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IE4521_HW6b_Solution

# IE4521_HW6b_Solution - 9.2.2 9.2.3 The diﬂerences 25 =.17...

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Unformatted text preview: 9.2.2 9.2.3 The diﬂerences 25 = .17.; 7 ya: have a sample mean 2 = 71.36 and a sample standard deviation 3 = 6.08. Consider the hypotheses Hg:p2pA—a320versusHA:p.=,u.A—,u.37é0. The test statistic is _ \F- _ «Em—1.3a) _ t— 1:; — T — 70.837. The p—value is 2 X P(t13 3 —0.837) = 0.418. There is not sufﬁcient evidence to conclude that the diﬂerent stimulation conditions aﬂect the adhesion of the red blood cells. With tn_025,13 = 2.160 a two-sided 95% conﬁdence level conﬁdence interval for u = MA — MB is ( 1.36 ”53:45-08, 1.36 1 2163303) —( 437,215). The differences 23' 2 xi — yé have a sample mean 2 = 0.570 and a sample standard deviation 3 = 0.813. Consider the hypotheses Ho=#:1tA—#B :10 versusHA =M:uA—MB > 0 Where the alternative hypothesis states that the new tires have a smaller average reduction in tread depth than the standard tires. The test statistic is x/HE @0570 t: 8 2—0313 =3.14. The p—value is 130519 2 3.14) = 0.003. There is sufﬁcient evidence to conclude that the new tires are better than the standard tires in terms of the average reduction in tread depth. With 5035,19 : 1.729 a one-sided 95% conﬁdence level conﬁdence interval fora=pAipBis (0.570 — \$30) = (0256,00). 9.2.5 The diHerences 22- = 517g — yi have a sample mean 2 = 2.20 and a sample standard deviation 3 = 147.8. Consider the hypotheses Ho :M:m—MB:0versus HA=M:MA—MB7§0- The test statistic is t: ﬁ5_w:0.063- s 147.8 The p—value is 2 X P(t17 2 0.063) = 0.95. There is not sufﬁcient evidence to conclude that the two laboratories are any diﬂerent in the datings that the}r provide. With @025,” = 2.110 a two-sided 95% conﬁdence level conﬁdence interval f0ru=uA—#B is 2.110x147.8 2.110X147.8 _ (2.20 m ,2.20: m ) (71.3,75.7). 9.3.3 (a) The pooled variance is (n—1)53+(m—1)s§ ((871)x44.76'~’)+((1771)><38.942) = 1664.6. 3 : n+m—2 3+17—2 1310 With \$030533 2 2.807 a 99% two-sided conﬁdence interval for MA — p. B is 675.1 — 702.4 : 2.807 x «1664.6 x 1/; + ﬁ : (—76.4,21.8). (b) Since 2 44.762 3:23.942 3 + 17 12 2 442m 33.04 _ ‘ six(s—1}+1?—‘x(17—1) the degrees of freedom are u = 12. Using a critical point human 2 3.055 a 99% two-sided conﬁdence interval for MA — mg is 675.1 — 702.4 : 3.055 x 44?” + 33%? = (—83.6,29.0). (c) The test statistic is _ i—g _ ﬁnal—702.4 _ 1.56. was mes The null hypothesis is accepted since lt| = 1.56 is smaller than the critical point tﬂooﬁ’gg = 2.807. The p—value is 2 X P(t23 2 1.56) = 0.132. t 9.3.5 (a) The pooled variance is 2 = (n—UsiHm—Usﬁ ((13—1)x0.00128'-’)+((15—1)x0.000969) P n+m—2 13+15—2 = 1.25 x 10—5. 8 With 13110535 2 1.706 a 95% one-sided conﬁdence interval for ILA — #3 is (—00, 0.0548 — 0.0569 + 1.706 x V1.25 x 10—Ei X1/1—13 + 1—15) 2 (—00, —0.0014). (b) The test statistic is t = 5—3; = 00543—00569 2 4 95. .sm/ﬁI +31 1.25x10- x T3+T5 ' The null hypothesis is rejected at size 0: = 0.01 since i = —4.95 < —tﬂ_01’26 = —2.479. The null hypothesis is consequently also rejected at size 0: = 0.05. The p—value is P(t25 S —4.95) = 0.000. 9.3.7 (a) Since 2 11.002 4.612 20 + 25 23 6 11.90 4.61 _ ' 209x (20—1)+ 25% (25—1) the degrees of freedom are U : 23. Consider the hypotheses Ha :J”:JUA—;UB 20versusHA=#:#A—#B<0 Where the alternative hypothesis states that the synthetic ﬁber bundles have an average breaking strength larger than the wool ﬁber bundles. The test statistic is t _ Egg—f 2 _ 435.5;45232 _ 5.788. af+sﬁy «11:50 +4.35]. The null hypothesis is rejected at size or = 001 since t = —5.788 < —to‘01,23 = —2.500. The p—value is P(t23 g —5.788) = 0.000. (b) With a critical point tomgg = 2.500 a 99% one-sided conﬁdence interval for ,uA — p B is (—oo,436.5 — 452.8 + 2.500 x % + %) = (—00, —9.3). (c) There is sufﬁcient evidence to conclude that the synthetic ﬁber bundles have an average breaking strength larger than the wool ﬁber bundles. 9.3.15 Using “1005,80 = 2.639 equal total sample sizes of 2 9 2 n _ m > 4 tom, (52+8y) _ 4x2.6392x(0.1242+0.13?2) _ L3 0.1;} = 95.1 should be sufﬁcient. Additional sample sizes of at least 96 — 41 = 55 from each population can be reconnnended. 9.3.19 There is sufﬁcient evidence to conclude that the damped feature is eHective in re- ducing the heel-strike force. ...
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IE4521_HW6b_Solution - 9.2.2 9.2.3 The diﬂerences 25 =.17...

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