Statistics: Informed Decisions Using Data (4th Edition)

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

Chapter 14 - Review - Test - Page 731: 2i

Answer

Confidence interval: $14.90\lt ŷ\lt18.46$ We are 90% confident that the number of chirps on a day when the temperature is 80.2°F is between 14.90 and 18.46.

Work Step by Step

$s_e=0.9715$ (item (b)) $∑(x_i-x ̅)^2=25.0964^2=629.8293$ (item (f)) $x ̅=\frac{88.6+93.3+80.6+69.7+69.4+79.6+80.6+76.3+71.6+84.3+75.2+82.0+83.3+82.6+83.5}{15}=80.04$ $n=15$, so: $d.f.=n-2=13$ $level~of~confidence=(1-α).100$% $90$% $=(1-α).100$% $0.90=1-α$ $α=0.1$ $t_{\frac{α}{2}}=t_{0.05}=1.771$ (According to Table VI, for d.f. = 13 and area in right tail = 0.05) $Lower~bound=ŷ -t_{\frac{α}{2}}.s_e\sqrt {1+\frac{1}{n}+\frac{(x^*-x ̅)^2}{∑(x_i-x ̅)^2}}=16.68-1.771\times0.9715\sqrt {1+\frac{1}{15}+\frac{(80.2-80.04)^2}{629.8293}}=14.90$ $Upper~bound=ŷ +t_{\frac{α}{2}}.s_e\sqrt {1+\frac{1}{n}+\frac{(x^*-x ̅)^2}{∑(x_i-x ̅)^2}}=16.68+1.771\times0.9715\sqrt {1+\frac{1}{15}+\frac{(80.2-80.04)^2}{629.8293}}=18.46$
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