Respuesta :
Answer:
a) We want to test the claim that 35% of adults have heard of the new electronic reader, then the system of hypothesis are.:
Null hypothesis:[tex]p=0.35[/tex]
Alternative hypothesis:[tex]p \neq 0.35[/tex]
And is a two tailed test
b) [tex]z=\frac{0.334 -0.35}{\sqrt{\frac{0.35(1-0.35)}{1562}}}=-1.326[/tex]
c) [tex]p_v =2*P(z<-1.326)=0.184[/tex]
d) Null hypothesis:[tex]p=0.35[/tex]
e) Fail to reject the null hypothesis because the P-value is greater than the significance level, alpha.
Step-by-step explanation:
Information provided
n=1562 represent the random sample selected
X=522 represent the people who have heard of a new electronic reader
[tex]\hat p=\frac{522}{1562}=0.334[/tex] estimated proportion of people who have heard of a new electronic reader
[tex]p_o=0.35[/tex] is the value to verify
[tex]\alpha=0.05[/tex] represent the significance level
z would represent the statistic
[tex]p_v[/tex] represent the p value
Part a
We want to test the claim that 35% of adults have heard of the new electronic reader, then the system of hypothesis are.:
Null hypothesis:[tex]p=0.35[/tex]
Alternative hypothesis:[tex]p \neq 0.35[/tex]
And is a two tailed test
Part b
The statistic for this case is given :
[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{0.334 -0.35}{\sqrt{\frac{0.35(1-0.35)}{1562}}}=-1.326[/tex]
Part c
We can calculate the p value using the laternative hypothesis with the following probability:
[tex]p_v =2*P(z<-1.326)=0.184[/tex]
Part d
The null hypothesis for this case would be:
Null hypothesis:[tex]p=0.35[/tex]
Part e
The best conclusion for this case would be:
Fail to reject the null hypothesis because the P-value is greater than the significance level, alpha.
