Question

Reserve Problems Chapter 13 Section 2 Problem 2 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). N= 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. 9.0h). MSError EL 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. P Source DF SS (to two decimal places) MS (to two decimal places) F (to two decimal places) (to three decimal places) Treatments

Answer 1

a)

count, ni =	13	19	18
mean , x̅ i =	8.400	8.80	6.20
std. dev., si =	3.1	2.6	2.9
sample variances, si^2 =	9.610	6.760	8.410
total sum	109.2	167.2	111.6				388	(grand sum)
grand mean , x̅̅ =	Σni*x̅i/Σni =	7.76

Grand mean =7.76

SS(between)= SSB = Σn( x̅ - x̅̅)² =

5.325

20.550

43.805

69.68

no. of treatment , k =   3  
df between = k-1 =    2  
N = Σn =   50  
df within = N-k =   47  
      
mean square between groups , MSB = SSB/k-1 =    69.68/2=   34.84

b)

SS(within ) = SSW = Σ(n-1)s² =

115.320

121.680

142.970

379.9700

mean square within groups , MSW = SSW/N-k =    379.97/47=   8.08

c)

	df	SS	MS	F	p-value
Between:	2	69.68	34.84	4.31	0.019
Within:	47	379.97	8.08
Total:	49	449.65

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