Skewed distributionsBMJ 2012; 345 doi: https://doi.org/10.1136/bmj.e7534 (Published 08 November 2012) Cite this as: BMJ 2012;345:e7534
- Philip Sedgwick, reader in medical statistics and medical education
- 1Centre for Medical and Healthcare Education, St George’s, University of London, London, UK
The effectiveness of early abdominopelvic computed tomography in patients with acute abdominal pain of unknown cause was evaluated using a randomised controlled trial study design. Computed tomography was delivered within 24 hours of admission. Control treatment was standard practice (radiological investigations as indicated). In total, 55 patients were randomised to early computed tomography and 55 to control treatment.1
The main outcome measures included length of hospital stay. The mean length of hospital stay was 6.6 (standard deviation 5.8) days for the early computed tomography arm and 9.2 (9.8) days for the standard practice arm.
Which of the following statements, if any, are true?
a) The distribution of length of hospital stay for the intervention was skewed to the right
b) The sample mean of length of hospital stay for the intervention was smaller than the sample median
c) Treatment groups would be compared in length of hospital stay using the independent samples t test
Statement a is true, whereas b and c are false.
The spread of sample measurements for the outcome measure length of hospital stay can be described by the sample mean and standard deviation. As described in a previous question, …
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