Question

For developing countries in Africa and the Americas, let p₁ and p₂ be the respective proportions of babies with a low birth weight (below 2500 grams). A random sample of n₁ = 2000 African women yielded y₁ = 750 with nutritional anemia anda random sample of n₂ = 2000 women from the Americas yielded y₂ = 650 women with nutritional anemia. We shall test H₀: p₁ = p₂ against the alternative hypothesis H₁: p₁ > p₂ at α = 0.05.

Answer 1

Here, we have given that,

n1= Number of African women yielded=2000

y1= Number of African women yielded with nutritional anaemia=750

=Sample proportion of African women yielded with nutritional anaemia

=

n2= Number of America's women yielded= 2000

y2= Number of America's women yielded with nutritional anaemia= 650

=Sample proportion of America's women yielded with nutritional anaemia

=

Claim: To check whether the population proportion of African women yielded with nutritional anaemia greater than the population proportion of America's women yielded with nutritional anaemia.

The null and alternative hypotheses are as follows:

v/s

Where p1 = The population proportion of African women yielded with nutritional anaemia

p2= The population proportion of America's women yielded with nutritional anaemia.

This is the Right one-tailed test.

Here, we are using the two-sample proportion test.

Now, we can find the test statistics

Z-statistics=

                 =

       =

= 3.32

The test statistics is 3.32.

Now, we can find the p-value

P-value = P( Z> z-statistics) as this is right one-tailed test

              =1- P (Z < 3.32)

              = 1- 0.99955 Using standard normal z table see the value corresponding to the z=-3.32

              =0.0004

we get the p-value is 0.0004

Decision:

=level of significance =0.05

Here P-value (0.0004) less than (<) 0.05

Conclusion:

Since the p-value is less than 0.05 , we can conclude that there is sufficient evidence to support the claim

(i.e. the population proportion of African women yielded with nutritional anaemia greater than the population proportion of America's women yielded with nutritional anaemia. )