Mathematics: A doctor at a local hospital is interested in estimating the birth weight of infants. How large a sample must she select if she desires to be 90% confident that her estimate is within 4 ounces of the true mean

A doctor at a local hospital is interested in estimating

Ответ:

freebyyy7032

18.07.2021

The minimum sample size needed is $n = (\frac{1.96\sqrt{\sigma}}{4})^2$ . If n is a decimal number, it is rounded up to the next integer. $\sigma$ is the standard deviation of the population.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.9}{2} = 0.05$

Now, we have to find z in the Z-table as such z has a p-value of $1 - \alpha$ .

That is z with a pvalue of 1 - 0.05 = 0.95 , so Z = 1.645.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

How large a sample must she select if she desires to be 90% confident that her estimate is within 4 ounces of the true mean?

A sample of n is needed, and n is found when M = 4. So

$M = z\frac{\sigma}{\sqrt{n}}$

$4 = 1.96\frac{\sigma}{\sqrt{n}}$

$4\sqrt{n} = 1.96\sqrt{\sigma}$

$\sqrt{n} = \frac{1.96\sqrt{\sigma}}{4}$

$(\sqrt{n})^2 = (\frac{1.96\sqrt{\sigma}}{4})^2$

$n = (\frac{1.96\sqrt{\sigma}}{4})^2$

The minimum sample size needed is $n = (\frac{1.96\sqrt{\sigma}}{4})^2$ . If n is a decimal number, it is rounded up to the next integer. $\sigma$ is the standard deviation of the population.