Respuesta :
Answer:
[tex]z=\frac{0.36 -0.39}{\sqrt{\frac{0.39(1-0.39)}{100}}}=-0.615[/tex]
The p value for this case would be:
[tex]p_v =2*P(z<-0.615)=0.539[/tex]
For this case since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true proportion is not different from 0.39
Step-by-step explanation:
Information given
n=100 represent the random sample taken
X=36 represent the number of people that take E supplement
[tex]\hat p=\frac{36}{100}=0.36[/tex] estimated proportion of people who take R supplement
[tex]p_o=0.39[/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
Hypothesis to test
We want to test if the true proportion is equatl to 0.39 or not, the system of hypothesis are.:
Null hypothesis:[tex]p=0.39[/tex]
Alternative hypothesis:[tex]p \neq 0.39[/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 we got:
[tex]z=\frac{0.36 -0.39}{\sqrt{\frac{0.39(1-0.39)}{100}}}=-0.615[/tex]
The p value for this case would be:
[tex]p_v =2*P(z<-0.615)=0.539[/tex]
For this case since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true proportion is not different from 0.39