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.

asked Oct 10, 2021 22.5k views

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Jazzyfresh · Answer 1 · 2021-10-16T15:45:37+0000

Answer:

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

For this case we know this:

n=37 , p=0.2

We can find the standard error like this:

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

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:

$z = (p -\mu_p)/(SE_p)$

And using this we got this:

P(p>0.4) = 1-P(z< (0.4-0.2)/(0.0658)) = 1-P(z<3.04) = 0.0012

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)

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 :