Answer:
He will need 4 newborns in the sample to be 95% confident.
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 Ztable as such z has a pvalue of [tex]1-\alpha[/tex].
So it is z with a pvalue of [tex]1-0.05 = 0.95[/tex], so [tex]z = 1.645[/tex]
Now, find the margin of error M as such
[tex]M = z*\sqrt{\sigma}{\sqrt{n}}[/tex]
In which [tex]\sigma[/tex] is the standard deviation of the population and n is the length of the sample.
So:
[tex]646 = 1.645*\frac{746}{\sqrt{n}}[/tex]
[tex]646\sqrt{n} = 1227.17[/tex]
[tex]\sqrt{n} = 1.90[/tex]
[tex]\sqrt{n}^{2} = (1.90)^{2}[/tex]
[tex]n = 3.61[/tex]
He will need 4 newborns in the sample to be 95% confident.
