Question: Suppose we wish to compare the length of stay in the hospital for patients at two different hospitals with the same diagnosis. Suppose the data was as follows:

Hospital 1: 21, 10, 32, 60, 8, 44, 29, 5, 13, 26, 33, 54, 45, 100

Hospital 2: 86, 27, 10, 68, 87, 76, 125, 60, 35, 73, 96, 44, 120.

Use the following output to conduct the appropriate test using α=0.05.

In addition to running the test, tell me why you choose the test you that you used – i.e. paired t-test, two independent sample assuming equal variances, two-independent sample not assuming equal variances, Wilcoxon Signed Rank test, or Wilcoxon Rank Sum test and why.

The TTEST Procedure

Statistics

Variable Hospital N Mean Mean Mean Std Dev Std Dev Std Dev Std Err Min Max

LengthStay 1 14 19.566 34.286 49.005 18.482 25.493 41.071 6.8134 5 100

LengthStay 2 13 48.949 69.769 90.589 24.706 34.453 56.873 9.5556 10 125

LengthStay Diff (1-2) -59.38 -35.48 -11.58 23.628 30.129 41.59 11.604

T-Tests

Variable Method Variances DF t Value Pr > |t|

LengthStay Pooled Equal 25 -3.06 0.0053

LengthStay Satterthwaite Unequal 22 -3.02 0.0062

Equality of Variances

Variable Method Num DF Den DF F Value Pr > F

LengthStay Folded F 12 13 1.83 0.2951

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The NPAR1WAY Procedure

Wilcoxon Scores (Rank Sums) for Variable LengthStay

Classified by Variable Hospital

Sum of Expected Std Dev Mean

Hospital N Scores Under H0 Under H0 Score

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1 14 141.50 196.0 20.598004 10.107143

2 13 236.50 182.0 20.598004 18.192308

Average scores were used for ties.

Wilcoxon Two-Sample Test

Statistic 236.5000

Normal Approximation

Z 2.6216

One-Sided Pr > Z 0.0044

Two-Sided Pr > |Z| 0.0088

t Approximation

One-Sided Pr > Z 0.0072

Two-Sided Pr > |Z| 0.0144

Z includes a continuity correction of 0.5.

Kruskal-Wallis Test

Chi-Square 7.0007

DF 1

Pr > Chi-Square 0.0081

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Solution: The downloadable solution consists of 3 pages
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