In a randomized controlled trial insecticidetreated bednets

In a randomized controlled trial, insecticide-treated bednets were tested as a way to reduce malaria. Among 321infants using bednets, 20 developed malaria. Among 289 infants not using bednets, 27 developed malaria. Use a 0.05 significance level to test the claim that the incidence of malaria is lower for infants using bednets. Do the bednets appear to be effective? Conduct the hypothesis test by using the results from the given display.

1. What are the null and alternative hypotheses?

A.

H0: p 1=p 2

H1: p 1>p 2

B.

H0: p 1=p 2

H1: p 1 p 2

C.

H0: p 1 p 2

H1: p 1 = p 2

D.

H0: p 1 is less than or equal to p 2

H1: p 1 > p 2

E.

H0: p 1=p 2

H1: p1 < p 2

F. H0: p1 p 2

H1: p 1 < p 2

2. Identify the test statistic.

t = ????

(Type an integer or decimal rounded to two decimal places as needed.)

Solution

Null, Ho: p1 > p2
Alternate, H1: p1 < p2
Test Statistic
Sample 1 : X1 =20, n1 =321, P1= X1/n1=0.062
Sample 2 : X2 =27, n2 =289, P2= X2/n2=0.093
Finding a P^ value For Proportion P^=(X1 + X2 ) / (n1+n2)
P^=0.077
Q^ Value For Proportion= 1-P^=0.923
we use Test Statistic (Z) = (P1-P2)/(P^Q^(1/n1+1/n2))
Zo =(0.062-0.093)/Sqrt((0.077*0.923(1/321+1/289))
Zo =-1.439
| Zo | =1.439
Critical Value
The Value of |Z | at LOS 0.05% is 1.645
We got |Zo| =1.439 & | Z | =1.645
Make Decision
Hence Value of |Zo | < | Z | and Here we Do not Reject Ho
P-Value: Left Tail - Ha : ( P < -1.4392 ) = 0.07505
Hence Value of P0.05 < 0.07505,Here We Do not Reject Ho

[ANSWERS]
1. Ho: p1 = p2 , H1: p1 < p2
2. Zo =-1.44