# Solved: A random sample of 487 nonsmoking women of normal weight (body mass index between 19.8 and 26.0) who had given birth at a large metropolitan medical center was selected

#### ByDr. Raju Chaudhari

Feb 23, 2021

A random sample of 487 nonsmoking women of normal weight (body mass index between 19.8 and 26.0) who had given birth at a large metropolitan medical center was selected ( The Effects of Cigarette Smoking and Gestational Weight Change on Birth Outcomes in Obese and Normal-Weight Women, Amer. J. Public Health, 1997: 591-596). It was determined that 7.2% of these births resulted in children of low birth weight (less than 2500 g). Calculate an upper confidence bound using a confidence level of 99% for the proportion of all such births that result in children of low birth weight.

#### Solution

Given that sample size $n = 487$. The sample proportion is $\hat{p}=0.072$.

Confidence level is $1-\alpha = 0.99$. Thus, the level of significance is $\alpha = 0.01$.

$100(1-\alpha)$% confidence interval for population proportion is

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

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

Thus $Z_{\alpha/2} = Z_{0.005} = 2.576$.

The margin of error for proportions is

 \begin{aligned} E & = Z_{\alpha/2} \sqrt{\frac{\hat{p}*(1-\hat{p})}{n}}\\ & = 2.576 \sqrt{\frac{0.072*(1-0.072)}{487}}\\ & =0.03. \end{aligned}

$99$% confidence interval estimate for population proportion is

 \begin{aligned} \hat{p} - E & \leq p \leq \hat{p} + E\\ 0.072 - 0.03 & \leq p \leq 0.072 + 0.03\\ 0.0418 & \leq p \leq 0.1022. \end{aligned}
Thus, an upper confidence bound using a confidence level of 99% for the proportion of all such births that result in children of low birth weight is $0.1022$.