Download Part 3 Estimation

Part 3
Estimation
Name___________________________________
SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.
Provide an appropriate response.
1) Explain the difference between descriptive and inferential statistics.
1)
2) Define margin of error. Explain the relation between the confidence interval and the error
2)
estimate. Suppose a confidence interval is 9.65 < μ < 11.35. Find the sample mean x and the
error estimate E.
3) Define a point estimate. What is the best point estimate for μ?
4) Under what circumstances can you replace σ with s in the formula E = z
3)
α/2
· σ
.
n
4)
5) How do you determine whether to use the z or t distribution in computing the margin of
s
σ
or E = t
?
error, E = z
· · α/2
α/2
n
n
5)
6) When determining the sample size needed to achieve a particular error estimate you need
to know σ. What are the two methods of estimating σ if σ is unknown?
6)
7) What assumption about the parent population is needed to use the t distribution to
compute the margin of error?
7)
8) Under what three conditions is it appropriate to use the t distribution in place of the
standard normal distribution?
8)
9) Interpret the following 95% confidence interval for mean weekly salaries of shift managers
at Guiseppeʹs Pizza and Pasta.
325.80 < μ < 472.30
9)
10) Identify the correct distribution (z, t, or neither) for each of the following.
10)
11) What is the best point estimate for the population proportion? Explain why that point
estimate is best.
11)
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Find the indicated critical z value.
12) Find the critical value z α/2 that corresponds to a 98% confidence level.
A) 2.33
B) 1.75
C) 2.05
1
12)
D) 2.575
13) Find the critical value z α/2 that corresponds to a 91% confidence level.
A) 1.34
B) 1.645
C) 1.75
13)
D) 1.70
14) Find the critical value z α/2 that corresponds to a 90% confidence level.
A) 2.33
B) 1.28
C) 1.75
14)
D) 1.645
Determine whether the given conditions justify using the margin of error E = z α/2 σ/ n when finding a confidence
interval estimate of the population mean μ.
15)
15) The sample size is n = 4, σ = 12.7, and the original population is normally distributed.
A) No
B) Yes
16) The sample size is n = 2 and σ is not known.
A) Yes
B) No
16)
17) The sample size is n = 200 and σ = 19.
A) No
B) Yes
17)
18) The sample size is n = 9, σ is not known, and the original population is normally distributed.
A) Yes
B) No
18)
Use the confidence level and sample data to find the margin of error E. Round your answer to the same number of
decimal places as the sample mean unless otherwise noted.
19) Weights of eggs: 95% confidence; n = 53, x = 1.34 oz, σ = 0.58 oz
A) 0.13 oz
B) 0.16 oz
C) 0.02 oz
19)
D) 0.36 oz
20) Replacement times for washing machines: 90% confidence; n = 45, x = 11.9 years, σ = 2.0 years
A) 2.9 yr
B) 0.1 yr
C) 0.4 yr
D) 0.5 yr
20)
21) The duration of telephone calls directed by a local telephone company: σ = 3.6 minutes, n = 560,
90% confidence. Round your answer to the nearest thousandth.
A) 0.250 min
B) 0.006 min
C) 0.092 min
D) 0.011 min
21)
Use the confidence level and sample data to find a confidence interval for estimating the population μ. Round your
answer to the same number of decimal places as the sample mean.
22)
22) Test scores: n = 92, x = 90.6, σ = 8.9; 99% confidence
A) 88.8 < μ < 92.4
B) 88.4 < μ < 92.8
C) 89.1 < μ < 92.1
D) 88.2 < μ < 93.0
23) Test scores: n = 75, x = 46.1, σ = 5.8; 98% confidence
A) 44.4 < μ < 47.8
B) 44.5 < μ < 47.7
C) 45.0 < μ < 47.2
D) 44.8 < μ < 47.4
23)
24) A random sample of 130 full-grown lobsters had a mean weight of 21 ounces and a standard
deviation of 3.0 ounces. Construct a 98% confidence interval for the population mean μ.
A) 19 oz < μ < 21 oz
B) 20 oz < μ < 22 oz
C) 21 oz < μ < 23 oz
D) 20 oz < μ < 23 oz
2
24)
25) A laboratory tested 83 chicken eggs and found that the mean amount of cholesterol was 233
milligrams with σ = 12.9 milligrams. Construct a 95% confidence interval for the true mean
cholesterol content, μ, of all such eggs.
