One particular morning, the length of time spent in the examination rooms is rec

Question

One particular morning, the length of time spent in the examination rooms is recorded for each patient seen by each physician at an orthopedic clinic.

Fill in the missing data. (Round your p-value to 4 decimal places, mean values to 1 decimal place, and other answers to 3 decimal places.)

Based on the given hypotheses, choose the correct option.

Calculate the F for one factor. (Round your answer to 2 decimal places.)

On the basis of the above findings, we reject the null hypothesis. Is the statement true?

One particular morning, the length of time spent in the examination rooms is recorded for each patient seen by each physician at an orthopedic clinic.

Explanation / Answer

Step 1

Null Hypothesis Ho : µ1 =µ2 =µ3 = µ4

Alternative Hypothesis : atleast one mean is diffrent

Step 2

Degrees of freedom between = k - 1 = 3

Degrees of freedom Within = n - k = 24

Degrees of freedom Total F( k-1,n - k,) at 0.05 is = F Crit = 3.31

Step 3

Grand Mean = G / N = 30.53571

SST = ( Xi - GrandMean)^2 = 782.96428

SS Within = (Xi - Mean of Xi ) ^2 =,553.785714285714

SS Between = SST - SS Within = 229.1785

Step 4

Mean Square Between = SS Between / df Between = 76.392857

Mean Square Within = SS Within / df Within = 23.074

Step 5

F Cal = MS Between / Ms Within = 35/19.667 = 1.78

We got |F cal| = 3.31 & |F Crit| =3.009

MAKE DECISION

Hence Value of |F cal| > |F Crit|and Here We reject Ho

[ANSWERS]

H1: Not all the means are equal
Reject the null hypothesis if F > 3.01
F for one factor is = 3.31
we reject the null hypothesis

Mean n Std. Dev 27.9 7 4.06 Physician 1 34.5 6 4.64 Physician 2 28.0 8 4.87 Physician 3 32.7 7 5.50 Physician 4 30.5 28 5.39 Total

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