Respuesta :
Answer:
[tex]\hat p \sim N (p ,\sqrt{\frac{p*(1-p)}{n}}[/tex]
The estimated standard error is given by:
[tex] SE= \sqrt{\frac{0.62*(1-0.62)}{150}}=0.040[/tex]
Step-by-step explanation:
For this case we have the following data:
n =150 represent the sample size selected
x = 93 people stated that they always wear their seatbelt when they travel in a car
For this case the the proportion estimated is :
[tex] \hat p = \frac{93}{150}= 0.62[/tex]
We can can check if we can use the normal approximation :
[tex] np =150*0.62= 93>10[/tex]
[tex]n(1-p) = 150*(1-0.62)= 57>10[/tex]
So then we can use the normal approximation and the distribution for the proportion is given:
[tex]\hat p \sim N (p ,\sqrt{\frac{p*(1-p)}{n}}[/tex]
The estimated standard error is given by:
[tex] SE= \sqrt{\frac{0.62*(1-0.62)}{150}}=0.040[/tex]
Answer:
p'=.62
Op'=.04
Step-by-step explanation:
To find p' divide 93 by 150:
93/150=.62
To find Op' put p' into the equation: [tex]\sqrt{.62(1-.62)/150[/tex] which is .0396 rounded to .04