A) 229 mg < μ < 235 mg
B) 229 mg < μ < 236 mg
C) 230 mg < μ < 236 mg
D) 231 mg < μ < 237 mg
25)
26) 37 packages are randomly selected from packages received by a parcel service. The sample has a
mean weight of 17.0 pounds and a standard deviation of 3.3 pounds. What is the 95% confidence
interval for the true mean weight, μ, of all packages received by the parcel service?
B) 15.6 lb < μ < 18.4 lb
A) 15.9 lb < μ < 18.1 lb
C) 16.1 lb < μ < 17.9 lb
D) 15.7 lb < μ < 18.3 lb
26)
27) A laboratory tested 73 chicken eggs and found that the mean amount of cholesterol was 203
milligrams with σ = 15.0 milligrams. Construct a 95% confidence interval for the true mean
cholesterol content, μ, of all such eggs.
A) 199 mg < μ < 206 mg
B) 199 mg < μ < 205 mg
C) 201 mg < μ < 207 mg
D) 200 mg < μ < 206 mg
27)
28) A random sample of 158 full-grown lobsters had a mean weight of 16 ounces and a standard
deviation of 3.5 ounces. Construct a 98% confidence interval for the population mean μ.
A) 15 oz < μ < 18 oz
B) 16 oz < μ < 18 oz
C) 14 oz < μ < 16 oz
D) 15 oz < μ < 17 oz
28)
29) A group of 66 randomly selected students have a mean score of 22.4 with a standard deviation of
2.8 on a placement test. What is the 90% confidence interval for the mean score, μ, of all students
taking the test?
A) 21.8 < μ < 23.0
B) 21.7 < μ < 23.1
C) 21.5 < μ < 23.3
D) 21.6 < μ < 23.2
29)
Do one of the following, as appropriate: (a) Find the critical value z α/2, (b) find the critical value tα/2, (c) state that
neither the normal nor the t distribution applies.
30) 90%; n = 10; σ is unknown; population appears to be normally distributed.
A) z α/2 = 2.262
B) tα/2 = 1.833
C) tα/2 = 1.812
30)
D) z α/2 = 1.383
31)
31) 95%; n = 11; σ is known; population appears to be very skewed.
A) z α/2 = 1.96
B) tα/2 = 2.228
C) z α/2 = 1.812
D) Neither the normal nor the t distribution applies.
Assume that a sample is used to estimate a population mean μ. Use the given confidence level and sample data to find
the margin of error. Assume that the sample is a simple random sample and the population has a normal distribution.
Round your answer to one more decimal place than the sample standard deviation.
_
32)
32) 95% confidence; n = 91; x = 16, s = 9.1
A) 1.63
B) 4.10
C) 1.90
D) 1.71
_
33) 99% confidence; n = 201; x = 276; s = 75
A) 12.4
B) 16.0
33)
C) 13.8
3
D) 10.5
_
34) 95% confidence; n = 12; x = 76.4; s = 5.9
A) 3.75
B) 3.38
34)
C) 2.81
D) 4.50
Use the given degree of confidence and sample data to construct a confidence interval for the population mean μ.
Assume that the population has a normal distribution.
35) A laboratory tested twelve chicken eggs and found that the mean amount of cholesterol was 225
35)
milligrams with s = 15.7 milligrams. Construct a 95% confidence interval for the true mean
cholesterol content of all such eggs.
B) 214.9 mg < μ < 235.1 mg
A) 216.9 mg < μ < 233.1 mg
C) 215.0 mg < μ < 235.0 mg
D) 215.1 mg < μ < 234.9 mg
36) Thirty randomly selected students took the calculus final. If the sample mean was 83 and the
standard deviation was 13.5, construct a 99% confidence interval for the mean score of all students.
A) 76.23 < μ < 89.77
B) 78.81 < μ < 87.19
C) 76.21 < μ < 89.79
D) 76.93 < μ < 89.07
36)
37) A sociologist develops a test to measure attitudes towards public transportation, and 27 randomly
selected subjects are given the test. Their mean score is 76.2 and their standard deviation is 21.4.
Construct the 95% confidence interval for the mean score of all such subjects.
