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?

inimigurhkg

inimigurhkg

Answered question

2023-02-22

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?

Answer & Explanation

Aleksy7860

Aleksy7860

Beginner2023-02-23Added 6 answers

The proportion of female babies born needs to be estimated.
If the following criteria are satisfied, we can use a one-proportion z interval with a confidence level of 99%:
Random: It is necessary to assume that the study draws a sample of babies from a population at random.
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 ( p ^ ) ( n ) and ( 1 - p ^ ) ( n ) are greater than or equal to 10:
( p ^ ) ( n ) = ( .85 ) ( 340 ) = 289 10
( 1 - p ^ ) ( n ) = ( 1 - .85 ) ( 340 ) = 51 10
p ^ = 289 340 = .85
The equation for a one-proportion z-interval is:
p ^ ± z * ( p ^ ) ( 1 - p ^ ) 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
Put values in the formula as replacements:
0.85 ± ( 2.5758 ) ( .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.

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