Question

An article in the American Journal of Emergency Medicine compared the ability to detect acute traumatic aortic injury (ATAI) on cervical x-ray images. The data consisted of 13 cases of ATAI, 19 cases with negative aortography (NAO) and 18 cases with multiple trauma (MT) without aortography. Measurements of the cervical soft-tissue width at the third cervical vertebrae generated the following results. Averages were 9.2, 8.1 and 7.8 mm and standard deviations were 2.2, 3.3 and 3 mm for the ATAI, NAO, and MT groups, respectively. Consider an ANOVA for these data. a) Use the sample sizes and group averages to compute the grand mean width and the mean square for treatments. Round your answer to two decimal places (e.g. 9.87). u= MS Treatments b) Use the sample sizes and standard deviations in the groups to compute the mean square error. Round your answer to two decimal places (e.g. 9.87). MSError = c) Use the results in the previous parts to conduct the ANOVA test and prove that the mean widths do not differ. Calculate the P-value for the test. F Source DF SS (to two decimal places) MS (to two decimal places) (to two decimal places) P (to three decimal places) Treatments Error Total

Answer 1

Group	ni	x̅i	nx̅i	ni*(Xi-Xgrand)2	SS=(ni-1)*s2
1	13	9.2	119.60	11.0511	58.0800
2	19	8.1	153.90	0.6020	196.0200
3	18	7.8	140.40	4.1127	153.0000
total=	50		413.90	15.7658	407.1000
	Grand mean=Σnx̅i/Σni=	8.278		SSTr	SSE

df(between)=groups(k)-1=		2
df(error)=N-k=	50-3=	47

a)

from above

µ =8.28

MS(Treatments) =SS(factor)/df(factor)=

7.88

b)

MS(Error)=SS(error)/df(error)=

8.66

c)

Source of variation	df	SS	MS	F	P value
groups	2	15.77	7.88	0.91	0.409
error	47	407.10	8.66
total	49	422.87	0.76