Answer:
Z= -4.33
Step-by-step explanation:
Hello!
The study variable is:
X: number of voters that support the candidate out of 200 surveyed voters.
This variable has a binomial distribution, where the "success" is that the voter supports the candidate and the "failure" is that the voter doesn't support the candidate. X~Bi(n;p)
Considering the sample size is large enough, you can apply the Central Limit Theorem and approximate the distribution of the sample proportion to normal: ^p ≈ N(p;[tex]\frac{p(1-p)}{n}[/tex])
Using this distribution you can an approximation to the standard normal distribution: [tex]Z= \frac{p'-p}{\sqrt{\frac{p(1-p)}{n} }}[/tex]≈N(0;1)
Where the sample proportion is ^p= p'= [tex]\frac{x}{n} = \frac{90}{200} = 0.45[/tex]
x= number of successes
n= sample size
The population proportion is p= 0.60
[tex]Z= \frac{0.45-0.60}{\sqrt{\frac{0.60*0.40}{200} }}= -4.33[/tex]
I hope this helps!
