Can someone please help me? If you could at least SHOW how to do the first probl

Question

Can someone please help me?

If you could at least SHOW how to do the first problem then I can figure it out hopefully!! I just cant figure out what the F A versus B thing is? I posted this question previously and it wasn't correct so I provided the answer options for that blank.

Please help me out. thank you

A clinical psychologist studies the effects of tonsillectomy and adenoidectomy on hyperactive behavior. Her quasi-experiment includes three groups of 9 children. The first group of children does not have sleep apnea, the second group has untreated sleep apnea, and the third group has sleep apnea treated by tonsillectomies and adenoidectomies. Hyperactivity was measured using parent reports on the Conners Rating Scale. The sample means and sums of squared deviations of the scores for each of the three groups are presented in the table that follows. Group Sample Mean Sum of Squares No sleep apnea 0.27 0.1568 0.2592 Untreated sleep apnea 0.54 0.1800 Treated sleep apnea 0.32 After collecting the data, the clinical psychologist analyzes the data using an ANOVA. The results of her analysis are presented in the ANOVA table that follows. ANOVA Table Source of Degrees of Sum of squares Freedom Mean Square Variation 0.3717 7.50 Between Treatments 0.1859 0.5960 24 Within Treatments 0.0248 0.9677 Total 26

Explanation / Answer

1. SSbetween A and B = ?

first calculate average of these 2 = 1/2 ( 0.27 + 0.54) = 0.405 and then calculate

SSbetween A and B = 9 [ (0.405 - 0.27)2 + ( 0.405 - 0.54)2 ]= 0.3280

so mean squarebetween A and B = 0..3280/1 = 0.3280 [Here degree of freedom =1 ]

SS within A and B = 0.1568 + 0.2592 = 0.416

degree of freedom = 16

so mean squarein A and B = 0.416/ 16 = 0.026

and FA versus B = 0.3280/ 0.026 = 12.615 ( no option is given correct, this is the right answer)

at alpha = 0.01 , F critical = 8.53 so we can say that F >  F critical , so we can reject the null hypothesis and say that the population mean for children without sleep apnea differ with children with sleap apnea untreated.

(2) SSbetween A and C = ?

first calculate average of these 2 = 1/2 ( 0.27 + 0.32) = 0.295 and then calculate

SSbetween A and C = 9 [ (0.295 - 0.27)2 + ( 0.295 - 0.32)2 ]= 0.01125

so mean squarebetween A and B = 0.01125/1 = 0.01125 [Here degree of freedom =1 ]

SS within A and C = 0.1568 + 0.1800 = 0.3368

degree of freedom = 16

so mean squarein A and C = 0.3368/ 16 = 0.02105

and FA versus C = 0.01125/ 0.02105 = 0.5344

at alpha = 0.01 , F critical = 8.53 so we can say that F <  F critical , so we cannot reject the null hypothesis and say that the population mean for children without sleep apnea is same with children with sleap apnea (treated).

Q.3

SSbetween B and C = ?

first calculate average of these 2 = 1/2 ( 0.54 + 0.32) = 0.43 and then calculate

SSbetween C and B = 9 [ (0.43 - 0.32)2 + ( 0.43 - 0.54)2 ]= 0.2178

so mean squarebetween C and B = 0.2178/1 = 0.2178 [Here degree of freedom =1 ]

SS within C and B = 0.2592 + 0.1800 = 0.4392

degree of freedom = 16

so mean squarein C and B = 0.4392/ 16 = 0.02745

and FC versus B = 0.2178/ 0.02745 = 7.93

at alpha = 0.01 , F critical = 8.53 so we can say that F <  F critical , so we cannot reject the null hypothesis and say that the population mean for children with sleep apnea treated is same with children with sleap apnea (untreated).

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