Let p1 and p2 be the respective proportions of women with iron-deficiency anemia in each of...

Let

and

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.)

The null hypothesis:	H0:
The alternative hypothesis:	H1:
The type of test statistic:	(Choose one)ZtChi squareF (include degree of freedom please)

The value of the test statistic: (Round to at least three decimal places.)
The p-value: (Round to at least three decimal places.)
Can we conclude that the proportion of women with anemia in the first country is greater than the proportion of women with anemia in the second country?		Yes	No

Solutions

Expert Solution

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

orchestra answered 1 year ago

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