# Clinical trial tests a method designed to increase the probability of conceiving a girl. In the study, 340 babies were born, and 289 of them were girls. Use the sample data to construct a 99​% confidence interval estimate of the percentage of girls born?

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Jun 6, 2018

We are 99% confident that the true proportion of female babies is captured by the interval from $0.8001$ to $0.8999$.

#### Explanation:

We want to estimate the proportion of female babies born.

We can use a one-proportion $z$ interval with confidence level 99% if the following conditions are met:

• Random: We must assume that the study uses a random sample of some population of babies born.
• Independence/10% condition: We must assume that there are more than $340 \cdot 10 = 3400 \text{ babies}$ in the sample.
• Large/Normal: We must check that both $\left(\hat{p}\right) \left(n\right)$ and $\left(1 - \hat{p}\right) \left(n\right)$ are greater than or equal to 10:
$\left(\hat{p}\right) \left(n\right) = \left(.85\right) \left(340\right) = 289 \ge 10$
$\left(1 - \hat{p}\right) \left(n\right) = \left(1 - .85\right) \left(340\right) = 51 \ge 10$

$\hat{p} = \frac{289}{340} = .85$

The equation for a one-proportion z-interval is:
$\hat{p} \pm z \text{*} \sqrt{\frac{\left(\hat{p}\right) \left(1 - \hat{p}\right)}{n}}$

Use the table of standard normal probabilities to find $z \text{*}$.
To find $z$, we want the area to the left to have a probability of $\frac{1 - .99}{2} = 0.005$

We get:
$z \text{*} = 2.5758$

Substitute values into the formula:
$0.85 \pm \left(2.5758\right) \sqrt{\frac{\left(.85\right) \left(.15\right)}{340}}$

$0.85 \pm 0.0499$

Confidence interval: $\left[0.8001 , 0.8999\right]$

We are 99% confident that the true proportion of female babies is captured by the interval from $0.8001$ to $0.8999$.