** Will rate once all questions are answered, thanks! 9. The Scheffe test Aa Aa

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** Will rate once all questions are answered, thanks!

9. The Scheffe test Aa Aa 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 A clinical psychologist studies the effects of tonsillectomy and adenoidectomy on inattentive 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. I nattentiveness 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 0.2888 No sleep apnea 0.26 Untreated sleep apnea 0.3872 0.62 Treated sleep apnea 0.2312 0.33 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 Variation Sum of Squares Freedom Mean Square 0.3281 Between Treatments 0.6561 8.66 0.0378 Within Treatments 0.9072 24 Total 1.5633 26

Explanation / Answer

1. SSbetween A and B = ?

first calculate average of averages of A and B = 1/2 ( 0.26 + 0.62) = 0.44 and then calculate

SSbetween A and B = 9 [ (0.44 - 0.26)2 + ( 0.44 - 0.62)2 ]= 0.5832

so mean squarebetween A and B = 0.5832/1 = 0.5832 [Here degree of freedom =1 ]

SS within A and B = 0.2888 + 0.3872 = 0.676

degree of freedom = 16

so mean squarein A and B = 0.676/ 16 = 0.04225

and FA versus B = 0.5832/ 0.04225 = 13.80

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 ( 0.26 + 0.33) = 0.295 and then calculate

SSbetween A and C = 9 [ (0.295 - 0.26)2 + ( 0.295 - 0.33)2 ]= 0.02205

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

SS within A and C = 0.2888 + 0.2312 = 0.52

degree of freedom = 16

so mean squarein A and C = 0.52/ 16 = 0.0325

and FA versus C = 0.02205/ 0.0325 = 0.6784

at alpha = 0.01 , F critical = 8.53 so we can say that FA versus C <  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 ( 0.62 + 0.33) = 0.475 and then calculate

SSbetween C and B = 9 [ (0.475 - 0.62)2 + ( 0.475 - 0.33)2 ]= 0.37845

so mean squarebetween C and B = 0.37845/1 = 0.37845 [Here degree of freedom =1 ]

SS within C and B = 0.3872 + 0.2312 = 0.6184

degree of freedom = 16

so mean squarein C and B = 0.6184/ 16 = 0.03865

and FC versus B = 0.37845/ 0.03865 = 9.79

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 with sleep apnea treated is stastically differrent with children with sleap apnea (untreated).

So, out of these 3 groups, The group, nameley, children with sleep apnea (untreated) is different.

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