Answer:
The number of weeks required = 216 weeks
Step-by-step explanation:
Given that:
Margin of Error E = 200
Confidence interval = 95% = 0.95
Level of SIgnificance = 1 - C.I
= 1 - 0.95
= 0.05
Standard deviation = 1500
The Critical value for Z :
[tex]Z_{\alpha/2} =Z_{0.05/2} \\ \\ = Z_{0.025} = 1.96[/tex]
The number of weeks( i.e the sample size (n) ) required is :
[tex]n = (\dfrac{Z_{\alpha/2} \times \sigma}{E})^2[/tex]
[tex]n = (\dfrac{1.96 \times 1500}{200})^2[/tex]
[tex]n = (14.7)^2[/tex]
n = 216.09
n ≅ 216 weeks
