Answer:
The result is significant.
Step-by-step explanation:
Here,
p = 30 ÷ 330 = 0.091
thus, q = 1 - p = 0.909
Also, [tex]\sigma=\sqrt{\frac{pq}{n} }[/tex]
⇒ [tex]\sigma=\sqrt{\frac{0.091\times0.909}{330} }=0.0158 [/tex]
The z-score is:
[tex]z=\frac{p-p_{0}}{\sigma}[/tex]
[tex]z=\frac{0.091-0.05}{0.0158}=2.595[/tex]
Thus, at 95% confidence interval, p-value = 0.005
The result is significant at p < .05.