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.
Time in Examination Rooms (minutes) | |||
Physician 1 | Physician 2 | Physician 3 | Physician 4 |
33 | 34 | 15 | 27 |
26 | 37 | 32 | 31 |
25 | 33 | 28 | 31 |
30 | 33 | 27 | 27 |
24 | 44 | 33 | 30 |
36 | 35 | 26 | 35 |
18 | 28 | 41 | |
32 | |||
Click here for the Excel Data File
Fill in the missing data. (Round your p-value to 4
decimal places, mean values to 1 decimal place, and other answers
to 2 decimal places.)
Treatment | Mean | n | Std. Dev |
Physician 1 | |||
Physician 2 | |||
Physician 3 | |||
Physician 4 | |||
Total | |||
One-Factor ANOVA | |||||
Source | SS | df | MS | F | p-value |
Treatment | |||||
Error | |||||
Total | |||||
(a) Based on the given hypotheses, choose the correct
option.
H_{0}: μ_{1} = μ_{2}
= μ_{3} = μ_{4}
H_{1}: Not all the means are equal
α = 0.05
Reject the null hypothesis if F < 3.01
Reject the null hypothesis if F > 3.01
(b) Calculate the F for one factor.
(Round your answer to 2 decimal places.)
F for one factor is
(c) On the basis of the above findings, we reject the null
hypothesis. Is the statement true?
No
Yes
Summary | |||||
---|---|---|---|---|---|
Groups | N | Mean | Std. Dev. | Std. Error | |
Group 1 | 7 | 27.4 | 6.05 | 2.29 | |
Group 2 | 6 | 36.0 | 4.19 | 1.71 | |
Group 3 | 8 | 27.6 | 5.73 | 2.03 | |
Group 4 | 7 | 31.7 | 4.92 | 1.86 |
ANOVA Summary | |||||
---|---|---|---|---|---|
Source | Degrees of Freedom DF |
Sum of Squares SS |
Mean Square MS |
F-Stat | P-Value |
Between Groups | 3 | 323.66 | 107.89 | 3.79 | 0.0234 |
Within Groups | 24 | 683.02 | 28.46 | ||
Total: | 27 | 1006.68 |
A) null and alternate hypothesis
H0: μ1 = μ2 = μ3 = μ4
H1: Not all the means are equal
α = 0.05
Reject the null hypothesis if F > 3.01
B) since F= MST/MSE = 3.79
C) yes,we reject the null hypothesis.
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