A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 320 babies were born, and 256 of them were girls. Use the sample data to construct a 99% confidence interval estimate of the percentage of girls born. Based on the results, does the method appear to be effective?
To construct a confidence interval estimate, we can use the formula:
Confidence Interval = Sample proportion ± Margin of error
First, we need to calculate the sample proportion, which is the number of girls divided by the total number of babies:
Sample proportion = Number of girls / Total number of babies
= 256 / 320
= 0.8
Next, we need to calculate the margin of error. The formula for the margin of error is:
Margin of error = Z * √((p * (1 - p)) / n)
Where:
Z is the Z-score corresponding to the desired confidence level (99% confidence level corresponds to a Z-score of 2.576)
p is the sample proportion
n is the sample size
In this case, the Z-score is 2.576, the sample proportion (p) is 0.8, and the sample size (n) is 320.
Margin of error = 2.576 * √((0.8 * (1 - 0.8)) / 320)
= 2.576 * √((0.16) / 320)
= 0.0519
Now we can construct the confidence interval:
Confidence Interval = Sample proportion ± Margin of error
Confidence Interval = 0.8 ± 0.0519
Therefore, the 99% confidence interval estimate of the percentage of girls born is (0.7481, 0.8519).
To determine if the method appears to be effective, we can check if the confidence interval includes the desired outcome of having more than 50% girls (0.5). In this case, the lower bound of the confidence interval is 0.7481, which is greater than 0.5. It suggests that the method may be effective in increasing the probability of conceiving a girl. However, further analysis and consideration of other factors are necessary to draw a definitive conclusion.