Consider a study comparing two medications for severe bladder infec- tions. The
ID: 3311129 • Letter: C
Question
Consider a study comparing two medications for severe bladder infec- tions. The variable x is the length of time (in days) to recovery. For the n1 = 15 patients who were given medication 1, we observed a mean recovery time of x1 = 16.87 days. The mean recovery time was x2 = 19.09 days for the n2 = 18 patients who were given medication.
a) Here are overlayed quantile-quantile plot for the two samples of recovery times. Is it reasonable to assume that both populations of recovery times are normally distributed with equal variances?
b) Based on the following R output, compute the value of the pooled standard deviation sp.
(c) Based on the R output in (b), give a 95% confidence interval for difference between the mean recovery time on medication 1 and the mean recovery time on medication 2.
(d) Based on the confidence interval from (c), which medication is best?
Explanation / Answer
Part a
From the given normal Q-Q plot, it is observed that all points for the medication 1 and medication 2 are approximately lies on straight line and also both lines have approximate similar slope. This indicates that both populations of recovery times are normally distributed with equal variances.
Part b
Here, we have to find pooled standard deviation.
The 95% confidence interval for difference between two population means is given as below:
CI = (-3.105940, -1.349615)
CI = (X1bar – X2bar) -/+ t* sqrt[Sp2*((1/n1)+(1/n2))]
Where Sp2 is pooled variance
Sp2 = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)
n1 = 15
n2 = 18
df = n1 + n2 – 2 = 15 + 18 – 2 = 31
c = 95% = 0.95
= 1 – c = 1 – 0.95 = 0.05
Critical t value = 2.039513
CI = (16.87 – 19.09) -/+ 2.039513* sqrt[Sp2*((1/15)+(1/18))]
CI = (16.87 - 19.09) -/+ 2.039513* sqrt[Sp2* 0.122222]
CI = -2.22 -/+ 2.039513* sqrt[Sp2* 0.122222]
CI = -2.22 -/+ 2.039513*Sp*sqrt(0.122222)
CI = -2.22 -/+ 2.039513*Sp* 0.349603
CI = -2.22 -/+ 0.71302*Sp
We are given lower limit = -3.105940
So,
-3.105940 = -2.22 - 0.71302*Sp
3.105940 = 2.22 + 0.71302*Sp
0.71302*Sp = 3.105940 - 2.22
0.71302*Sp = 0.88594
Sp = 0.88594/0.71302
Sp = 1.242518
Pooled standard deviation = 1.242518
Part c
The 95% confidence interval for difference between two population means is given as below:
CI = (-3.105940, -1.349615)
Part d
This confidence interval has both negative values which indicate that medication 1 requires less recovery days. This means medication 1 is best.