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Example 3

T-Test — Hub A vs Hub B Turnaround Time

Is Hub A (mean = 2.20 days) genuinely faster than Hub B (mean = 3.81 days), or is the difference just random variation?

Hub A Mean

2.20days

Hub A SD

0.18days

Hub B Mean

3.81days

Hub B SD

0.24days

Hub A — Observations

2.022.052.102.142.182.222.252.282.352.41

Hub B — Observations

3.453.523.603.683.753.823.903.984.054.35

Hub Comparison — Mean Turnaround Time

Hub AHub B025Days
Hub A (μ = 2.2 days)
Hub B (μ = 3.81 days)
Step-by-Step

Independent Samples T-Test Calculation

We calculate the Standard Error of the Mean (SEM) for each hub, then the Standard Error of the Difference (SED), and finally the t-statistic.

Step 1 — SEM for each hub

1.SEM_A = SD_A / √n = 0.18 / √10
2.SEM_A = 0.18 / 3.162 = 0.057
3.
4.SEM_B = SD_B / √n = 0.24 / √10
5.SEM_B = 0.24 / 3.162 = 0.076

Step 2 — Standard Error of Difference

1.SED = √(SEM_A² + SEM_B²)
2.SED = √(0.057² + 0.076²)
3.SED = √(0.003240 + 0.005760)
4.SED = √0.009000
ResultSED = 0.095

Step 3 — T-Statistic

1.t = (X̄_A − X̄_B) / SED
2.t = (2.2 − 3.81) / 0.095
3.t = -1.61 / 0.095
Resultt = -16.97
1

Calculate Standard Error of Mean (SEM) for Hub A

SEM_A = 0.18 / √10 = 0.057
2

Calculate SEM for Hub B

SEM_B = 0.24 / √10 = 0.076
3

Combine into Standard Error of Difference (SED)

SED = √(0.057² + 0.076²) = 0.095
4

Calculate the t-statistic

t = (2.2 − 3.81) / 0.095 = -16.97
5

Compare |t| to critical value (df = 18, α = 0.05)

|-16.97| = 16.97 >> 2.101 ✓
Decision

Hypothesis Test Result

With df = 18 and α = 0.05, the critical value is ±2.101.

t-statistic

-16.97

Critical Value

±2.101

Degrees of Freedom

18

Decision Rule Visualization

-2.101
+2.101
0
t = -16.97
Reject H₀Fail to Reject H₀Reject H₀
Since |t| = 16.97 is far greater than the critical value of 2.101, we reject the null hypothesis.
Hub A is significantly faster than Hub B. This is a real performance gap, not random noise.
Actionable insight: This evidence can justify vendor contract renegotiation or hub reallocation to optimize patient outcomes.