Lawson and Jick 1976 compare drug prescription in the United

Lawson and Jick [1976] compare drug prescription in the United States and Scotland.

(a) In patients with congestive heart failure, two or more drugs were prescribed in 257 of 437 U.S. patients. In Scotland, 39 of 179 patients had two or more drugs prescribed. Test the null hypothesis of equal proportions giving the resulting p- value. Construct a 95% confidence interval for the difference in proportions.

(b) Patients with dehydration received two or more drugs in 55 of 74 Scottish cases as compared to 255 of 536 in the United States. Answer the questions of part (a).

Solution

a). let p1 be the proportions of patients in U.S. with two or more drugs.

         p2 be the proportions of patients in scotland with two or more drugs.

let f1 be the number of patients in U.S. with two or more drugs out of n1 persons.

    f2 be the number of patients in scotland with two or more drugs out of n2 persons.

f1 and f2 are independently distributed with f1~Bin(n1,p1)   f2~Bin(n2,p2)

E(f1/n1)=p1 , E(f2/n2)=p2 , V(f1/n1)=p1(1-p1)/n1    , V(f2/n2)=p2(1-p2)/n2

therefore E[f1/n1-f2/n2]=p1-p2     V(f1/n1-f2/n2)=(p1(1-p1)/n1)+ (p2(1-p2)/n2)

since here n1=437 and n2=179 is quite large we can use large sample approximation

therefore f1/n1-f2/n2~N(p1-p2,(p1(1-p1)/n1)+ (p2(1-p2)/n2))

here H₀:p1=p2 vs H₁:p1 is not equal to p2

therefore under H₀   f1/n1-f2/n2~N(0,(p(1-p)/n1)+ (p(1-p)/n2))   where p is the common value of p1 and p2.

now p is unknown. it is estimated by the pooled estimator phat=(f1+f2)/(n1+n2)

an appropriate test statistic is then given by---

T= (f1/n1-f2/n2)/squareroot((phat(1-phat)/n1)+ (phat(1-phat)/n2))~N(0,1) under H₀

now f1=257 f2=39 n1=437 n2=179 phat=(257+39)/(437+179)=0.48

observed value of T is t=8.350

taking alpha=0.05 the desired level of significance we have p-value=2*min{P[T<t| H0],P[T>t| H0]}

now P[T<8.350]=1         and P[T>8.350]=0

hence P-value=0<0.05

hence we reject the null hypothesis based on the given data at hand and conclude that the two proportions are not equal at 5% level of significance.

95% C.I for the difference p1-p2 is given as--

[f1/n1-f2/n2-1.96*squareroot(f1/n1(1-f1/n1)/n1+f2/n2(1-f2/n2)/n2),f1/n1-f2/n2+1.96*squareroot(f1/n1(1-f1/n1)/n1+f2/n2(1-f2/n2)/n2)]

[0.370-0.075,0.370+0.075]=[0.295,0.445] [answer]

b) comparing with a) with we have f1=55 f2=255 n1=74 n2=536 phat=(f1+f2)/(n1+n2)=0.508

hence exactly following the same procedure we have the test statistic as

T=(f1/n1-f2/n2)/squareroot((phat(1-phat)/n1)+ (phat(1-phat)/n2))~N(0,1) under H₀

observed value of T is t=4.314

taking alpha=0.05 the desired level of significance we have p-value=2*min{P[T<t| H0],P[T>t| H0]}

now P[T<4.314]=1         and P[T>4.314]=0

hence P-value=0<0.05

hence we reject the null hypothesis based on the given data at hand and conclude that the two proportions are not equal at 5% level of significance.

95% C.I for the difference p1-p2 is given as--

[f1/n1-f2/n2-1.96*squareroot(f1/n1(1-f1/n1)/n1+f2/n2(1-f2/n2)/n2),f1/n1-f2/n2+1.96*squareroot(f1/n1(1-f1/n1)/n1+f2/n2(1-f2/n2)/n2)]

[0.267-0.108,0.267+0.108]=[0.159,0.375] [answer]