A healthcare consultant wants to compare the patient satisfaction ratings of two hospitals. The consultant performs a randomization test for 2sample means to determine whether there is a difference in the patient ratings between the hospitals.
 Open the sample data, HospitalComparison.MTW.
 Open the randomization test for 2sample means dialog box.
 In Samples, enter Rating.
 In Sample IDs, enter Hospital.
 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.
 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 pvalue 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.

μ₁: mean of Rating when Hospital = A  μ₂: mean of Rating when Hospital = B  Difference: μ₁  μ₂ 


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 


Null hypothesis  H₀: μ₁  μ₂ = 0  Alternative hypothesis  H₁: μ₁  μ₂ ≠ 0 



