In a study to find the proportion of children with asthma in a certain population, we desire a 95% confidence interval with standard error no more than 0.05. Without any prior knowledge of the proportion, what should be our minimum sample size

Respuesta :

Answer:

The minimum sample size should be of 381.

Step-by-step explanation:

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 zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].

The margin of error is given by:

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

95% confidence level

So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].

Without any prior knowledge of the proportion, what should be our minimum sample size

We dont know the population proportion, so we use [tex]\pi = 0.5[/tex], which is when the largest sample size is needed. We have to find n for which [tex]M = 0.05[/tex]. So

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

[tex]0.05 = 1.95\sqrt{\frac{0.5*0.5}{n}}[/tex]

[tex]0.05\sqrt{n} = 1.95*0.5[/tex]

[tex]\sqrt{n} = \frac{1.95*0.5}{0.05}[/tex]

[tex](\sqrt{n})^2 = (\frac{1.95*0.5}{0.05})^2[/tex]

[tex]n = 380.25[/tex]

Rounding up

The minimum sample size should be of 381.