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Two Independent Samples t-test You want to find out whether a particular form of

ID: 3225163 • Letter: T

Question

Two Independent Samples t-test You want to find out whether a particular form of cognitive behavioral therapy decreases depressive symptoms in patients. You use a symptom score that ranges from 0 to 50 (with 50 being the most severe depression). You do a small study in which five randomly sampled patients are assigned to a treatment group and five randomly sampled patients are assigned to control group, which receives no treatment. We are assuming that the two populations have the same standard deviation. Alpha is set to .05. 1. What is your null and altemative hypothesis (in words AND symbols)? (Remember: You need to state these hypotheses in tems of population means!) (5pt) 2. The mean for the control group sample is 35 and the mean for the treatment group sample is 25. Find the pooled estimate of the population standard deviation using the table and formulas below. (15pt) Group Symptom Group mean Deviation from mean Squared deviation (score -mean) (score-mean) 35 Control Control 37 Control 26 39 Control 38 Control Treatment 32 28 Treatment Treatment 19 Treatment 24 Treatment Sum of squared deviations

Explanation / Answer

Q-1

Null hypothesis H0: µ1 = µ2 is to be tested against the alternative Ha: µ1 < µ2 where µ1 = µ2 are the true means of treatment and control

Null hypothesis H0: There is not difference in mean depression score of treatment and control.

Alternative hypothesis Ha: The mean depression score of treatment is lower than control.

Q-2

Group

Symptom Score

Group Mean

Deviation from Mean=Score-Mean

Squared deviations =(score-Mean)^2

Control

35

35

0

0

Control

37

35

2

4

Control

26

35

-9

81

Control

39

35

4

16

Control

38

35

3

9

Treatment

32

25

7

49

Treatment

28

25

3

9

Treatment

22

25

-3

9

Treatment

19

25

-6

36

Treatment

24

25

-1

1

Sum of Sqaured Deviations=

214

SD(pooled)hat = sqrt(sum of squared dviations/(n1+n2-2)=sqrt(214/(5+5-2)) =5.172

SE(difference)hat= SD(pooled)hat *sqrt(1/n1+1/n2) =sqrt( 5.172*sqrt(1/5+1/5) =3.271

Q-3

Test statistic t= (35-25)/3.271=3.0572

Group

Symptom Score

Group Mean

Deviation from Mean=Score-Mean

Squared deviations =(score-Mean)^2

Control

35

35

0

0

Control

37

35

2

4

Control

26

35

-9

81

Control

39

35

4

16

Control

38

35

3

9

Treatment

32

25

7

49

Treatment

28

25

3

9

Treatment

22

25

-3

9

Treatment

19

25

-6

36

Treatment

24

25

-1

1

Sum of Sqaured Deviations=

214

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