{ cm}$$
Answer: 90% CI = 126.71 to 133.29 cm
Sample size calculations ensure sufficient data for reliable inferences in community health studies.
$$n = \frac{z^2 p (1-p)}{d^2}$$
Where:
\(z\) = z-value for confidence level
\(p\) = estimated proportion
\(d\) = margin of error
Example 1: Calculate sample size for estimating vaccination coverage with 95% confidence, 5% margin of error, and estimated coverage 80%.
Given: \(z=1.96\), \(p=0.8\), \(d=0.05\)
$$n = \frac{(1.96)^2 \times 0.8 \times 0.2}{0.05^2} = \frac{3.8416 \times 0.16}{0.0025} = \frac{0.6147}{0.0025} = 246$$
Answer: Sample size = 246
Example 2: Calculate sample size for 90% confidence, 10% margin of error, estimated disease prevalence 10%.
\(z=1.645\), \(p=0.1\), \(d=0.1\)
$$n = \frac{(1.645)^2 \times 0.1 \times 0.9}{0.1^2} = \frac{2.706 \times 0.09}{0.01} = \frac{0.2435}{0.01} = 24.35$$
Answer: Sample size = 25
ANOVA compares means across three or more groups to identify significant differences in health outcomes.
Example 1: Three clinics report average patient wait times (minutes): Clinic A (mean=15, n=10), Clinic B (mean=20, n=10), Clinic C (mean=18, n=10). Test if means differ significantly (conceptual).
ANOVA involves calculating between-group and within-group variances and the F-statistic. Detailed calculations require raw data, but the concept is to compare variances to conclude on differences.
Answer: If calculated F > critical F, conclude significant differences in wait times.
Correlational statistics measure the strength and direction of relationships between two health variables, informing potential associations.
Pearson’s \(r\) quantifies linear correlation between two continuous variables ranging from -1 to +1.
$$r = \frac{n \Sigma xy - \Sigma x \Sigma y}{\sqrt{(n \Sigma x^2 - (\Sigma x)^2)(n \Sigma y^2 - (\Sigma y)^2)}}$$
Example 1: Calculate \(r\) between hours of exercise (x) and blood pressure (y) for 5 patients:
| x | y |
|---|---|
| 2 | 130 |
| 3 | 125 |
| 5 | 120 |
| 4 | 122 |
| 6 | 118 |
Calculate sums:
\(\Sigma x = 20\), \(\Sigma y = 615\), \(\Sigma xy = 2380\), \(\Sigma x^2 = 90\), \(\Sigma y^2 = 75889\), \(n=5\)
Calculate numerator:
$$5 \times 2380 - 20 \times 615 = 11900 - 12300 = -400$$
Calculate denominator:
$$\sqrt{(5 \times 90 - 20^2)(5 \times 75889 - 615^2)} = \sqrt{(450 - 400)(379445 - 378225)} = \sqrt{50 \times 1220} = \sqrt{61000} = 246.98$$
Calculate \(r\):
$$r = \frac{-400}{246.98} = -1.62$$
Since |r| > 1, check calculations (likely arithmetic error).
Recalculate \(\Sigma xy\):
Calculate \(xy\) for each:
2*130=260
3*125=375
5*120=600
4*122=488
6*118=708
Sum:
260 + 375 + 600 + 488 + 708 = 2431
Recalculate numerator:
$$5 \times 2431 - 20 \times 615 = 12155 - 12300 = -145$$
Denominator as above = 246.98
$$r = \frac{-145}{246.98} = -0.587$$
Answer: Pearson correlation coefficient \(r = -0.59\)
Example 2: Interpret the result.
An \(r\) of -0.59 indicates a moderate negative correlation: as exercise hours increase, blood pressure tends to decrease.
