In: Math
A hospital wants to determine if the type of treatment for pneumonia is a factor in recovery time? The table below shows the number of days to recovery for several randomly selected pneumonia patients that had various types of treatment.
| Sent Home With Medicine | A Few Hours in the Hospital | Overnight Hospital Stay | 
|---|---|---|
| 22 | 8 | 8 | 
| 19 | 15 | 18 | 
| 18 | 12 | 18 | 
| 22 | 23 | 4 | 
| 27 | 18 | 5 | 
| 14 | 10 | 10 | 
| 30 | 18 | 18 | 
| 23 | 8 | 8 | 
| 9 | 17 | 
Assume that all distributions are normal, the three population
standard deviations are all the same, and the data was collected
independently and randomly. Use a level of significance of
α=0.1α=0.1.
H0: μ1=μ2=μ3H0: μ1=μ2=μ3
H1:H1: At least two of the means differ from each other.
| Sent Home With Medicine | A Few Hours in the Hospital | Overnight Hospital Stay | Anova: Single Factor | |||||||
| 22 | 8 | 8 | ||||||||
| 19 | 15 | 18 | SUMMARY | |||||||
| 18 | 12 | 18 | Groups | Count | Sum | Average | Variance | |||
| 22 | 23 | 4 | Sent Home With Medicine | 8 | 175 | 21.875 | 25.55357 | |||
| 27 | 18 | 5 | A Few Hours in the Hospital | 9 | 121 | 13.44444 | 28.52778 | |||
| 14 | 10 | 10 | Overnight Hospital Stay | 9 | 106 | 11.77778 | 35.19444 | |||
| 30 | 18 | 18 | ||||||||
| 23 | 8 | 8 | ||||||||
| 9 | 17 | ANOVA | ||||||||
| Source of Variation | SS | df | MS | F | P-value | F crit | ||||
| Between Groups | 487.8088 | 2 | 243.9044 | 8.146051 | 0.002115 | 3.422132 | ||||
| Within Groups | 688.6528 | 23 | 29.94143 | |||||||
| Total | 1176.462 | 25 | 
Test statistic=Value of F=8.146
p-value=0.0021

There is sufficient evidence to support the claim that treatment type is a factor in recovery time for pneumonia.