Example of Randomization test for 2-sample means

A healthcare consultant wants to compare the patient satisfaction ratings of two hospitals. The consultant performs a randomization test for 2-sample means to determine whether there is a difference in the patient ratings between the hospitals.

  1. Open the sample data, HospitalComparison.MTW.
  2. Choose Calc > Resampling > Randomization Test for 2-Sample Means.
  3. In Samples, enter Rating.
  4. In Sample IDs, enter Hospital.
  5. Click Options. Enter 1 in Base for random number generator.

    Using Base for random number generator ensures that your results match the example.

  6. Click OK in each dialog box.

Interpret the results

The null hypothesis states that the difference in the patient ratings between the hospitals is equal to 0. Because the p-value is less than 0.002, which is less than the significance level of 0.05, the consultant rejects the null hypothesis and concludes that the difference in patient ratings between the hospitals is not equal to 0. The histogram shows that the bootstrap distribution appears to be normal, so the consultant can trust the results.

The difference in observed means is 21, indicating that hospital A has higher patient satisfaction ratings than hospital B.

Randomization Test for Difference in Means: Rating by Hospital

Randomization Test Histogram for Rating by Hospital

Method μ₁: mean of Rating when Hospital = A µ₂: mean of Rating when Hospital = B Difference: μ₁ - µ₂
Observed Samples Hospital N Mean StDev Variance Minimum Median Maximum A 20 80.30 8.18 66.96 62.00 79.00 98.00 B 20 59.30 12.43 154.54 35.00 58.50 89.00
Difference in Observed Means Mean of A - Mean of B = 21.000
Randomization Test Null hypothesis H₀: μ₁ - µ₂ = 0 Alternative hypothesis H₁: μ₁ - µ₂ ≠ 0
Number of Resamples Average StDev P-Value 1000 -0.185 4.728 < 0.002