Elementary Statistics (12th Edition)

Published by Pearson
ISBN 10: 0321836960
ISBN 13: 978-0-32183-696-0

Chapter 7 - Estimates and Sample Sizes - Review - Review Exercises: 5

Answer

a)543 b)247 c)543

Work Step by Step

a)$\alpha=1-0.98=0.02.$ Then, by using the table, the critical value: $z_{\alpha/2}=z_{0.01}=2.33.$ Hence the minimum sample size: $\frac{z_{\alpha/2}^2\cdot \sigma}{E^2}=\frac{2.33^2\cdot 0.25}{0.05^2}\approx543$. b)$\alpha=1-0.98=0.02.$ Using the table: $z_{\alpha/2}=z_{0.01}=2.33.$ Hence the minimum sample size:$\left (\frac{z_{\alpha/2}\cdot \sigma}{E}\right)^2=\left (\frac{2.33\cdot 337}{50}\right)^2\approx247$. c) We need to use the maximum out of a) and b), which is 543.
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