Answer :
Answer:
a. The 98% confidence interval for the mean monthly rent for all studio apartments in this city is between $481.85 and $515.67
b. A sample size of 102 is needed.
Step-by-step explanation:
Question a:
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].
That is z with a pvalue of [tex]1 - 0.01 = 0.99[/tex], so Z = 2.327.
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.
[tex]M = 2.327\frac{65}{\sqrt{80}} = 16.91[/tex]
The lower end of the interval is the sample mean subtracted by M. So it is 498.76 - 16.91 = $481.85
The upper end of the interval is the sample mean added to M. So it is 498.76 + 16.91 = $515.67
The 98% confidence interval for the mean monthly rent for all studio apartments in this city is between $481.85 and $515.67
Question b:
This is n for which M = 15. So
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]15 = 2.327\frac{65}{\sqrt{n}}[/tex]
[tex]15\sqrt{n} = 2.327*65[/tex]
[tex]\sqrt{n} = \frac{2.327*65}{15}[/tex]
[tex](\sqrt{n})^2 = (\frac{2.327*65}{15})^2[/tex]
[tex]n = 101.68[/tex]
Rounding up:
A sample size of 102 is needed.