Question

In a recent survey of 150 teenagers, 93 stated that they always wear their seatbelt when they travel in a car. Assuming the distribution is approximately normal, find the point estimate and standard error for the proportion of teenagers that always wear a seatbelt when traveling in a car. Round your answers to three decimal places, as needed.

Answer 1

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: $\sqrt{.62(1-.62)/150$ which is .0396 rounded to .04

Answer 2

Answer:

$\hat p \sim N (p ,\sqrt{\frac{p*(1-p)}{n}}$

The estimated standard error is given by:

$SE= \sqrt{\frac{0.62*(1-0.62)}{150}}=0.040$

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 :

$\hat p = \frac{93}{150}= 0.62$

We can can check if we can use the normal approximation :

$np =150*0.62= 93>10$

$n(1-p) = 150*(1-0.62)= 57>10$

So then we can use the normal approximation and the distribution for the proportion is given:

$\hat p \sim N (p ,\sqrt{\frac{p*(1-p)}{n}}$

The estimated standard error is given by:

$SE= \sqrt{\frac{0.62*(1-0.62)}{150}}=0.040$