Paper Example Undergraduate 335 words

Mann-Whitney U Test the Mean

Last reviewed: June 29, 2010 ~2 min read

¶ … Mann-Whitney U Test

The mean length of stay will be equal to 14 days.

The hypothesis is rejected. Based on the one-sample t-test, the mean length of stay in this sample is not 14, nor does 14 days fall within the range of the 95% confidence intervals.

A Mann-Whitney U test was conducted using gender as a nominal grouping variable. The results indicate that there is not a significant difference in the length of stay based on gender (p. 229).

Kruskal-Wallis H Test

A One-Way ANOVA was conducted using Age (3 levels) and number of cardiac risk factors. No significant difference was found between the three age groups in the number of cardiac risk factors.

The median number of risk factors will differ based on age group.

The hypothesis is rejected. The results of the Kruskal-Wallis Test indicate that there is no significant difference between the median number of cardiac risk factors between the various age groups, X2 (2, 84) = 2.611, p = .271.

Chi Square Test

Hypothesis: The proportion of men and women alive at discharge will be the same.

They hypothesis is rejected. The Chi-Square was significant X2 (1, 84) = 4.403, p < .05. The proportion of females alive at discharge was much greater than the proportion of males alive at discharge. This can be seen in the barchart below.

alive discharge * gender Crosstabulation

gender

Total

male female alive discharge no

Count

7

5

12

Expected Count

3.9

8.1

12.0

yes

Count

20

52

72

Expected Count

23.1

48.9

72.0

Total

Count

27

57

84

Expected Count

27.0

57.0

84.0

4) Correlation & Regression

Variables: Hours in Operating Room & Number of Preop Risk Factors

Scatter Plot:

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PaperDue. (2010). Mann-Whitney U Test the Mean. PaperDue. https://paperdue.com/essay/mann-whitney-u-test-the-mean-9995

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