Statistics: Informed Decisions Using Data (4th Edition)

Published by Pearson
ISBN 10: 0321757270
ISBN 13: 978-0-32175-727-2

Chapter 14 - Section 14.2 - Assess Your Understanding - Applying the Concepts - Page 698: 12d

Answer

Confidence interval: $0.6420\lt y\lt1.1560$ We are 95% confident that the nicotine content of a randomly selected cigarette whose tar content is 12 mg is between 0.6420 and 1.1560 mg.

Work Step by Step

From problem 16 from Section 14.1: $s_e=0.1121$ $∑(x_i-x ̅)^2=24.366^2=593.702$ $x ̅=\frac{5+16+24+24+8+9+9+24+15+18+17+5+10}{13}=14.15$ $n=13$, so: $d.f.=n-2=11$ $level~of~confidence=(1-α).100$% $95$% $=(1-α).100$% $0.95=1-α$ $α=0.05$ $t_{\frac{α}{2}}=t_{0.025}=2.201$ (According to Table VI, for d.f. = 11 and area in right tail = 0.025) $Lower~bound=ŷ -t_{\frac{α}{2}}.s_e\sqrt {1+\frac{1}{n}+\frac{(x^*-x ̅)^2}{∑(x_i-x ̅)^2}}=0.899-2.201\times0.1121\sqrt {1+\frac{1}{13}+\frac{(12-14.15)^2}{593.702}}=0.6420$ $Upper~bound=ŷ +t_{\frac{α}{2}}.s_e\sqrt {1+\frac{1}{n}+\frac{(x^*-x ̅)^2}{∑(x_i-x ̅)^2}}=0.899+2.201\times0.1121\sqrt {1+\frac{1}{13}+\frac{(12-14.15)^2}{593.702}}=1.1560$
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