In a comparative double-blind randomized study of single dose fosfomycin trometa
ID: 3158492 • Letter: I
Question
In a comparative double-blind randomized study of single dose fosfomycin trometamol with trimethoprim in the treatment of uncomplicated urinary infections, 24 of 32 were cured with fosfomycin trometamol and 20 of 38 were cured with the reference drug trimethoprim.
Denote the eradication rates for fosfomycin trometamol and the reference drug trimethoprim by p1and p2 respectively.
(a) Evaluate the test statistic (using Z=)
=_________
(b) Use the R function pnorm() and the value of the test statistic from (a) to find the p-value for the test H0:p1=p2
versus HA:p1p2.
=______
(c) Calculate a 98% two-sided confidence interval for the difference p1p2
using the formula
p^1p^2±zp^1(1p^1)n1+p^2(1p^2)n2
and the critical value z=2.3263.
The interval ranges from:_______to_______
(d) Find the p-value for the test H0:p1=p2
versus HA:p1p2
using the R function prop.test().
=_________
(e) Are the eradication rates for the two drugs significantly different? Use the result of (d) and significance level =0.02
to answer this question.
=yes or no?
(f) Calculate a 98% two-sided confidence interval for the difference p1p2 using the R function prop.test().
The interval ranges from:
_______to_______
**Could you please work through this problem step by step. Using the exact numbers from the problems in the formulas would be helpful.**
Explanation / Answer
Null, Ho: p1 = p2
Alternate, H1: p1 != p2
Test Statistic
Sample 1 : X1 =24, n1 =32, P1= X1/n1=0.75
Sample 2 : X2 =20, n2 =38, P2= X2/n2=0.526
Finding a P^ value For Proportion P^=(X1 + X2 ) / (n1+n2)
P^=0.629
Q^ Value For Proportion= 1-P^=0.371
we use Test Statistic (Z) = (P1-P2)/(P^Q^(1/n1+1/n2))
Zo =(0.75-0.526)/Sqrt((0.629*0.371(1/32+1/38))
Zo =1.929
| Zo | =1.929
Critical Value
The Value of |Z | at LOS 0.02% is 2.326
We got |Zo| =1.929 & | Z | =2.326
Make Decision
Hence Value of |Zo | < | Z | and Here we Do not Reject Ho
P-Value: Two Tailed ( double the one tail ) -Ha : ( P != 1.9295 ) = 0.0537
Hence Value of P0.02 < 0.0537,Here We Do not Reject Ho
a.
Zo =1.929
b.
( P != 1.9295 ) = 0.0537
c.
Confidence Interval for Diffrence of Proportion
CI = (p1 - p2) ± Z a/2 Sqrt(p1(1-p1)/n1 + p2(1-p2)/n2 )
Proportion 1
No. of chances( X1 )=24
No.Of Observed (n1)=32
P1= X1/n1=0.75
Proportion 2
No. of chances(X2)=20
No.Of Observed (n2)=38
P2= X2/n2=0.526
C.I = (0.75-0.526) ±Z a/2 * Sqrt( (0.75*0.25/32) + (0.526*0.474/38) )
=(0.75-0.526) ± 2.33* Sqrt(0.012)
=0.224-0.26,0.224+0.26
=[-0.036,0.483]