A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 624 babies were born, and 312 of them were girls. Use the sample data to construct a 99% confidence interval estimate of the percentage of girls born. Based on the result, does the method appear to be effective?
__ < p < ___
To construct a confidence interval estimate for the percentage of girls born, you can use the formula:
CI = p̂ ± Z * sqrt((p̂*(1-p̂))/n)
Where:
- p̂ (p-hat) is the sample proportion of girls born (312/624 = 0.5)
- Z is the Z-score corresponding to your desired confidence level (99% confidence level corresponds to a Z-score of 2.576)
- n is the sample size (624)
Let's substitute these values into the formula:
CI = 0.5 ± 2.576 * sqrt((0.5*(1-0.5))/624)
Calculating the values inside the square root:
CI = 0.5 ± 2.576 * sqrt(0.25/624)
Completing the calculation inside the square root:
CI = 0.5 ± 2.576 * sqrt(0.0004)
CI = 0.5 ± 2.576 * 0.02
Calculating the product of Z and the square root:
CI = 0.5 ± 0.05152
Finally, calculate the upper and lower bounds by adding and subtracting the result of the previous calculation from the sample proportion:
Lower bound = 0.5 - 0.05152 = 0.44848
Upper bound = 0.5 + 0.05152 = 0.55152
Therefore, the confidence interval for the percentage of girls born is:
0.44848 < p < 0.55152
Since the confidence interval does not contain 50% (0.5), which represents an equal chance of having a girl or a boy, the method appears to be effective in increasing the probability of conceiving a girl.