Question

A random sample of medical files is used to estimate the proportion p of all people who have blood type B. (a) If you have no pre-liminary estimate for p, how many medical files should you include in a random sample in order to be 90% sure that the point estimate will be within a distance of 0.03 from p?(b) Answer part (a) if you use the pre-liminary estimate that about 13 out of 90 people have blood type B.

Answer 1

Answer:

a) 752 medical files should be included.

b) 372 medical files should be included.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$.

The margin of error is of:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

90% confidence level

So $\alpha = 0.1$, z is the value of Z that has a p-value of $1 - \frac{0.1}{2} = 0.95$, so $Z = 1.645$.

Question a:

This is n for which M = 0.03. We have no estimate, so we use $\pi = 0.5$. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.03 = 1.645\sqrt{\frac{0.5*0.5}{n}}$

$0.03\sqrt{n} = 1.645*0.5$

$\sqrt{n} = \frac{1.645*0.5}{0.03}$

$(\sqrt{n})^2 = (\frac{1.645*0.5}{0.03})^2$

$n = 751.67$

Rounding up:

752 medical files should be included.

Question b:

Now we have that:

$\pi = \frac{13}{90} = 0.1444$

So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.03 = 1.645\sqrt{\frac{0.1444*0.8556}{n}}$

$0.03\sqrt{n} = 1.645\sqrt{0.1444*0.8556}$

$\sqrt{n} = \frac{1.645\sqrt{0.1444*0.8556}}{0.03}$

$(\sqrt{n})^2 = (\frac{1.645\sqrt{0.1444*0.8556}}{0.03})^2$

$n = 371.5$

Rounding up:

372 medical files should be included.