A) 74.6 < μ < 77.8
B) 64.2 < μ < 88.2
C) 69.2 < μ < 83.2
D) 67.7 < μ < 84.7
37)
38) A savings and loan association needs information concerning the checking account balances of its
local customers. A random sample of 14 accounts was checked and yielded a mean balance of
$664.14 and a standard deviation of $297.29. Find a 98% confidence interval for the true mean
checking account balance for local customers.
B) $493.71 < μ < $834.57
A) $453.59 < μ < $874.69
C) $492.52 < μ < $835.76
D) $455.65 < μ < $872.63
38)
39) The principal randomly selected six students to take an aptitude test. Their scores were:
76.5 85.2 77.9 83.6 71.9 88.6
Determine a 90% confidence interval for the mean score for all students.
A) 85.74 < μ < 75.49
B) 75.49 < μ < 85.74
C) 75.39 < μ < 85.84
D) 85.84 < μ < 75.39
39)
40) The amounts (in ounces) of juice in eight randomly selected juice bottles are:
15.2 15.5 15.9 15.5
15.0 15.7 15.0 15.7
Construct a 98% confidence interval for the mean amount of juice in all such bottles.
A) 15.00 oz < μ < 15.87 oz
B) 15.87 oz < μ < 15.00 oz
C) 15.77 oz < μ < 15.10 oz
D) 15.10 oz < μ < 15.77 oz
40)
41) The football coach randomly selected ten players and timed how long each player took to perform
a certain drill. The times (in minutes) were:
7.2 10.5 9.9 8.2 11.0
7.3 6.7 11.0 10.8 12.4
Determine a 95% confidence interval for the mean time for all players.
A) 10.85 min < μ < 8.15 min
B) 8.15 min < μ < 10.85 min
D) 8.05 min < μ < 10.95 min
C) 10.95 min < μ < 8.05 min
41)
4
Express the confidence interval using the indicated format.
^
42) Express the confidence interval 0.047 < p < 0.507 in the form of p ± E.
A) 0.277 - 0.23
B) 0.277 ± 0.5
C) 0.277 ± 0.23
42)
D) 0.23 ± 0.5
Solve the problem.
43) The following confidence interval is obtained for a population proportion, p: (0.505, 0.545). Use
43)
^
these confidence interval limits to find the point estimate, p.
A) 0.527
B) 0.525
C) 0.545
D) 0.505
Assume that a sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the
given statistics and confidence level. Round the margin of error to four decimal places.
44)
44) 95% confidence; n = 320, x = 60
A) 0.0428
B) 0.0514
C) 0.0385
D) 0.0449
45) In a random sample of 184 college students, 97 had part-time jobs. Find the margin of error for the
95% confidence interval used to estimate the population proportion.
A) 0.0649
B) 0.00266
C) 0.126
D) 0.0721
45)
Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p.
46) n = 125, x = 72; 90% confidence
46)
A) 0.506 < p < 0.646
B) 0.503 < p < 0.649
C) 0.502 < p < 0.650
D) 0.507 < p < 0.645
47) A survey of 865 voters in one state reveals that 408 favor approval of an issue before the
legislature. Construct the 95% confidence interval for the true proportion of all voters in the state
who favor approval.
A) 0.438 < p < 0.505
B) 0.444 < p < 0.500
C) 0.471 < p < 0.472
D) 0.435 < p < 0.508
47)
48) A study involves 669 randomly selected deaths, with 31 of them caused by accidents. Construct a
98% confidence interval for the true percentage of all deaths that are caused by accidents.
A) 2.54% < p < 6.73%
B) 3.04% < p < 6.23%
C) 3.29% < p < 5.97%
D) 2.74% < p < 6.53%
48)
49) Of 140 adults selected randomly from one town, 35 of them smoke. Construct a 99% confidence
interval for the true percentage of all adults in the town that smoke.
A) 15.6% < p < 34.4%
B) 17.8% < p < 32.2%
C) 16.5% < p < 33.5%
D) 19.0% < p < 31.0%
49)
50) Of 234 employees selected randomly from one company, 12.82% of them commute by carpooling.
Construct a 90% confidence interval for the true percentage of all employees of the company who
carpool.
A) 8.54% < p < 17.1%
B) 7.18% < p < 18.5%
C) 9.23% < p < 16.4%
D) 7.73% < p < 17.9%
50)
51) Of 346 items tested, 12 are found to be defective. Construct the 98% confidence interval for the
proportion of all such items that are defective.
