PRACTICE TEST #2 (SOLUTIONS)
32.
how ’s distributed if (note 55 is in the middle corresponding to z=0)
YES
33. The p-value is the area to the right of z = 2.13 multiplied by 2 since it is a two-tail test.. Looking in the z table we see the area from the middle to 2.13 is .4834, so the tail area is .5 - .4834 or .0166 and the p-value is .0166(2) = .0332.
If the mean of all possible days is 55 the chance we would get evidence as strong as or stronger than we got that it is not 55 is the .0332 or 3.32%. This is assuming all conditions were met and the data were gathered properly.
34. 5% or less
35. unknown
36.
how ’s distributed if (note 55 is in the middle corresponding to z=0)
YES
37. The p-value is the area to the right of z = 2.13. Looking in the z table we see the area from the middle to 2.13 is .4834, so the p-value is .5 - .4834 or .0166.
If the mean of all possible days is not over 55 the chance we would get evidence as strong as or stronger than we got that it is over 55 is the .0166 or 1.66%. This is assuming all conditions were met and the data were gathered properly.
40. A) no B) no C) yes D) no
42. From #22, and , since we don’t know , we use t with df=91-1=90. Since our table doesn’t have 90 df, we will use df=80.
which is so basically we are 90% sure the mean of all possible years is on May 3rd, May 4th, or May 5th.
43.
how ’s distributed if (note 33 is in the middle corresponding to t=0)
NO
Note on this problem it is good that we did not say YES because actually if was 33.8 that would still be on May 3rd. If we would have said YES it would be time to get creative with Ho and Ha. (You won’t such a problem on a HW or exam, but perhaps the real world)
44. If the mean of all possible days is 33, the chance we would get evidence as strong as or stronger than we got that it is not 33 is the p-value. This is assuming all conditions were met and the data were gathered properly.
The p-value is the area to the right of t = 2.230 multiplied by 2 (since it is a 2-tailed test). 2.230 is between 2.088 and 2.374 on the df=80 row. So the area in one tail is between 1% and 2%. So the p-value is between 2% and 4%.
45.
df=16
how ’s distributed if (note 24 is in the middle corresponding to t=0)
NO
46. If the mean age is not less than 24, the chance we would get evidence as strong as or stronger than we got that it is less than 24 is the p-value. This is assuming all conditions were met and the data were gathered properly.
The p-value is the area to the left of t =-1.596. 1.596 is between 1.337 and 1.746 on the df=16 row. So the p-value is between 5% and 10%.
48. or
or
This is a matched pairs we will subtract “new – old” for each secretary:
Secretary / A / B / C / D / E / F / G / H / I / JOld Keyboard / 73 / 82 / 95 / 107 / 92 / 111 / 83 / 72 / 68 / 85
New Keyboard / 78 / 90 / 105 / 103 / 90 / 116 / 92 / 88 / 75 / 93
New-Old / 5 / 8 / 10 / -4 / -2 / 5 / 9 / 16 / 7 / 8
how would be distributed if (using t with df = 9)
YES
49. 3.398 is in between 3.250 and 3.690 on the row with df = 9. So the area of the tail is between .0025 and .005.
If the new keyboard is not better on average the chance we would get evidence as strong as or stronger than we got that it is better is between .0025 and .005. This is assuming all conditions were met and the data were gathered properly.
50. unknown
51. 10% or less
52. or
53. or
or df = 5
This is a matched pairs we will subtract “ 1st – 2nd ” for each team:
Team / Fruita / Canon City / Ft Lupton / Liberty / Delta / AspenPoints in first game / 35 / 44 / 6 / 41 / 25 / 65
Point in last game / 7 / 3 / 27 / 51 / 52 / 13
1st – 2nd / 28 / 41 / -21 / -10 / -27 / 52
how would be distributed if (using t with df = 5)
NO
54. .757 is between .727 and .920 on the row with df = 5, so the area to the right of .757 is between 20% and 25%. There is just as strong evidence on the left hand side since it is a 2-tail test, so we double this to get between 40% and 50%.
If there is no difference on average the chance we would get evidence as strong as or stronger than we got that there is a difference is between 40% and 50%. This is assuming all conditions were met and the data were gathered properly.
55. This month summary: Last month summary:
or
or
how would be distributed if (using t with df = 4)
YES
56. 5.822 is between 5.598 and 7.173 so the p-value is between .001 and .0025.
If the price now is not lower on average the chance we would get evidence as strong as or stronger than we got that it is lower is under between .001 and .0025. This is assuming all conditions were met and the data were gathered properly.
57. (2.4833-2.166) or , note the .054499 came from work in #55.
58. or
or df = 50
how would be distributed if (using t with df = 50)
NO
59. .856 is between .849 and .1.047 on the row with df = 50, so the tail area is between 15% and 20%. There is just as good evidence against Ho on the other side since it is a two-tail test so we double to get a p-value of between 30% and 40%.
If there is no difference on average then the chance we would get as strong or stronger evidence than we got that there is a difference is between 30% and 40%. This is assuming all conditions were met and the data were gathered properly.
60.
61.
62.
or or
63.
Picture of how all p’ s would be distributed if . The best evidence that Ha is true is in the shaded part that is in the right tail. The shaded area is .05. Note that p’ = .60 is in the middle corresponding to the z = 0.
YES
64.
Picture of how all p’ s would be distributed if . The best evidence that Ha is true is in the shaded part that is in both tails. The shaded area is .05. with .025 in each tail. Note that p’ = .60 is in the middle corresponding to the z = 0.
YES
65. The p-value is the area to the right of z = 5.042 doubled (since it’s a two-tail test). The area to the right of 5.042 is less than .5 - .499999 = .000001, so the p-value is less than .000002. If the population percentage was 60% the chance we would get evidence as strong or stronger than we got that is not 60% is <.000002. This is assuming all conditions were met and the data were gathered properly.
66. from #62,
67. or
or
Picture of how all ’ s would be distributed if . The best evidence that Ha is true is in the shaded part that is in both tails. The shaded area is .01. with .005 in each tail.
NO
68. or
or
Picture of how all ’ s would be distributed if . The best evidence that Ha is true is in the shaded part in the right tail. The shaded area is .01.
NO
69. The area to the right of z = 1.799 is .5000 - .4641 = .0359.
If the percent wearing seatbelts is not higher on the first road, the chance we would get evidence as strong as or stronger than we got that it is higher is .0359. This is assuming all conditions were met and the data were gathered properly.