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 8.4, 8.8 and 6.2 mm and standard deviations were 3.1, 2.6 and 2.9 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). W= 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). MS Exor=L 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. Source DF SS (to two decimal places) MS (to two decimal places) F (to two decimal places) р (to three decimal places) Treatments Error Total

Answer 1

Mean =

Σ. .. n

= Mean * n

T

SD =

Σ (Στ)2 ✓Var η η –1

Σ Σ Υ
= Var * (n-1) + (
T
)² / n

Σ Σ Υ
=(SD)² * (n-1) + (
T
)² / n

	n	Mean	T =Mean * n	SD	Σ Σ Υ =(SD)2 * (n-1) + ( T )2 / n
ATAI	13	8.4	109.2	3.1	1032.6
NAO	19	8.8	167.2	2.6	1593.04
MT	18	6.2	111.6	2.9	834.89
Total	50		388		3460.53

ANOVA

This is ANOVA - Single Factor. We first need to find the
LLSS
and
SSTS
. They are 'Total variance' and 'Variance in between the treatments' respectively.

We know

SST, = SSTT – SSTB

LLSS
=
Grand total Sum of individual squares n

= 449.65

LLSS

SSTS
=
Grand total square of total of individual treatment score no.of obs in that treatment n

Note: We have to separately sum for all the treatment.

Square of total of treatment obs

=69.68

SSTS

379.97

LSS

Df (Bet) = k -1 ………..where k = no. of treatments

=2

Df (res) = N – k

=47

Test Stat = MSB / MSE

	SS	df	MS  (SS /df)	F-stat
Between	69.68	2	34.84	4.3095
Error	379.97	47	8.0845
Total	449.65	49

a)

Grand mean= Grand total / N

$\mu$ = 388 /50

= 7.76

MS-Treatments (MS between)

Ans: 34.84

b)

MS error = 8.0845

c)

	SS	df	MS  (SS /df)	F-stat
Between	69.68	2	34.84	4.3095
Error	379.97	47	8.0845
Total	449.65	49

p-value = P( > F-stat)

Fofo,df.

=P(F_2,47 > 4.31)

p-value = 0.0191........................found using f-dist tables with df1 = 2 df2=47