Answer:
<em>a)Sample size would be required to obtain a margin of error of 1 days is </em>
<em>n = 179</em>
<em>b) sample size would be required to obtain a margin of error of 2.5 days is n = 20</em>
<u>Step-by-step explanation</u>:
<u><em>step(i):</em></u>-
Given Population standard deviation = 6.83 days
a) Given margin of error = 1 day
<em>The margin of error is determined by</em>
![M.E = \frac{Z_{\alpha } S.D }{\sqrt{n} }](https://tex.z-dn.net/?f=M.E%20%3D%20%5Cfrac%7BZ_%7B%5Calpha%20%7D%20S.D%20%7D%7B%5Csqrt%7Bn%7D%20%7D)
<u><em>Step(ii):</em></u>-
<em>Given 95 % of confidence level</em>
<em>Now the critical value Z₀.₀₅ = 1.96</em>
<em></em>
<em></em>
<em>√n = 13.38</em>
<em>Squaring on both sides, we get</em>
<em>n = 179.206</em>
<em>b)</em>
<em>step(i):-</em>
a) Given margin of error = 2.5 day
<em>The margin of error is determined by</em>
![M.E = \frac{Z_{\alpha } S.D }{\sqrt{n} }](https://tex.z-dn.net/?f=M.E%20%3D%20%5Cfrac%7BZ_%7B%5Calpha%20%7D%20S.D%20%7D%7B%5Csqrt%7Bn%7D%20%7D)
<u><em>Step(ii):</em></u>-
<em>Given 90 % of confidence level</em>
<em>Now the critical value Z₀.₁₀ = 1.645</em>
<em></em>
<em></em>
Cross multiplication , we get
![\sqrt{n} = \frac{1.645 X 6.83 }{2.5 }](https://tex.z-dn.net/?f=%5Csqrt%7Bn%7D%20%3D%20%5Cfrac%7B1.645%20X%206.83%20%7D%7B2.5%20%7D)
√n = 4.494
<em>Squaring on both sides, we get</em>
<em>n = 20.19</em>
<em></em>
<u><em>Final answer</em></u><em>:-</em>
<em>a)Sample size would be required to obtain a margin of error of 1 days is </em>
<em>n = 179</em>
<em>b) sample size would be required to obtain a margin of error of 2.5 days is n = 20</em>