Respuesta :
Answer:
155 women must be randomly selected.
Step-by-step explanation:
We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1 - 0.9}{2} = 0.05[/tex]
Now, we have to find z in the Z-table as such z has a p-value of [tex]1 - \alpha[/tex].
That is z with a pvalue of [tex]1 - 0.05 = 0.95[/tex], so Z = 1.645.
Now, find the margin of error M as such
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
In which [tex]\sigma[/tex] 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 [tex]\sigma = 28[/tex]
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
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]3.7 = 1.645\frac{28}{\sqrt{n}}[/tex]
[tex]3.7\sqrt{n} = 1.645*28[/tex]
[tex]\sqrt{n} = \frac{1.645*28}{3.7}[/tex]
[tex](\sqrt{n})^2 = (\frac{1.645*28}{3.7})^2[/tex]
[tex]n = 154.97[/tex]
Rounding up:
155 women must be randomly selected.
