Respuesta :
Answer:
[tex]z=\frac{0.3725 -0.4}{\sqrt{\frac{0.4(1-0.4)}{400}}}=-1.123[/tex]
The p value for a left tailed test would be:
[tex]p_v =P(z<-1.123)=0.131[/tex]
Since the p value is very higher we can conclude that the true proportion of teenagers who floss twice a day is NOT less than 40%.
Step-by-step explanation:
Information given
n=400 represent the random sample given
X=149 represent the floss twice a day
[tex]\hat p=\frac{149}{400}=0.3725[/tex] estimated proportion of floss twice a day
[tex]p_o=0.4[/tex] is the value the proportion that we want to check
z would represent the statistic
[tex]p_v[/tex] represent the p value
System of hypothesis
We want to check proportion of teenagers who floss twice a day is less than 40%, so then the system of hypothesis are.:
Null hypothesis:[tex]p \geq 0.4[/tex]
Alternative hypothesis:[tex]p < 0.4[/tex]
For the one sample proportion test the statistic is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
If we replace the info given we got:
[tex]z=\frac{0.3725 -0.4}{\sqrt{\frac{0.4(1-0.4)}{400}}}=-1.123[/tex]
The p value for a left tailed test would be:
[tex]p_v =P(z<-1.123)=0.131[/tex]
Since the p value is very higher we can conclude that the true proportion of teenagers who floss twice a day is NOT less than 40%.
