please show all step PART 1 The effect of a new antidepressant drug on reducing

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please show all step

PART 1 The effect of a new antidepressant drug on reducing the severity of depression was studied in manic-depressive patients at three mental hospitals. In each hospital all such patients were randomly assigned to either a treatment (new drug) or a control (old drug) group with different doses. The results of this experiment are summarized in the following tables; a high mean score indicates more of a lowering in depression level than does a low mean score. Table 1: The summary statistics of scores for the 6 groups Group Hospital Drnu DoseAverage S.D 1.195 1.049 1.354 1.146 1.317 1.643 0.5 0.5 8.0 5.5 5.0 New Old New Old New Old 10 0.3 0.3 6.4 4.8 4 Table2: The ANOVA table for the scores for the 6 groups ANOVA Sum of Squaresdf Mean Square Between Groups Within Groups Total 58.500 Table 3: The ANOVA table for the comparison of average scores of patients using new drug versus old drug (ignoring hospitals and doses): ANOVA Sum of Squaresdf Mean Square Between Groups Within Groups Total 123.415 171.636 Table 4: The ANOVA table for the comparison of average scores of patients between hospitals (ignoring drugs and doses) ANOVA Sum of Squaresdf Mean Square Between Groups Within Groups Total 104.565

Explanation / Answer

To be able to complete the table, remember the following:

SCT = SCbg + SCwg

DFT = DFbg + DFwg

MEANS SQUARE bw= SCbg / DFbg

MEANS SQUARE wg = SCwg / DFwg

F = MEANS SQUARE bg / MEANS SQUAREwg

then, we clear each unknown term from these equations

for the table ANAVA 1 we have

SCT = 171,636 DFT= n-1= 41 the SCT will be the same in all the ANOVA tables, since the data set is the same.

SCbg= 117,136 DFbg= 36 , means squarebg = 117,136/36

SCwg = 58,500 DFwg = 5, means squarewg = 58,500/ 5

for the second anova table

SCT = 171,636 DFT = 41

SCbg = 48,22 DFbg = 1, meanssquarebg = 48,22/1

SCwg = 123, 415 DFwg = 40, means squarewg = 123,415/40

F=48,22/3,08 = 15,65

For the last table of anova

SCT = 171,636 DFT = 41

SCbg = 67,071 DFbg = 2, means squarebg = 67,071/2

SCwg = 104, 565 DFwg = 39 means squarewg = 104,565/39

items from 5 to 8

Meannew= (8+5+6,4)/3

SDnew = (1,195+1,354+1,317) / (8+1+4)

Meanold = (5,5+3+4,8) / 3

SDold = (1,049+1,146+1,64) / (6+9+5)

5. the estimated value for the difference of means is

6,4 - 4,3 = 2

6. The standard error of this estimator is

0,1575 + 0,1919 = 0,34 approx. 0,4

7. the statistic used to prove the hypothesis is

t ob = (meannew - meanold) / SDnew/nnew + SDold/nold

the t statistic has n-2 degrees of freedom = 42-2 = 40

tob = (6,4-4,3) / (0,1575 + 0,1919) = 6,1

p-valor = (probability of obtaining a value of t equal to or greater than that obtained. being Ho certain)*2

multiplies by 2 because the hypothesis test is bilateral

the probability of obtaining a value of t higher than that obtained is 2.36 E-7 (value obtained by means of a soft).

2,36E-7*2=4,72E-7 this value is less than 0.001

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