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.

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

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

z is the z-score that has a p-value of [tex]1 - \frac{\alpha}{2}[/tex].

The margin of error is of:

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

90% confidence level

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

Question a:

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

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

[tex]0.03 = 1.645\sqrt{\frac{0.5*0.5}{n}}[/tex]

[tex]0.03\sqrt{n} = 1.645*0.5[/tex]

[tex]\sqrt{n} = \frac{1.645*0.5}{0.03}[/tex]

[tex](\sqrt{n})^2 = (\frac{1.645*0.5}{0.03})^2[/tex]

[tex]n = 751.67[/tex]

Rounding up:

752 medical files should be included.

Question b:

Now we have that:

[tex]\pi = \frac{13}{90} = 0.1444[/tex]

So

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

[tex]0.03 = 1.645\sqrt{\frac{0.1444*0.8556}{n}}[/tex]

[tex]0.03\sqrt{n} = 1.645\sqrt{0.1444*0.8556}[/tex]

[tex]\sqrt{n} = \frac{1.645\sqrt{0.1444*0.8556}}{0.03}[/tex]

[tex](\sqrt{n})^2 = (\frac{1.645\sqrt{0.1444*0.8556}}{0.03})^2[/tex]

[tex]n = 371.5[/tex]

Rounding up:

372 medical files should be included.