Respuesta :
Answer:
[tex]z=\frac{0.176 -0.2}{\sqrt{\frac{0.2(1-0.2)}{500}}}=-1.342[/tex]
We are conducting a bilateral test so then the p value would be:
[tex]p_v =2*P(z<-1.342)=0.180[/tex]
For this case since the p value is higher than the significance level we FAIL to reject the null hypothesis and we can conclude that true proportion of customers have at least two telephone lines is not diffeent from 0.2 or 20%
Step-by-step explanation:
Information given
n=500 represent the random sample taken
X=88 represent the people with two or more telephone lines
[tex]\hat p=\frac{88}{500}=0.176[/tex] estimated proportion of people with two or more lines
[tex]p_o=0.2[/tex] is the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level
z would represent the statistic
[tex]p_v[/tex] represent the p value
System of hypothesis
We want to check if the true proportion of people in a company with at least two telephone lines is 0.2 or no .:
Null hypothesis:[tex]p=0.2[/tex]
Alternative hypothesis:[tex]p \neq 0.2[/tex]
The statistic is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
Replacing the info given we got:
[tex]z=\frac{0.176 -0.2}{\sqrt{\frac{0.2(1-0.2)}{500}}}=-1.342[/tex]
We are conducting a bilateral test so then the p value would be:
[tex]p_v =2*P(z<-1.342)=0.180[/tex]
For this case since the p value is higher than the significance level we FAIL to reject the null hypothesis and we can conclude that true proportion of customers have at least two telephone lines is not diffeent from 0.2 or 20%