A) 0.0118 < p < 0.0576
B) 0.0110 < p < 0.0584
C) 0.0154 < p < 0.0540
D) 0.0345 < p < 0.0349
51)
5
52) A survey of 300 union members in New York State reveals that 112 favor the Republican candidate
for governor. Construct the 98% confidence interval for the true population proportion of all New
York State union members who favor the Republican candidate.
A) 0.301 < p < 0.445
B) 0.316 < p < 0.430
D) 0.308 < p < 0.438
C) 0.304 < p < 0.442
52)
53) When 319 college students are randomly selected and surveyed, it is found that 120 own a car.
Find a 99% confidence interval for the true proportion of all college students who own a car.
A) 0.323 < p < 0.429
B) 0.332 < p < 0.421
C) 0.306 < p < 0.446
D) 0.313 < p < 0.439
53)
54) Of 88 adults selected randomly from one town, 69 have health insurance. Find a 90% confidence
interval for the true proportion of all adults in the town who have health insurance.
A) 0.671 < p < 0.897
B) 0.698 < p < 0.870
C) 0.682 < p < 0.886
D) 0.712 < p < 0.856
54)
Assume that a sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the
given statistics and confidence level. Round the margin of error to four decimal places.
55) In a clinical test with 3300 subjects, 660 showed improvement from the treatment. Find the margin
55)
of error for the 99% confidence interval used to estimate the population proportion.
A) 0.0156
B) 0.0136
C) 0.0102
D) 0.0179
56) 90% confidence; n = 480, x = 120
A) 0.0387
B) 0.0325
C) 0.0406
D) 0.0348
56)
57) 99% confidence; n = 6500, x = 1950
A) 0.0128
B) 0.00833
C) 0.0146
D) 0.0111
57)
Solve the problem.
58) The following confidence interval is obtained for a population proportion, p: 0.537 < p < 0.563. Use
58)
^
these confidence interval limits to find the point estimate, p.
A) 0.550
B) 0.537
C) 0.555
D) 0.545
59) The following confidence interval is obtained for a population proportion, p: 0.689 < p < 0.723. Use
these confidence interval limits to find the margin of error, E.
A) 0.034
B) 0.017
C) 0.706
D) 0.018
59)
Express the confidence interval using the indicated format.
^
60) Express the confidence interval 0.66 < p < 0.8 in the form of p ± E.
A) 0.66 ± 0.07
B) 0.73 ± 0.14
C) 0.66 ± 0.14
60)
D) 0.73 ± 0.07
^
61) Express the confidence interval (0.668, 0.822) in the form of p ± E.
A) 0.745 ± 0.077
B) 0.668 ± 0.154
C) 0.668 ± 0.077
^
61)
D) 0.745 ± 0.154
^
62) Express the confidence interval 0.457 ± 0.044 in the form of p - E < p < p + E.
A) 0.413 < p < 0.501
B) 0.457 < p < 0.501
C) 0.435 < p < 0.479
D) 0.413 < p < 0.457
6
62)
Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p.
63) Of 82 adults selected randomly from one town, 69 have health insurance. Find a 90% confidence
63)
interval for the true proportion of all adults in the town who have health insurance.
B) 0.747 < p < 0.935
A) 0.762 < p < 0.921
C) 0.738 < p < 0.945
D) 0.775 < p < 0.908
64) Of 367 randomly selected medical students, 30 said that they planned to work in a rural
community. Find a 95% confidence interval for the true proportion of all medical students who
plan to work in a rural community.
A) 0.0449 < p < 0.119
B) 0.0537 < p < 0.110
C) 0.0582 < p < 0.105
D) 0.0484 < p < 0.115
64)
65) Of 87 adults selected randomly from one town, 63 have health insurance. Find a 90% confidence
interval for the true proportion of all adults in the town who have health insurance.
A) 0.645 < p < 0.803
B) 0.630 < p < 0.818
C) 0.612 < p < 0.836
D) 0.601 < p < 0.848
65)
66) Of 253 employees selected randomly from one company, 18.97% of them commute by carpooling.
Construct a 90% confidence interval for the true percentage of all employees of the company who
carpool.
