In: Statistics and Probability
Let
p1
and
p2
be the respective proportions of women with iron-deficiency anemia in each of two developing countries. A random sample of
2000
women from the first country yielded
447
women with iron-deficiency anemia, and an independently chosen, random sample of
2300
women from the second country yielded
467
women with iron-deficiency anemia. Can we conclude, at the
0.01
level of significance, that the proportion of women with anemia in the first country is greater than the proportion of women with anemia in the second country?
Perform a one-tailed test. Then fill in the table below.
Carry your intermediate computations to at least three decimal places and round your answers as specified in the table. (If necessary, consult a list of formulas.)
|
Given that,
sample one, x1 =447, n1 =2000, p1= x1/n1=0.224
sample two, x2 =467, n2 =2300, p2= x2/n2=0.203
null, Ho: p1 = p2
alternate, H1: p1 > p2
level of significance, α = 0.01
from standard normal table,right tailed z α/2 =2.326
since our test is right-tailed
reject Ho, if zo > 2.326
we use test statistic (z) = (p1-p2)/√(p^q^(1/n1+1/n2))
zo =(0.224-0.203)/sqrt((0.213*0.787(1/2000+1/2300))
zo =1.635
| zo | =1.635
critical value
the value of |z α| at los 0.01% is 2.326
we got |zo| =1.635 & | z α | =2.326
make decision
hence value of |zo | < | z α | and here we do not reject
Ho
p-value: right tail - Ha : ( p > 1.6354 ) = 0.05098
hence value of p0.01 < 0.05098,here we do not reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 > p2
test statistic: 1.635
critical value: 2.326
decision: do not reject Ho
p-value: 0.05098
we do not have enough evidence to support the claim that t the
proportion of women with anemia in the first country is greater
than the proportion of women with anemia in the second
country
No,
the proportion of women with anemia in the first country is greater
than the proportion of women with anemia in the second country