Question

In a randomized controlled​ trial, insecticide-treated bednets were tested as a way to reduce malaria. Among 322 infants using​ bednets, 13 developed malaria. Among

276 infants not using​ bednets, 29 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.

please show me how to do on ti-84

Difference=​p(1)minus−​p(2)

Estimate for​ difference: - 0.0646998

95% upper bound for​ difference: - 0.02261687

Test for difference=0 ​(vs <​ 0): Z=- 3.09  

​P-Value=0.001

Answer 1

TI 84 METHOD

Press STAT and the right arrow twice to select TESTS.
Use the down arrow to select
2-PropZTest…
Press ENTER.
Enter number of favorable outcomes and sample size of samples 1 and 2. Select the alternate hypothesis.

X1 -> 13, n1 -> 322, X2 -> 29, n2 = 276, choose <P2,
Use down arrow to select Calculate and press ENTER.

RESULT

HAND WRITTEN METHOD

Given that,
sample one, x1 =13, n1 =322, p1= x1/n1=0.04
sample two, x2 =29, n2 =276, p2= x2/n2=0.105
finding a p^ value for proportion p^=(x1 + x2 ) / (n1+n2)
p^=0.07
q^ Value For Proportion= 1-p^=0.93
null, Ho: p1 = p2
alternate, H1: p1 < p2
level of significance, α = 0.01
from standard normal table,left tailed z α/2 =2.326
since our test is left-tailed
reject Ho, if zo < -2.326
we use test statistic (z) = (p1-p2)/√(p^q^(1/n1+1/n2))
zo =(0.04-0.105)/sqrt((0.07*0.93(1/322+1/276))
zo =-3.087
| zo | =3.087
critical value
the value of |z α| at los 0.01% is 2.326
we got |zo| =3.087 & | z α | =2.326
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value: left tail - Ha : ( p < -3.0866 ) = 0.00101
hence value of p0.01 > 0.00101,here we reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 < p2
test statistic: -3.087
critical value: -2.326
decision: reject Ho
p-value: 0.00101

we have have incidence of malaria is lower for infants using bednets