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

155 women must be randomly selected.

Explanation:

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

`\alpha = (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(\sigma)/(√(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(\sigma)/(√(n))`

`3.7 = 1.645(28)/(√(n))`

`3.7√(n) = 1.645*28`

`√(n) = (1.645*28)/(3.7)`

`(√(n))^2 = ((1.645*28)/(3.7))^2`

`n = 154.97`

Rounding up:

155 women must be randomly selected.