Answer:
Based on the confidence interval, the method appears to not bring expected results
Step-by-step explanation:
The given parameters are;
The number of babies in the study, n = 620 babies
The number of girls in the study sample = 341 girls
The proportion of girls in the study sample, [tex]\hat p[/tex] = 341/620 = 0.55
At 99% confidence level, we have the critical z = 2.576;
The confidence interval is given by the following formula;
[tex]CI=\hat{p}\pm z\times \sqrt{\dfrac{\hat{p} \cdot (1-\hat{p})}{n}}[/tex]
Plugging in the values of the variables, we get;
[tex]CI=0.55\pm 2.576\times \sqrt{\dfrac{0.55 \times (1-0.55)}{620}} \approx 0.55 \ \pm 5.14680 \times 10^{-2}[/tex]
Therefore, 0.498532 < p < 0.601468
Based on the result of the confidence interval, we can be 99% sure that the true mean is between 0.498532 and 0.601468, therefore, given that the range includes a true mean of 0.5 or 50% probability of conceiving a girl, which is the percentage predicted by genetics, the method appears ineffective
