Answer:
The right answer is "z = 3.68".
Step-by-step explanation:
Given that,
p = 52%
or,
= 0.52
n = 200
[tex]\hat{p}= \frac{130}{200}[/tex]
[tex]= 0.65[/tex]
Now
The test statistic will be:
⇒ [tex]z=\frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n} } }[/tex]
On substituting all the given values, we get
⇒ [tex]=\frac{0.65-0.52}{\sqrt{\frac{0.52(1-0.52)}{200} } }[/tex]
⇒ [tex]=\frac{0.13}\sqrt{{\frac{0.52\times 0.48}{200} }}[/tex]
⇒ [tex]=\frac{0.13}{\sqrt{0.001248} }[/tex]
⇒ [tex]=3.68[/tex]
