Question

Let P, and pą be the respective proportions of women with iron-deficiency anemia in each of two developing countries. A random sample of 2100 women from the first country yielded 418 women with iron deficiency anemia, and an independently chosen, random sample of 2400 women from the second country yielded 560 women with iron-deficiency anemia. Can we conclude, at the 0.05 level of significance, that the proportion of women with anemia in the first country affers from the proportion of women with anemia in the second country? Perform a two-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.

Answer 1

Ho : p1 = p2
Ha : p1  $\neq$ p2

Level of Significance(l.o.s.) :  $\alpha$ = 0.05

Decision Criteria : Reject Ho at 5% l.o.s. if | Z cal | > Z tab,
where Z tab = Z ($\alpha$/2) = Z (0.025) = 1.96

Calculation :  $\hat{p}_{1}$ = 418/2100 = 0.1991,  $\hat{p}_{2}$ = 560/2400 = 0.2333 , n1 = 2100 & n2 = 2400

p = $\frac{\hat{p}_{1}*n_{1}+\hat{p}_{2}*n_{2}}{n_{1}+n_{2}}$ =  $\frac{418+560}{2100+2400}$ = 0.2173

Z cal =  $\frac{\hat{p}_{1}-\hat{p}_{2}}{\sqrt{\frac{p*(1-p)}{n}}}$
  =  $\frac{0.1991-0.2333}{\sqrt{{0.2173*(1-0.2173)}*(\frac{1}{2100}+\frac{1}{2400})}}$
= -2.775
p-value = P( |Z| < -2.775 ) = 0.006

Conclusion : Since | Z cal | > Z tab, we reject Ho at 5% l.o.s. & thus, conclude that the proportion of women with anemia in the first country differs from the proportion of women with anemia in the second country.

Hope this answers your query!