Answer:Yes
Step-by-step explanation:
Given
n=390 x=312
[tex]\hat{p}=\frac{312}{390}=0.8[/tex]
Confidence level=99 %
[tex] Z_{\frac{\alpha }{2}}=2.575[/tex]
Standard error(S.E.)=[tex]\sqrt{\frac{\hat{p}\left ( 1-\hat{p}\right )}{n}}[/tex]
S.E.=[tex]\sqrt{\frac{0.8\times 0.2}{390}}[/tex]
S.E.=0.0202
Confidence interval
[tex]p\pm \left [ z_{\frac{\alpha }{2}}\cdot S.E.\right ][/tex]
[tex]0.8 \pm 0.0521[/tex]
[tex]\left ( 0.7479,0.8521 \right )[/tex]
Since 0.5 does not lie in interval therefore method appear to be effective
