Answer:
267 observations are needed.
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 - 0.95}{2} = 0.025[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1 - \alpha[/tex].
That is z with a pvalue of [tex]1 - 0.025 = 0.975[/tex], so Z = 1.96.
Now, find the margin of error M as such
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
Let us assume that the population standard deviation is $12.5.
This means that [tex]\sigma = 12.5[/tex]
Willing to accept only a $1.5 margin of errors in our estimations. How many observations do we need in order to estimate the population mean for the above cost?
This is n for which M = 1.5. So
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]1.5 = 1.96\frac{12.5}{\sqrt{n}}[/tex]
[tex]1.5\sqrt{n} = 1.96*12.5[/tex]
[tex]\sqrt{n} = \frac{1.96*12.5}{1.5}[/tex]
[tex](\sqrt{n})^2 = (\frac{1.96*12.5}{1.5})^2[/tex]
[tex]n = 266.8[/tex]
Rounding up:
267 observations are needed.
