Answer: a) 1082, b) wider.
Step-by-step explanation:
Since we have given that
Margin of error = 0.1 millimeter = E
Standard deviation = 2 mm = σ
Critical value would be
[tex]z_{\frac{\alpha}{2}}=z_{0.05}=1.645[/tex]
a) Sample size would be
[tex]n=(\dfrac{z_{\frac{\alpha }{2}}\times \sigma}{E})^2\\\\n=(\dfrac{1.645\times 2}{1})^2\\\\n=1082.41\\\\n=1082[/tex]
b) Sample size, n = 100
So, the margin of error would be
[tex]E=z_{\frac{\alpha }{2}}\times \dfrac{\sigma}{\sqrt{n}}\\\\E=1.645\times \dfrac{2}{\sqrt{100}}\\\\E=1.645\times \dfrac{2}{10}\\\\E=0.329[/tex]
Since the margin of error in b part is more than a part, as we know that the higher the margin of error, the wider the confidence interval.
So, it would have wider confidence interval.
Hence, a) 1082, b) wider.