Answer:
The 99% confidence interval estimate of the percentage of girls born is (0.8, 0.9).
b)Yes, the proportion of girls is significantly different from 0.5.
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of
, and a confidence interval
, 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:
In the study 340 babies were born, and 289 of them were girls. This means that 
Use the sample data to construct a 99% confidence interval estimate of the percentage of girls born.
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:

The 99% confidence interval estimate of the percentage of girls born is (0.8, 0.9).
Does the method appear to be effective?
b)Yes, the proportion of girls is significantly different from 0.5.