Question

Blood pressure: High blood pressure has been identified as a risk factor for heart attacks and strokes. The proportion of U.S. adults with high blood pressure is 0.2. A sample of 37 U.S. adults is chosen. Use the TI-84 Plus Calculator as needed. Round the answer to at least four decimal places.

Answer 1

Answer:

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

For this case we know this:

$n=37 , p=0.2$

We can find the standard error like this:

$SE = \sqrt{\frac{\hat p (1-\hat p)}{n}}= \sqrt{\frac{0.2*0.8}{37}}= 0.0658$

So then our random variable can be described as:

$p \sim N(0.2, 0.0658)$

Let's suppose that the question on this case is find the probability that the population proportion would be higher than 0.4:

$P(p>0.4)$

We can use the z score given by:

$z = \frac{p -\mu_p}{SE_p}$

And using this we got this:

$P(p>0.4) = 1-P(z< \frac{0.4-0.2}{0.0658}) = 1-P(z$

And we can find this probability using the Ti 84 on this way:

2nd> VARS> DISTR > normalcdf

And the code that we need to use for this case would be:

1-normalcdf(-1000, 3.04; 0;1)

Or equivalently we can use:

1-normalcdf(-1000, 0.4; 0.2;0.0658)

Step-by-step explanation:

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

$np = 37 *0.2 = 7.4 \geq 5$

$n(1-p) = 37*0.8 = 29.6\geq 5$

We assume independence on each event and a random sampling method so we can conclude that we can use the normal approximation and then ,the population proportion have the following distribution :

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

For this case we know this:

$n=37 , p=0.2$

We can find the standard error like this:

$SE = \sqrt{\frac{\hat p (1-\hat p)}{n}}= \sqrt{\frac{0.2*0.8}{37}}= 0.0658$

So then our random variable can be described as:

$p \sim N(0.2, 0.0658)$

Let's suppose that the question on this case is find the probability that the population proportion would be higher than 0.4:

$P(p>0.4)$

We can use the z score given by:

$z = \frac{p -\mu_p}{SE_p}$

And using this we got this:

$P(p>0.4) = 1-P(z< \frac{0.4-0.2}{0.0658}) = 1-P(z$

And we can find this probability using the Ti 84 on this way:

2nd> VARS> DISTR > normalcdf

And the code that we need to use for this case would be:

1-normalcdf(-1000, 3.04; 0;1)

Or equivalently we can use:

1-normalcdf(-1000, 0.4; 0.2;0.0658)