Respuesta :
Answer:
[tex]z=\frac{0.667 -0.5}{\sqrt{\frac{0.5(1-0.5)}{165}}}=4.29[/tex]
The p value for this case is given by:
[tex]p_v =P(z>4.29)=8.93x10^{-6}[/tex]
Since the p value is lower than the significance level we have enough evidence to conclude that the true proportion is significantly higher than 0.5 at 5% of significance.
Step-by-step explanation:
Information given
n=165 represent the random sample selected
55 represent the students indicated a belief that such software unfairly targets students
X =165-55= 110 represent students who NOT belief that such software unfairly targets students
[tex]\hat p=\frac{110}{165}=0.667[/tex] estimated proportion of students who NOT belief that such software unfairly targets students
[tex]p_o=0.5[/tex] is the value that we want to check
[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 majority of students at the university do not believe that it unfairly targets them, and the system of hypothesis are:
Null hypothesis:[tex]p\leq 0.5[/tex]
Alternative hypothesis:[tex]p > 0.5[/tex]
The statistic for this case is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
After replace we got:
[tex]z=\frac{0.667 -0.5}{\sqrt{\frac{0.5(1-0.5)}{165}}}=4.29[/tex]
The p value for this case is given by:
[tex]p_v =P(z>4.29)=8.93x10^{-6}[/tex]
Since the p value is lower than the significance level we have enough evidence to conclude that the true proportion is significantly higher than 0.5 at 5% of significance.
