# 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. Open the randomization test for 2-sample means dialog box.
• Mac: Statistics > Resampling > Randomization Test for 2-Sample Means
• PC: STATISTICS > Resampling > Randomization Tests > 2-Sample Means
3. In Samples, enter Rating.
4. In Sample IDs, enter Hospital.
5. On the Options tab, enter 1 in Base for random number generator.
###### Note

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

6. Click OK.

## 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.

 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 80.300 8.183 66.958 62.000 79.000 98.000 B 20 59.300 12.431 154.537 35.000 58.500 89.000
 Difference in Observed Means
 Mean of A - Mean of B = 21
 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 <0.0020
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