Answer:
No, Method is not effective.
Step-by-step explanation:
Let x be the number of girls = 319
and n be the total number of babies = 580
Then, proportion
[tex]\hat{p}=\frac{p}{n}\\\\=\frac{310}{580}\\\\=0.55[/tex]
Standard Error,
[tex]SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\\= \sqrt{\frac{0.55(1-0.55)}{580}}\\= 0.02066[/tex]
The level of significance is:
[tex]Z_{\frac{\alpha}{2}}=Z_{0.005}=2.575[/tex]
Confidence Interval:
[tex]p \pm [Z_{\frac{\alpha}{2}} \times SE]\\=0.550 \pm [2.575 \times 0.02066\\=(0.4968, 0.6032)[/tex]
Since 0.50 lies in the interval hence we fail to reject
[tex]H_0: p = 0.50[/tex]
Thus, the method does not appear to be effective.
