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Bookmarks Window Help courses aplia.com Saved Homes I Zllow Aa Aa 11. The Scheffe test sleep apnea is a sleep disorder characterized by pauses in breathing during sleep. Children with sleep apnea have behavior problems, including hyperactivity, inattention, and aggression, as well as impaired learning and diminished academic performance. The removal of tonsils and adenoids that are enlarged, causing the obstruction of the airways, is one of the most common treatments for pediatric sleep apnea l psychologist studies the effects of tonsilectomy and adenoidectomy on inattentive behavior. Her the quasi-experiment includes three groups of 21 children. The first group of children does not have sleep apnea, second group has untreated sleep apnea, and the third group has sleep apnea treated by tonsillectomies a Inattentiveness was measured using teacher 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 sample Mean sum of squares Group 0.5120 0.41 No sleep apnea 0.8820 0.72 Untreated sleep apnea 0.7220 Treated sleep apnea 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 Degrees of Source of Sum of Squares Freedom Mean square Variation 0.5555 15.74 1.1109 Between Treatments 0.0353 2.1160 Within Treatments 3.2269 The critical value of F when a 05 is 3.150, meaning the critical region consists of all F-ratios greater than 3.150.

Explanation / Answer

1. SSbetween A and B = ?

first calculate average of these 2 = 1/2* ( 0.41 + 0.72) = 0.565 and then calculate

SSbetween A and B = 9 [ (0.565 - 0.41)2 + ( 0.565 - 0.72)2 ]= 0.43245

so mean squarebetween A and B = 0.43245/1 = 0.43245 [Here degree of freedom =1 ]

SS within A and B = 0.5120 + 0.8820 = 1.3940

degrees of freedom = 40

so mean squarein A and B = 1.3940/ 40 = 0.03485

and FA versus B = 0.43245/ 0.03485 = 12.41

at alpha = 0.05 and dF = 40 , F critical = 4.0847 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 * ( 0.41 + 0.48) = 0.445 and then calculate

SSbetween A and C = 9 [ (0.445 - 0.41)2 + ( 0.445 - 0.48)2 ]= 0.02205

so mean squarebetween A and C = 0.02205/1 = 0.02205 [Here degree of freedom =1 ]

SS within A and C = 0.5120 + 0.7220 = 1.234

degree of freedom = 40

so mean squarein A and C = 1.234/ 40 = 0.03085

and FA versus C = 0.02205/ 0.03085 = 0.7147

at alpha = 0.05 , F critical = 4.0847 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* ( 0.48 + 0.72) = 0.60 and then calculate

SSbetween C and B = 9 [ (0.60 - 0.48)2 + ( 0.60 - 0.72)2 ]= 0.2592

so mean squarebetween C and B = 0.2592/1 = 0.2592 [Here degree of freedom =1 ]

SS within C and B = 0.8820 + 0.7220 = 1.604

degree of freedom = 40

so mean squarein C and B = 1.604/40 = 0.04010

and FC versus B = 0.2592/ 0.0401 = 6.46

at alpha = 0.05 , F critical = 4.0847 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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