Question

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?

.........< p <..........born?

Answer 1

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*10 = 3400 " babies"` in the sample.
• Large/Normal: We must check that both `(hatp)(n)` and `(1- hatp)(n)` are greater than or equal to 10:
  `(hatp)(n) = (.85)(340)=289 >=10`
  `(1- hatp)(n) = (1-.85)(340) = 51 >=10`

`hatp = frac{289}{340} = .85`

The equation for a one-proportion z-interval is:
`hatp +- z"*" sqrt(frac{(hatp)(1- hatp)}{n})`

Use the table of standard normal probabilities to find `z"*"`.
To find `z`, we want the area to the left to have a probability of `(1-.99)/2 = 0.005`

We get:
`z"*" = 2.5758`

Substitute values into the formula:
`0.85 +- (2.5758) sqrt(frac{(.85)(.15)}{340})`

`0.85 +- 0.0499`

Confidence interval: `[0.8001, 0.8999]`

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