Answer :
Answer:
[tex]z=\frac{0.116 -0.125}{\sqrt{\frac{0.125(1-0.125)}{2000}}}=-1.217[/tex]
We are conducting a bilateral test then the p value would is:
[tex]p_v =2*P(z<-1.217)=0.224[/tex]
Since the p value obtained is higher than the significance level used of 0.05 we have enough evidence to FAIL to reject the null hypothesis and there is no evidence to conclude that percentage of 18-25-year-olds who currently use marijuana or hashish has changed from the 1997 percentage of 12.5%
Step-by-step explanation:
Information given
n=2000 represent the random sample of people
X=232 represent the people between 18-25-year-old who currently use marijuana or hashish
[tex]\hat p=\frac{232}{2000}=0.116[/tex] estimated proportion of people between 18-25-year-old who currently use marijuana or hashish
[tex]p_o=0.125[/tex] is the value that we want to verify
[tex]\alpha=0.05[/tex] represent the significance level
Confidence=95% or 0.95
z would represent the statistic
[tex]p_v[/tex] represent the p value
Hypothesis to test
We want to verify if the percentage of 18-25-year-olds who currently use marijuana or hashish has changed from the 1997 percentage of 12.5%.:
Null hypothesis:[tex]p=0.125[/tex]
Alternative hypothesis:[tex]p \neq 0.125[/tex]
The statistic for a proportion z test is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
Replacing into the previous formula we got:
[tex]z=\frac{0.116 -0.125}{\sqrt{\frac{0.125(1-0.125)}{2000}}}=-1.217[/tex]
P value
We are conducting a bilateral test then the p value would is:
[tex]p_v =2*P(z<-1.217)=0.224[/tex]
Since the p value obtained is higher than the significance level used of 0.05 we have enough evidence to FAIL to reject the null hypothesis and there is no evidence to conclude that percentage of 18-25-year-olds who currently use marijuana or hashish has changed from the 1997 percentage of 12.5%