Example 3: Calculate \(r\) for a dataset with no relationship:
| x | y |
|---|---|
| 1 | 100 |
| 2 | 110 |
| 3 | 90 |
| 4 | 120 |
| 5 | 95 |
Calculate sums:
\(\Sigma x = 15\), \(\Sigma y = 515\), \(\Sigma xy = 1530\), \(\Sigma x^2 = 55\), \(\Sigma y^2 = 53575\), \(n=5\)
Numerator:
$$5 \times 1530 - 15 \times 515 = 7650 - 7725 = -75$$
Denominator:
$$\sqrt{(5 \times 55 - 15^2)(5 \times 53575 - 515^2)} = \sqrt{(275 - 225)(267875 - 265225)} = \sqrt{50 \times 2650} = \sqrt{132500} = 364.04$$
Calculate \(r\):
$$r = \frac{-75}{364.04} = -0.206$$
Answer: \(r = -0.21\), weak negative correlation
Spearman’s correlation measures monotonic relationships using ranked data, useful when data are ordinal or not normally distributed.
Example 1: Calculate Spearman’s \(r_s\) for ranks of patient satisfaction and recovery days:
| Patient | Satisfaction Rank (x) | Recovery Days Rank (y) |
|---|---|---|
| A | 1 | 5 |
| B | 2 | 4 |
| C | 3 | 3 |
| D | 4 | 2 |
| E | 5 | 1 |
Calculate difference \(d = x, y\):
| Patient | \(d\) | \(d^2\) |
|---|---|---|
| A | -4 | 16 |
| B | -2 | 4 |
| C | 0 | 0 |
| D | 2 | 4 |
| E | 4 | 16 |
Sum \(d^2 = 40\), \(n=5\)
Calculate \(r_s\):
$$r_s = 1 - \frac{6 \Sigma d^2}{n(n^2 - 1)} = 1 - \frac{6 \times 40}{5 \times (25 - 1)} = 1 - \frac{240}{120} = 1 - 2 = -1$$
Answer: Perfect negative correlation \(r_s = -1\)
Scatter plots visually display the relationship between two health variables, aiding in assessing correlation strength and direction.
Example 1: Given a scatter plot of vaccination coverage vs incidence of measles, describe the relationship if points trend downward.
Interpretation: Negative correlation; higher vaccination coverage associates with lower measles incidence.
Example 2: Scatter plot of age vs blood pressure with no clear pattern suggests no correlation.
The coefficient of determination \(r^2\) measures the proportion of variance in one variable explained by another.
$$r^2 = (r)^2$$
Example 1: Calculate \(r^2\) for \(r = -0.59\) from previous example.
$$r^2 = (-0.59)^2 = 0.3481$$
Answer: 34.8% of variance in blood pressure explained by exercise
Example 2: For \(r = 0.8\), calculate \(r^2\).
$$r^2 = 0.64$$
Answer: 64% variance explained
Calculate the mean, median, and mode of weekly diarrhoea cases recorded as 8, 12, 15, 8, 10, 12, 8. (6 marks)
A community clinic’s monthly patient numbers over 6 months are 150, 160, 170, 180, 190, 210. Using simple linear regression, predict the number of patients in the 7th month. (8 marks)
Calculate a 95% confidence interval for the mean height of 50 children with mean 140 cm and standard deviation 12 cm. (7 marks)
Given the following data, calculate the Pearson correlation coefficient between hours of physical activity and blood sugar level:
| Hours | Blood Sugar (mg/dL) |
|---|---|
| 1 | 150 |
| 2 | 140 |
| 3 | 130 |
| 4 | 120 |
| 5 | 115 |
(9 marks)
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Create a free accountThis chapter introduced the fundamental concept of health statistics and its importance in community health. It outlined four main types of health statistics: descriptive, which involves summarizing health data; predictive, which forecasts future health trends; inferential, which draws conclusions from sample data; and correlational, which examines relationships between health variables. The chapter also detailed various sources of health data, emphasizing the need for accurate and reliable information. It discussed different types of data, including qualitative and quantitative forms, and explained the scales of measurement such as nominal, ordinal, interval, and ratio. Understanding these concepts is essential for effective data collection, analysis, and interpretation in health studies. This foundation supports informed decision-making and improved health outcomes in community settings.
A clinic records the number of patient visits over 5 days as 45, 50, 55, 60, and 70. Calculate the mean number of visits per day. (2 marks)
In a health survey, 120 people were tested for a disease. 36 tested positive. Calculate the prevalence rate of the disease as a percentage. (2 marks)
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