Answer:
We need to sample at least 37 weeks of data.
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{1-0.98}{2} = 0.01[/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]1-0.01 = 0.99[/tex], so [tex]z = 2.327[/tex]
Now, find the margin of error M as such
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
We want 98% confidence that the sample mean is within $500 of the population mean, and the population standard deviation is known to be $1300
This is at least n weeks, in which n is found when [tex]M = 500, \sigma = 1300[/tex]
So
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]500 = 2.327*\frac{1300}{\sqrt{n}}[/tex]
[tex]500\sqrt{n} = 2.327*1300[/tex]
[tex]\sqrt{n} = \frac{2.327*1300}{500}[/tex]
[tex](\sqrt{n})^{2} = (\frac{2.327*1300}{500})^{2}[/tex]
[tex]n = 36.6[/tex]
Rounding up
We need to sample at least 37 weeks of data.
