let $p_1$ and $p_2$ be the respective proportions of women with nutritional anemia in each of two developing…

let $p_1$ and $p_2$ be the respective proportions of women with nutritional anemia in each of two developing countries. if a random sample of 300 women from the first country yielded 123 women with nutritional anemia, and an independently chosen, random sample of 350 women from the second country yielded 136 women with nutritional anemia, find a 99% confidence interval for $p_1 - p_2$. then find the lower limit and upper limit of the 99% confidence interval.\ncarry your intermediate computations to at least three decimal places. round your responses to at least three decimal places. (if necessary, consult a list of formulas.)\nlower limit: \nupper limit:

let $p_1$ and $p_2$ be the respective proportions of women with nutritional anemia in each of two developing countries. if a random sample of 300 women from the first country yielded 123 women with nutritional anemia, and an independently chosen, random sample of 350 women from the second country yielded 136 women with nutritional anemia, find a 99% confidence interval for $p_1 - p_2$. then find the lower limit and upper limit of the 99% confidence interval.\ncarry your intermediate computations to at least three decimal places. round your responses to at least three decimal places. (if necessary, consult a list of formulas.)\nlower limit: \nupper limit:

Answer

Explanation:

Step1: Calculate sample proportions

For the first sample, $n_1 = 300$ and $x_1=123$, so $\hat{p}_1=\frac{x_1}{n_1}=\frac{123}{300}=0.41$. For the second sample, $n_2 = 350$ and $x_2 = 136$, so $\hat{p}_2=\frac{x_2}{n_2}=\frac{136}{350}\approx0.389$.

Step2: Calculate the standard - error

The formula for the standard - error of $\hat{p}_1-\hat{p}_2$ is $SE=\sqrt{\frac{\hat{p}_1(1 - \hat{p}_1)}{n_1}+\frac{\hat{p}_2(1 - \hat{p}_2)}{n_2}}$. Substitute the values: [ \begin{align*} SE&=\sqrt{\frac{0.41\times(1 - 0.41)}{300}+\frac{0.389\times(1 - 0.389)}{350}}\ &=\sqrt{\frac{0.41\times0.59}{300}+\frac{0.389\times0.611}{350}}\ &=\sqrt{\frac{0.2419}{300}+\frac{0.2377}{350}}\ &=\sqrt{0.0008063 + 0.0006791}\ &=\sqrt{0.0014854}\ &\approx0.0385 \end{align*} ]

Step3: Find the z - value

For a 99% confidence interval, the significance level $\alpha=1 - 0.99 = 0.01$, and $\alpha/2=0.005$. The z - value $z_{\alpha/2}=z_{0.005}=2.576$.

Step4: Calculate the confidence interval

The confidence interval for $p_1 - p_2$ is $(\hat{p}_1-\hat{p}2)\pm z{\alpha/2}\times SE$. [ \begin{align*} (\hat{p}_1-\hat{p}2)-z{\alpha/2}\times SE&=(0.41 - 0.389)-2.576\times0.0385\ &=0.021- 0.0982\ &=- 0.077 \end{align*} ] [ \begin{align*} (\hat{p}_1-\hat{p}2)+z{\alpha/2}\times SE&=(0.41 - 0.389)+2.576\times0.0385\ &=0.021 + 0.0982\ &=0.119 \end{align*} ]

Answer:

Lower limit: $-0.077$ Upper limit: $0.119$