A) 13.2% < p < 24.7%
B) 12.6% < p < 25.3%
C) 14.1% < p < 23.8%
D) 14.9% < p < 23.0%
66)
Use the given information to find the minimum sample size required to estimate an unknown population mean μ.
67) How many women must be randomly selected to estimate the mean weight of women in one age
67)
group. We want 90% confidence that the sample mean is within 3.7 lb of the population mean, and
the population standard deviation is known to be 28 lb.
B) 221
C) 155
D) 156
A) 153
68) How many students must be randomly selected to estimate the mean weekly earnings of students
at one college? We want 95% confidence that the sample mean is within $ 5 of the population
mean, and the population standard deviation is known to be $63.
A) 430
B) 610
C) 537
D) 862
68)
69) How many business students must be randomly selected to estimate the mean monthly earnings
of business students at one college? We want 95% confidence that the sample mean is within $128
of the population mean, and the population standard deviation is known to be $536.
A) 47
B) 68
C) 59
D) 95
69)
70) How many commuters must be randomly selected to estimate the mean driving time of Chicago
commuters? We want 98% confidence that the sample mean is within 4 minutes of the population
mean, and the population standard deviation is known to be 12 minutes.
A) 35
B) 49
C) 25
D) 60
70)
71) How many weeks of data must be randomly sampled to estimate the mean weekly sales of a new
line of athletic footwear? We want 99% confidence that the sample mean is within $200 of the
population mean, and the population standard deviation is known to be $1100.
A) 117
B) 82
C) 165
D) 201
71)
7
Use the given data to find the minimum sample size required to estimate the population proportion.
^
^
72) Margin of error: 0.004; confidence level: 95%; p and q unknown
A) 50,024
B) 60,148
C) 60,018
72)
D) 60,025
^
73) Margin of error: 0.01; confidence level: 95%; from a prior study, p is estimated by the decimal
equivalent of 52%.
A) 9589
B) 8630
C) 19,976
D) 16,551
Solve the problem.
74) A newspaper article about the results of a poll states: ʺIn theory, the results of such a poll, in 99
cases out of 100 should differ by no more than 5 percentage points in either direction from what
would have been obtained by interviewing all voters in the United States.ʺ Find the sample size
suggested by this statement.
B) 544
C) 27
D) 664
A) 385
75) In a certain population, body weights are normally distributed with a mean of 152 pounds and a
standard deviation of 26 pounds. How many people must be surveyed if we want to estimate the
percentage who weigh more than 180 pounds? Assume that we want 96% confidence that the error
is no more than 4 percentage points.
B) 317
C) 501
D) 232
A) 658
73)
74)
75)
Use the given data to find the minimum sample size required to estimate the population proportion.
^
76) Margin of error: 0.05; confidence level: 99%; from a prior study, p is estimated by 0.15.
A) 17
B) 339
C) 407
D) 196
^
77) Margin of error: 0.04; confidence level: 95%; from a prior study, p is estimated by the decimal
equivalent of 89%.
A) 708
B) 236
C) 9
D) 209
^
^
78) Margin of error: 0.015; confidence level: 96%; p and q unknown
A) 6669
B) 4519
C) 4670
^
77)
78)
D) 3667
^
79) Margin of error: 0.04; confidence level: 94%; p and q unknown
A) 587
B) 486
C) 553
76)
79)
D) 572
Construct the indicated confidence interval for the difference between the two population means. Assume that the two
samples are independent simple random samples selected from normally distributed populations. Do not assume that
the population standard deviations are equal.
80) Independent samples from two different populations yield the following data. x1 = 236, x2 = 905,
s 1 = 88, s 2 = 13. The sample size is 381 for both samples. Find the 85% confidence interval for
μ1 - μ2 .
A) -676 < μ1 - μ2 < -662
B) -683 < μ1 - μ2 < -655
C) -670 < μ1 - μ2 < -668
D) -677 < μ1 - μ2 < -661
8
80)
81) Two types of flares are tested and their burning times are recorded. The summary statistics are
given below.
Brand X
Brand Y
n = 35
n = 40
x = 19.4 min
s = 1.4 min
x = 15.1 min
s = 0.8 min
81)
Construct a 95% confidence interval for the differences between the mean burning time of the
brand X flare and the mean burning time of the brand Y flare.
