Question

How many women must be randomly selected to estimate the mean weight of women in one age group? We want 90% confidence that the sample mean is within 3.7 lbs of the populations mean, and population standard deviation is known to be 28 lbs.

Answer 1

Answer:

155 women must be randomly selected.

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.

The population standard deviation is known to be 28 lbs.

This means that $\sigma = 28$

We want 90% confidence that the sample mean is within 3.7 lbs of the populations mean. How many women must be sampled?

This is n for which M = 3.7. So

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

$3.7 = 1.645\frac{28}{\sqrt{n}}$

$3.7\sqrt{n} = 1.645*28$

$\sqrt{n} = \frac{1.645*28}{3.7}$

$(\sqrt{n})^2 = (\frac{1.645*28}{3.7})^2$

$n = 154.97$

Rounding up:

155 women must be randomly selected.