Elementary Statistics: A Step-by-Step Approach with Formula Card 9th Edition

Published by McGraw-Hill Education
ISBN 10: 0078136334
ISBN 13: 978-0-07813-633-7

Chapter 7 - Confidence Intervals and Sample - 7-3 Confidence Intervals and Sample Size for Proportions - Exercises 7-3 - Page 396: 15

Answer

385 and 601

Work Step by Step

At a 95% confidence the critical z-value is $z_{\alpha/2}=1.96 $ With $E=0.04, \hat p=\frac{40}{200}=0.20, \hat q=1-\hat p=0.80$ The equation $E=z_{\alpha/2}\times\sqrt {\frac{\hat p\hat q}{n}}$ becomes $1.96\times\sqrt {\frac{0.2\times0.8}{n}}=0.04$ Thus, $n=(\frac{1.96}{0.04})^2\times0.2\times0.8=384.16\approx385$ (round up to the next integer here) With $E=0.04, $ and no estimate of the sample proportion, we use $\hat p= \hat q=0.5$ The equation $E=z_{\alpha/2}\times\sqrt {\frac{\hat p\hat q}{n}}$ becomes $1.96\times\sqrt {\frac{0.5\times0.5}{n}}=0.04$ Thus, $n=(\frac{1.96}{0.04})^2\times0.5\times0.5=600.25\approx601$ (round up to the next integer here)
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