A) 3.8 min < μX - μY < 4.8 min
B) 3.2 min < μX - μY < 5.4 min
D) 3.6 min < μX - μY < 5.0 min
C) 3.5 min < μX - μY < 5.1 min
Construct the indicated confidence interval for the difference between population proportions p1 - p2 . Assume that the
samples are independent and that they have been randomly selected.
82) x1 = 22, n1 = 38 and x2 = 31, n2 = 52; Construct a 90% confidence interval for the difference
82)
between population proportions p1 - p2 .
B) -0.190 < p1 - p2 < 0.156
D) 0.785 < p1 - p2 < 0.373
A) 0.406 < p1 - p2 < 0.752
C) 0.373 < p1 - p2 < 0.785
83) A marketing survey involves product recognition in New York and California. Of 558 New
Yorkers surveyed, 193 knew the product while 196 out of 614 Californians knew the product.
Construct a 99% confidence interval for the difference between the two population proportions.
A) 0.0247 < p1 - p2 < 0.0286
B) -0.0443 < p1 - p2 < 0.0976
83)
D) -0.0034 < p1 - p2 < 0.0566
C) -0.0177 < p1 - p2 < 0.1243
84) In a random sample of 300 women, 49% favored stricter gun control legislation. In a random
sample of 200 men, 28% favored stricter gun control legislation. Construct a 98% confidence
interval for the difference between the population proportions p 1 - p2 .
A) 0.126 < p1 - p2 < 0.294
B) 0.122 < p1 - p2 < 0.298
C) 0.099 < p1 - p2 < 0.321
D) 0.110 < p1 - p2 < 0.310
85) In a random sample of 500 people aged 20 -24, 22% were smokers. In a random sample of 450
people aged 25-29, 14% were smokers. Construct a 95% confidence interval for the difference
between the population proportions p 1 - p2 .
B) 0.035 < p1 - p2 < 0.125
D) 0.048 < p1 - p2 < 0.112
A) 0.025 < p1 - p2 < 0.135
C) 0.032 < p1 - p2 < 0.128
9
84)
85)
Answer Key
Testname: PP3
1) Descriptive statistics
summarizes or describes
important characteristics of
known population data.
Inferential statistics uses sample
data to make inferences or
generalizations about a
population.
2) The margin of error is the
maximum likely difference
between the observed sample
8) 1) n ≤ 30
2) σ is unknown, and
3) the parent population is
essentially normal
9) We are 95% sure that the
interval contains the true
population value for mean
weekly salaries of shift
managers at Guiseppeʹs Pizza
and Pasta.
10)
mean x and the true value for
the population mean μ. The
confidence interval is found by
taking the sample mean x and
adding the margin of error E to
find the high value and
subtracting E to find the low
value of the interval. In the
interval 9.65 < μ < 11.35, the
sample mean x is 10.5 and the
error estimate E is 0.85.
3) A point estimate is a single
value used to approximate a
population parameter. The
4)
5)
6)
7)
sample mean x is the best point
estimate of μ.
Provided n > 30, s can be used
in place of σ. If n ≤ 30, the
population must be normal and
σ must be known to use the
formula.
Provided n > 30, the standard
normal distribution is the one to
use. If n ≤ 30, the population
must be normal and σ must be
known to use the formula.
1) Use the range rule of
thumb.
2) Conduct a pilot study and
base your estimate of σ on the
first collection of at least 31
randomly selected values.
The parent population must be
essentially normal; that is, it
must have one mode and be
basically symmetric.
^
11) The sample proportion p.
^
1)p is unbiased (does not
consistently overestimate or
underestimate p).
^
2) p is most consistent (has the
least variation of all the
measures of central tendency).
12) A
13) D
14) D
15) B
16) B
17) B
18) B
19) B
20) D
21) A
22) D
23) B
24) B
25) C
26) A
27) D
28) D
29) A
30) B
31) D
32) C
33) C
34) A
35) C
10
36) C
37) D
38) A
39) B
40) D
41) B
42) C
43) B
44) A
45) D
46) B
47) A
48) D
49) A
50) C
51) A
52) D
53) C
54) D
55) D
56) B
57) C
58) A
59) B
60) D
61) A
62) A
63) D
64) B
65) A
66) D
67) C
68) B
69) B
70) B
71) D
72) D
73) A
74) D
75) A
76) B
77) B
78) C
79) C
80) A
81) A
82) B
83) B
84) D
85) C