A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 290 babies were born, and 261 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?

The 99% confidence interval estimate of the percentage of girls born is between 85.46% and 94.54%.

Usually, babies are equally as likely to be boys or girls. Here, it is desired to increase the probability of conceiving a girl, which is achieved, considering the lower bound of the confidence interval is considerably above 50%.

Step-by-step explanation:

Confidence interval for the proportion:

In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of , we have the following confidence interval of proportions.

In which

z is the zscore that has a pvalue of .

For this problem, we have that:

99% confidence level

So , z is the value of Z that has a pvalue of , so .

The lower limit of this interval is:

The upper limit of this interval is:

Percentage:

Proportion multplied by 100.

0.8546*100 = 85.46%

0.9454*100 = 94.54%

The 99% confidence interval estimate of the percentage of girls born is between 85.46% and 94.54%.

Based on the result, does the method appear to be effective?

Usually, babies are equally as likely to be boys or girls. Here, it is desired to increase the probability of conceiving a girl, which is achieved, considering the lower bound of the confidence interval is considerably above 50%.

10a^3+277a^2-132a-792

step-by-step explanation:

(2a+11)(5a-12)(foil)

=10a^2-24a+55a-132 =10a^2+31a-132

(10a^2+31a-132)(a+6)

=10a^3+60a^2+31a^2+186a^2-132a-792 =10a^3+277a^2-132a-792

i hope this is what you are looking for : )