67180713cp826.067)919.0)(081.0(180)919.0)(081.0(10448.007222.0zDo not reject the null. We cannot determine the white board markers from the new supplier are inferior to those in stock. 38. Ho: p1- p2 ≤ 0H1: p1– p2 > 0If z> 1.645, reject Ho.Chapter 108
07194.03073881139cp277.3307)92806.0)(07194.0(388)92806.0)(07194.0(03583.010052.0zReject the null.The Banking Services division has significantly more female top executives. (LO10-2)39.Ho: n- s= 0H1: n- s≠ 0Reject Hoif t< 2.086 or t> 2.086222(101)(10.5)(121)(14.25)161.296910122ps83.5578.80.87411161.29691012tThere is not enough evidence to reject Ho. There is no difference in the mean number of hamburgers sold at the two locations. The p-value is > than 0.20, which is > than the significance level, and so the decision is supported. (Computer p-value = .3928)(LO10-3)40.Ho: 1- 2= 0H1: 1- 2≠ 0If t is between 2.086 and 2.086, reject Ho.222(101)(4.44)(121)(2.68)12.8210122ps27.4625.691.151112.821012tThere is not enough evidence to reject Ho. The mean waiting times are not different.P-value = .2617(LO10-3)41.Ho: 1- 20H1: 1- 2> 0Reject Hoif t> 2.567222(81)(2.2638)(111)(2.4606)5.6728112ps10.3755.6364.28115.672811tReject Ho.The mean number of transaction by the young adults is more than for the senior citizens. P-value = .0003(LO10-3)42.Ho: 1- 2 = 0H1: 1- 2≠ 0Reject Hoif t> 2.086 or t< 2.0861X=12.17s1= 1.05632X= 14.875s2= 2.2079222(101)(1.0563)(121)(2.2079)3.183210122ps12.1714.8753.541113.18321012tReject Ho. There is a difference in the mean race times. p-value = .0021(LO10-3)43.Ho: 1- 20H1: 1- 2> 0Reject Hoif t> 2.6501X=125.125s1= 15.0942X= 117.714s2= 19.914222(81)(15.094)(71)(19.914)305.708872ps125.125117.7140.81911305.70887tThere is not enough evidence to reject Ho. There is no increase in the mean number sold at the regular price and the mean number sold at reduced price. P-value = .2138(LO10-3)Chapter 109
44.Ho: d0H1: d< 0Reject Hoif t< 2.998d= -2.5 sd= 2.9288928.25.2t= -2.415There is not enough evidence to reject Ho. The mean number of accidents has not been reduced. P-value = .0232(LO10-3)45.Ho: d0H1: d> 0Reject Hoif t> 1.895d= 1.75sd= 2.91551.751.6982.9155/8tThere is not enough evidence to reject Ho. There is no difference in the mean number of absences. The p-value is greater than 0.05. (actual p-value = .0667)(LO10-5)46.Ho: d= 0H1: d0Reject Hoif z < 1.761 or z> 1.761d= 246sd= 547247.671.742548.04/15tThere is not enough evidence to reject Ho. There is no difference in the mean insurance price.