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

a) 752 medical files should be included.

b) 372 medical files should be included.

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{(\pi(1-\pi))/(n)}`

In which

z is the z-score that has a p-value of
`1 - (\alpha)/(2)`.

The margin of error is of:

`M = z\sqrt{(\pi(1-\pi))/(n)}`

90% confidence level

So
`\alpha = 0.1`, z is the value of Z that has a p-value of
`1 - (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{(\pi(1-\pi))/(n)}`

`0.03 = 1.645\sqrt{(0.5*0.5)/(n)}`

`0.03√(n) = 1.645*0.5`

`√(n) = (1.645*0.5)/(0.03)`

`(√(n))^2 = ((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 = (13)/(90) = 0.1444`

So

`M = z\sqrt{(\pi(1-\pi))/(n)}`

`0.03 = 1.645\sqrt{(0.1444*0.8556)/(n)}`

`0.03√(n) = 1.645√(0.1444*0.8556)`

`√(n) = (1.645√(0.1444*0.8556))/(0.03)`

`(√(n))^2 = ((1.645√(0.1444*0.8556))/(0.03))^2`

`n = 371.5`

Rounding up:

372 medical files should be included.