# Solved (Free): Let p1 and p2 be the respective proportions of women with nutritional anemia in each of two developing countries. If a random sample

#### ByDr. Raju Chaudhari

Mar 29, 2021

Let p1 and p2 be the respective proportions of women with nutritional anemia in each of two developing countries. If a random sample of 200 women from the first country yielded 50 women with nutritional anemia, and an independently chosen, random sample of 350 women from the second country yielded 143 women with nutritional anemia, find a 90% confidence interval for p1-p2.

What is the lower limit of the 90% confidence interval?
What is the upper limit of the 90% confidence interval?

#### Solution

Given that $X_1 = 50$, $X_2 = 143$, $n_1 = 200$, $n_2 = 350$.

We want to determine $90$% confidence interval estimate of the difference between two population proportions $(p_1-p_2)$.

Specify the confidence level $(1-\alpha)$

Confidence level is $1-\alpha = 0.9$. Thus, the level of significance is $\alpha = 0.1$.

The estimate of the population proportions $p_1$ is $\hat{p}_1 =\frac{X_1}{n_1} =\frac{50}{200}=0.25$

and

the estimate of the population proportion $p_2$ is $\hat{p}_2 =\frac{X_2}{n_2} =\frac{143}{350}=0.409$.

Specify the formula

$100(1-\alpha)$% confidence interval estimate of the difference between two population proportions $(p_1-p_2)$ is

 \begin{aligned} (\hat{p}_1-\hat{p}_2) - E \leq (p_1 -p_2) \leq (\hat{p}_1 -\hat{p}_2)+ E. \end{aligned}
where $E = Z_{\alpha/2} \sqrt{\frac{\hat{p}_1*(1-\hat{p}_1)}{n_1}+\frac{\hat{p}_2*(1-\hat{p}_2)}{n_2}}$ and $Z_{\alpha/2}$ is the $Z$ value providing an area of $\alpha/2$ in the upper tail of the standard normal probability distribution.

Determine the critical value

The critical value of $Z$ for given level of significance is $Z_{\alpha/2}$. Z-critical 0.1

Thus $Z_{\alpha/2} = Z_{0.05} = 1.64$.

Compute the margin of error

The margin of error for the difference $(p_1-p_2)$ is

 \begin{aligned} E &= Z_{\alpha/2} \sqrt{\frac{\hat{p}_1*(1-\hat{p}_1)}{n_1}+\frac{\hat{p}_2*(1-\hat{p}_2)}{n_2}}\\ &= 1.64 \sqrt{\frac{0.25*(1-0.25)}{200}+\frac{0.409*(1-0.409)}{350}}\\ &= 0.066. \end{aligned}

Determine the confidence interval

$90$% confidence interval estimate of the difference between two population proportions $(p_1-p_2)$ is

 \begin{aligned} (\hat{p}_1-\hat{p}_2) - E &\leq (p_1-p_2) \leq (\hat{p}_1-\hat{p}_2) + E\\ (0.25-0.409) - 0.066 & \leq (p_1-p_2) \leq (0.25-0.409) + 0.066\\ -0.225 & \leq (p_1-p_2) \leq -0.092 \end{aligned}

Thus, $90$% confidence interval estimate of the difference between two population proportions $(p_1-p_2)$ is $(-0.225,-0.092)$.

The lower limit of the $90$% confidence interval is $-0.225$.

The upper limit of the $90$% confidence interval is $-0.092$.