You assess 3 different treatments for middle-age male impotence. Tx 1 is a new t

Question

You assess 3 different treatments for middle-age male impotence. Tx 1 is a new trophy wife. Tx 2 is a satelite dish with Showtime and Cinemax. Tx 3 is Viagra. Each group keeps track of their marital relations over a two month period and here are the result:

Tx 1 Tx 2    Tx3

10    09 17

11    04 23

16    02 10

13    15 10

a. Compute an ANOVA be sure to use an ANOVA table.

b. If you have a signifcant tx effect, do a Tukey's HSD to see if there is any difference between the treatments.

Please show the steps you took to get the answer.

Explanation / Answer

Step 1

Null Hypothesis Ho : µ1 =µ2 =µ3

Alternative Hypothesis : atleast one mean is diffrent

Step 2

Degrees of freedom between = k - 1 = 3 - 1 = 2

Degrees of freedom Within = n - k = 12 - 3 = 9

Degrees of freedom Total F( k-1,n - k,) at 0.05 is = F Crit = 4.256

Step 3

Grand Mean = G / N = 12.5+7.5+15 / 3 = 11.667

SST = ( Xi - GrandMean)^2 = (10-11.667)^2 + (11-11.667)^2 + (16-11.667)^2 + ……..& so on = 356.667

SS Within = (Xi - Mean of Xi ) ^2 =,(10-12.5)^2 + (11-12.5)^2 + (16-12.5)^2 + ……..& so on = 240

SS Between = SST - SS Within = 356.667 - 240 = 116.667

Step 4

Mean Square Between = SS Between / df Between = 116.667/2 = 58.334

Mean Square Within = SS Within / df Within = 240/9 = 26.667

Step 5

F Cal = MS Between / Ms Within = 58.334/26.667 = 2.188

We got |F cal| = 2.188 & |F Crit| =4.256

MAKE DECISION

Hence Value of |F cal| < |F Crit|and Here We Accept Ho

no diffrence we idntified b/w mean rate

b.

Tukey simultaneous comparison t-values (d.f. = 9)

Treatments ONE WAY ANOVA Mean = X /n TX1 10 11 16 13 12.5 TX2 9 4 2 15 7.5 TX3 17 23 10 10 15

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