67 180 7 13 c p 081 180 919 081 10448 07222 z Do not reject the null

67 180 7 13 c p 081 180 919 081 10448 07222 z do not

This preview shows page 8 - 11 out of 17 pages.

67 180 7 13 c p 826 . 0 67 ) 919 . 0 )( 081 . 0 ( 180 ) 919 . 0 )( 081 . 0 ( 10448 . 0 07222 . 0 z Do not reject the null. We cannot determine the white board markers from the new supplier are inferior to those in stock. 38. H o : p 1 - p 2 ≤ 0 H 1 : p 1 p 2 > 0 If z > 1.645, reject H o . Chapter 10 8
07194 . 0 307 388 11 39 c p 277 . 3 307 ) 92806 . 0 )( 07194 . 0 ( 388 ) 92806 . 0 )( 07194 . 0 ( 03583 . 0 10052 . 0 z Reject the null. The Banking Services division has significantly more female top executives. (LO10-2) 39. H o : n - s = 0 H 1 : n - s ≠ 0 Reject H o if t < 2.086 or t > 2.086 2 2 2 (10 1)(10.5) (12 1)(14.25) 161.2969 10 12 2 p s 83.55 78.8 0.874 1 1 161.2969 10 12 t There is not enough evidence to reject H o . 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. H o : 1 - 2 = 0 H 1 : 1 - 2 ≠ 0 If t is between 2.086 and 2.086, reject H o . 2 2 2 (10 1)(4.44) (12 1)(2.68) 12.82 10 12 2 p s 27.46 25.69 1.15 1 1 12.82 10 12 t There is not enough evidence to reject H o . The mean waiting times are not different. P- value = .2617 (LO10-3) 41. H o : 1 - 2 0 H 1 : 1 - 2 > 0 Reject H o if t > 2.567 2 2 2 (8 1)(2.2638) (11 1)(2.4606) 5.672 8 11 2 p s 10.375 5.636 4.28 1 1 5.672 8 11 t Reject H o. The mean number of transaction by the young adults is more than for the senior citizens. P- value = .0003 (LO10-3) 42. H o : 1 - 2 = 0 H 1 : 1 - 2 ≠ 0 Reject H o if t > 2.086 or t < 2.086 1 X =12.17 s 1 = 1.0563 2 X = 14.875 s 2 = 2.2079 2 2 2 (10 1)(1.0563) (12 1)(2.2079) 3.1832 10 12 2 p s 12.17 14.875 3.541 1 1 3.1832 10 12  t Reject H o . There is a difference in the mean race times. p- value = .0021 (LO10-3) 43. H o : 1 - 2 0 H 1 : 1 - 2 > 0 Reject H o if t > 2.650 1 X =125.125 s 1 = 15.094 2 X = 117.714 s 2 = 19.914 2 2 2 (8 1)(15.094) (7 1)(19.914) 305.708 8 7 2 p s 125.125 117.714 0.819 1 1 305.708 8 7 t There is not enough evidence to reject H o. 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 10 9
44. H o : d 0 H 1 : d < 0 Reject H o if t < 2.998 d = -2.5 s d = 2.928 8 928 . 2 5 . 2 t = -2.415 There is not enough evidence to reject H o . The mean number of accidents has not been reduced. P- value = .0232 (LO10-3) 45. H o : d 0 H 1 : d > 0 Reject H o if t > 1.895 d = 1.75 s d = 2.9155 1.75 1.698 2.9155/ 8 t There is not enough evidence to reject H o . 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. H o : d = 0 H 1 : d 0 Reject H o if z < 1.761 or z > 1.761 d = 246 s d = 547 247.67 1.742 548.04/ 15  t There is not enough evidence to reject H o . There is no difference in the mean insurance price.

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture