Problem: To study the effectiveness of possible treatments for insomnia, a sleep

Question

Problem: To study the effectiveness of possible treatments for insomnia, a sleep researcher conducted a study in which four participants were instructed to count sheep (the Sheep Condition), four were told to concentrate on their breathing (the Breathing Condition), and four were not given any special instructions. The average number of minutes taken for each participant to fall asleep over the next seven days were 14, 28, 27, and 31 minutes for the Sheep condition; 25, 22, 17, and 14 minutes for those in the Breathing condition; and 45, 33, 30, and 41 for the Control condition. Using the.05 significance level, did the different techniques have different effects? Step 1: Restate the question as a research hypothesis and a null hypothesis about the populations Pop 1:??? Pop 2: ??? Null hypothesis: ??? Research hypothesis: ??? Step 2: Determine the characteristics of the comparison distribution What type of distribution is this??? What are the df for the between-groups variance estimate: ??? What are the df for the within-groups variance estimate:??? hypothesis should be rejected Use the F table (table A-3) in the text book to look up the cutoff score using the degrees of freedom we just stated in Step 2. Cutoff score -??? This is where you determine the sample's F ratio (the numerator and denominator). To do that, you have to find the between-groups variance estimate and the within-groups varnance estimate Wthin-groups variance estimate This is the number that goes on the bottom of your F ratio.

Explanation / Answer

> pop1=c(14,28,27,31)

> pop2=c(25,22,17,14)

> pop3=c(45,33,30,41)

> pop=c(pop1,pop2,pop3)

> type=as.factor(c(1,1,1,1,2,2,2,2,3,3,3,3))

> df=data.frame(pop,type)

> df

   pop type

1   14    1

2   28    1

3   27    1

4   31    1

5   25    2

6   22    2

7   17    2

8   14    2

9   45    3

10 33    3

11 30    3

12 41    3

> model=aov(pop~type,data=df)

> summary(model)

            Df Sum Sq Mean Sq F value Pr(>F)

type         2 660.5   330.2   7.665 0.0114 *

Residuals    9 387.7    43.1                

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Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Step 1 -> Here, our null hypothesis is H0 : mean(pop1)=mean(pop2)=mean(pop3) and our alternative hypothesis is H1 : not H0.

Step 2 -> We assume the distribution to be normal distribution.
DF (between groups) = 3-1 = 2
DF( within groups) = 12-3 = 9

Step 3 -> Cutoff score = F 0.05; 2,9 = 4.256495

Step 4 -> mean(pop1)= 25

sd(pop1) = 7.527727

mean(pop2)= 19.5

sd(pop2)= 4.932883

mean(pop3)= 37.25

sd(pop3)= 6.946222

S2within = 387.7
MSE within = 43.1

Sample score = 7.665
Since sample score > cutoff score, we reject the null hypothesis and conclude that the mean number of minutes taken under different conditions were significantly different.

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