A clinic has four doctors. They wish to compare the amount of time that doctors spend with their patients. It is suspected that times may not be normally distributed but are otherwise assumed similar except in center, so a nonparametric test was chosen. The Kruskal-Wallis test results are shown below. $\text { Kruskal-Wallis Test }$
$\begin{array}{ccrl}
\hline \text { Wedian } & n & \text { Avg. Rank } & \
\hline 27.00 & 11 & 19.18 & \text { Doc A } \
28.50 & 14 & 15.93 & \text { Doc B } \
46.00 & 9 & 36.39 & \text { Doc C } \
31.00 & 11 & 24.86 & \text { Doc D } \
\hline 31.00 & 45 & & \text { Total }
\end{array}$
14.599 $ \mathrm{H} $ (corrected for ties)
3 d.f.
, 0022 p-value Which is the best conclusion?
A)The doctors have the same median times at ? = .01.
B)The doctors have different median times at ? = .01.
C)The sample size is insufficient for this kind of test.
D)A one-factor ANOVA would be a better kind of test.

Question

Quizplus · Accepted Answer

The Kruskal-Wallis test was used to compare the amount of time that doctors spend with their patients, and the p-value of 0.0022 indicates that there is a significant difference in median times between the doctors. Therefore, choice B is the best conclusion. Choice A is incorrect as it states that the doctors have the same median times, which is contradicted by the results of the Kruskal-Wallis test. Choice C is incorrect as the sample size of 45 is sufficient for the Kruskal-Wallis test. Choice D is incorrect as a one-factor ANOVA assumes that the data is normally distributed, while the prompt states that times may not be normally distributed.

A Clinic Has Four Doctors Kruskal-Wallis Test \text { Kruskal-Wallis Test } Kruskal-Wallis Test

A Clinic Has Four Doctors $\text { Kruskal-Wallis Test }$