Respuesta :
Answer:
a) Null hypothesis:[tex]p=0.09[/tex]
Alternative hypothesis:[tex]p \neq 0.09[/tex]
b) [tex] z_{crit} =\pm 1.96[/tex]
And we can see on the figure attached that the rejection zone is in red
c) [tex]z=\frac{0.0995 -0.09}{\sqrt{\frac{0.09(1-0.09)}{221}}}=0.493[/tex]
d) [tex]p_v =2*P(z>0.493)=0.622[/tex]
e) Since the p value is higher than the significance level we have don't have enough evidence to conclude that the true proportion of people left handed is different from 9%
Step-by-step explanation:
Information given
n=221 represent the sample selected
X=22 represent the number of people that are left handed
[tex]\hat p=\frac{22}{221}=0.0995[/tex] estimated proportion of people left handed
[tex]p_o=0.09[/tex] is the value that we want to test
[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
Part a
We need to conduct a hypothesis in order to test the claim that the true proportion of people left handed is 0.09.:
Null hypothesis:[tex]p=0.09[/tex]
Alternative hypothesis:[tex]p \neq 0.09[/tex]
Part b
For this case we need to find a critical value who accumulates on each tail [tex]\alpha/2=0.025[/tex] of the area and we got:
[tex] z_{crit} =\pm 1.96[/tex]
And we can see on the figure attached that the rejection zone is in red
Part c
The statistic is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
[tex]z=\frac{0.0995 -0.09}{\sqrt{\frac{0.09(1-0.09)}{221}}}=0.493[/tex]
On the second figure we have the calculated value in green
Part d
Since is a bilateral test the p value is:
[tex]p_v =2*P(z>0.493)=0.622[/tex]
Part e
Since the p value is higher than the significance level we have don't have enough evidence to conclude that the true proportion of people left handed is not different from 9%