Answer :
Answer:
At least 5263 adults need to be surveyed.
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*\frac{\sigma}{\sqrt{n}}[/tex]
How many parents do you have to sample?
We need to survey at least n parents.
n is found when [tex]M = 1, \sigma = 44.1[/tex]
So
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]1 = 1.645*\frac{44.1}{\sqrt{n}}[/tex]
[tex]\sqrt{n} = 1.645*44.1[/tex]
[tex](\sqrt{n})^{2} = (1.645*44.1)^{2}[/tex]
[tex]n = 5262.7[/tex]
Rounding up
At least 5263 adults need to be surveyed.