Respuesta :
Answer:
[tex]z=\frac{0.276 -0.3}{\sqrt{\frac{0.3(1-0.3)}{735}}}=-1.42[/tex]
Now we can claculate the p value with this formula:
[tex]p_v =P(z<-1.42)=0.0778[/tex]
If we use a signifiacn level of 5% we see that the p value is higher than 0.05 so then we have enough evidence to fail to reject the null hypothesis and we can't conclude that the true proportion is significantly higher than 0.3 at 5% of significance.
Step-by-step explanation:
Information to given
n=735 represent the random sample taken
X=203 represent the number of people who have their prostate regularly examined
[tex]\hat p=\frac{203}{735}=0.276[/tex] estimated proportion of people who have their prostate regularly examined
[tex]p_o=0.3[/tex] is the value to verify
z would represent the statistic
[tex]p_v[/tex] represent the p value
System of hypothesis
We want to test if the true proportion is less than 0.3, the ystem of hypothesis are.:
Null hypothesis:[tex]p \geq 0.3[/tex]
Alternative hypothesis:[tex]p < 0.3[/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{0.276 -0.3}{\sqrt{\frac{0.3(1-0.3)}{735}}}=-1.42[/tex]
Now we can claculate the p value with this formula:
[tex]p_v =P(z<-1.42)=0.0778[/tex]
If we use a signifiacn level of 5% we see that the p value is higher than 0.05 so then we have enough evidence to fail to reject the null hypothesis and we can't conclude that the true proportion is significantly higher than 0.3 at 5% of significance.
