Respuesta :
Answer:
[tex] P(\hat p>0.14)[/tex]
And using the z score given by:
[tex] z = \frac{\hat p -\mu_p}{\sigma_p}[/tex]
Where:
[tex]\mu_{\hat p} = 0.12[/tex]
[tex]\sigma_{\hat p}= \sqrt{\frac{0.12*(1-0.12)}{474}}= 0.0149[/tex]
If we find the z score for [tex]\hat p =0.14[/tex] we got:
[tex]z = \frac{0.14-0.12}{0.0149}= 1.340[/tex]
So we want to find this probability:
[tex] P(z>1.340)[/tex]
And using the complement rule and the normal standard distribution and excel we got:
[tex] P(Z>1.340) = 1-P(Z<1.340) = 1-0.9099= 0.0901[/tex]
Step-by-step explanation:
For this case we have the proportion of interest given [tex] p =0.12[/tex]. And we have a sample size selected n = 474
The distribution of [tex]\hat p[/tex] is given by:
[tex] \hat p \sim N (p , \sqrt{\frac{p(1-p)}{n}}) [/tex]
We want to find this probability:
[tex] P(\hat p>0.14)[/tex]
And using the z score given by:
[tex] z = \frac{\hat p -\mu_p}{\sigma_p}[/tex]
Where:
[tex]\mu_{\hat p} = 0.12[/tex]
[tex]\sigma_{\hat p}= \sqrt{\frac{0.12*(1-0.12)}{474}}= 0.0149[/tex]
If we find the z score for [tex]\hat p =0.14[/tex] we got:
[tex]z = \frac{0.14-0.12}{0.0149}= 1.340[/tex]
So we want to find this probability:
[tex] P(z>1.340)[/tex]
And using the complement rule and the normal standard distribution and excel we got:
[tex] P(Z>1.340) = 1-P(Z<1.340) = 1-0.9099= 0.0901[/tex]
