Two new drugs are being tested for their effect on the number of days a patient must remain hospitalized after surgery. A control group is receiving a placebo, and two treatment groups are each receiving one of two new drugs, both developed to promote recovery. The null hypothesis is that there is no difference between the means. The results of an analysis of variance used to analyze the data are shown here.
One-way ANOVA: Days versus Group
Source DF SS MS F P
Group 2 11.00 5.50 2.11 0.159
Error 14 36.53 2.61
Total 16 47.53
(a) How many patients were there?
16+1=17 .
(b) How does the table verify that there was one control group and two test groups?
df(Group) = 2.
(c) Using the SS values, verify the two mean square values.
MSG = 5.50/2=2.75 .
MSE = 36.53/14=2.61 .
(d) Using the MS values, verify the F-value. not sure on formula to find this one??
(e) Verify the p-value.
0.159 .
To verify the F-value using the MS values, you can use the formula:
F-value = (MSG / MSE)
In this case, the values are:
MSG = 2.75
MSE = 2.61
So, the F-value can be calculated as:
F-value = (2.75 / 2.61) ≈ 1.05
Therefore, the F-value for this analysis is approximately 1.05.
To verify the p-value, you can compare the obtained F-value with the critical F-value for the chosen significance level. The p-value corresponds to the probability of obtaining an F-value as extreme as the one observed, assuming the null hypothesis is true.
However, the p-value is not provided in the given information. You can refer to a statistical table or use statistical software to determine the p-value associated with an F-value of 1.05 for the specific degrees of freedom.
Alternatively, if you have access to statistical software, you can directly input the provided data to calculate the p-value.