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Stat 3615
Test 3, Version A KEY
4.
Read this: You must use a number 2 pencil and press hard, or
you may not get credit. Mark the one best answer for each
question. There is no penalty for guessing wrong. Use no
notes, books, scratch paper, or cell phones. You many neither
give nor receive help on this test, in accordance with the honor
system code.
1.
The two types of statistical inference are
5.
A. accuracy and precision.
B. estimation and significance testing.
C. population mean and population proportion.
D. qualitative and quantitative.
Weed Control Study: A crop scientist wants to estimate the
mean biomass (kg/acre) of weeds in corn fields using a new
method of tilling. His methodology is to plant n experimental
plots (each 20 meters by 20 meters), and, after 8 weeks,
measure the weed biomass (kg/acre) of each plot
2.
3.
6.
C. the mean weed biomass (kg/acre) of all corn fields
using the new tilling method.
In the Weed control Study, the parameter of interest is
A. the weed biomass (kg/acre) of the ith experimental
plot, i = 1, 2, …, n.
B. the mean weed biomass (kg/acre) of all n
experimental plots.
C. the mean weed biomass (kg/acre) of all corn fields
using the new tilling method
7.
In the Weed control Study, a pilot study was conducted on 4
experimental plots for the purpose of determining the sample
size, i.e., the number of experimental plots, n, required to
estimate the parameter of interest, with 95% confidence, to
within ±2 kg/acre. The purpose of the pilot study is to gage
dispersion. The sample from the pilot study follows.
Test3-13f-A-KEY.docx
the sample standard deviation (kg/acre) of the pilot
study is
A. 2.0
You must learn how to calculate the
sample standard deviation, s,
B. 3.5
automatically and quickly on your
C. 4.1
calculator. To check, enter the sample [-1,
D. 12.5
0, 1], for which s = 1 exactly.
E. 16.7
Using the Pilot Study, compute the sample size required
to estimate the parameter of interest with 95% confidence
and to within ±2 kg/acre.
A. 9
B. 12
n = (z * 𝛔 / m)2
C. 17
= (1.96 * 4.08 / 2) = 16.007
D. 28
Gender-Specific Birth-Rate Study: The Secretary of Health
and Human Services wants to estimate the proportion of live
USA births that is female.
In the Weed control Study, the random variable of
interest is
A. the weed biomass (kg/acre) of the ith experimental
plot, i = 1, 2, …, n.
B. the mean weed biomass (kg/acre) of all n
experimental plots.
Results of the Pilot Study
1
2
3
Experimental Plot
24
18
17
Biomass (kg/acre)
Fall 2013
8.
4
25
1
In the Gender-Specific Birth-Rate Study, the random
variable of interest is
A. the proportion of females among live USA births
B. the proportion of females in the sample
C. the gender (female, male) of the ith randomly sampled
USA live birth.
D. the gender (female, male) of all USA live births.
In the Gender-Specific Birth-Rate Study, what sample
size is required to achieve, with 95% confidence, a margin
of error of three percentage points, i.e., a margin of error
of ±0.03?
2
2
A. 1
  1     
0.5  0.5  

  1.96

nz
B. 28

 

m
0.03

 

C. 33
D. 752
E. 1068  1067.11
End of Estimation Section, Beginning of
Significance/Hypothesis Testing Section
A hypothesis is a proposition about
A. the sample
A proposition is a statement.
B. sampling
C. the population
D. statistical inference
11/22/2013
Stat 3615
9.
Test 3, Version A KEY
Suppose that X is distributed Normally with an unknown
mean and with a population standard deviation of 20.
Then a significance test about the mean of X would use
the test criterion
A.
Z   X  0  
B.
T n 1   X  0  s
C.
Z   p  0 
13. Using Weed control Study 2 Data, calculate the
observed value of the test statistic
A. −1.62
B. 1.62
t = (20.9 – 23) / 1.297 = -1.6197
C. 20.9
D. 23.0
n
n
0 1  0  n
14. Using Weed control Study 2 Data, calculate the P-value.
A. P > 0.10
B. 0.05 < P < 0.10
C. 0.01 < P < 0.05
D. P < 0.01
Weed Control Study 2: It is well known that using the
standard method of tilling, the mean biomass of weeds in
corn fields is 23 kg/acre. A crop scientist wants to show that
the mean biomass (kg/acre) of weeds in corn fields is
decreased using a new method of tilling. His methodology is
to plant n experimental plots (each 20 meters by 20 meters)
using the new method of tilling, and then, after 8 weeks,
measure the weed biomass (kg/acre) of each plot.
10. In Weed control Study 2, the assumptions we have to
make about the random variable of interest is /are
A. the weed biomass (kg/acre) has a known population
mean.
B. the weed biomass (kg/acre) has a known population
standard deviation.
C. the weed biomass (kg/acre) is normally distributed.
D. a and b
This is marked correct
E. b and c
for each student.
F. None of the above.
11. In Weed control Study 2, using µ to denote the
parameter of interest, the alternative hypothesis is
A. 𝛍 < 23
B. 𝛍 = 23
C. 𝛍 ≠ 23
D. 𝛍 > 23
Gender-Specific Birth-Rate Study 2: The Secretary of
Health and Human Services wants to show that 50% of USA
live births are female. Let φ represent the proportion of
females among live USA births.
15. In Gender-Specific Birth-Rate Study 2, using φ to
denote the parameter of interest, the alternative
hypothesis is
A.
B.
C.
D.
𝜑 < 0.50
𝜑 = 0.50
𝜑 ≠ 0.50
𝜑 < 0.50
In Gender-Specific Birth-Rate Study 2, 100,000 live USA
birth records were examined, and it was found that 51,142
females occurred among the 100,000 live USA births.
16. The observed value of the test statistic is
A. −7.2
B. 0.01142
p = 0.51142,
C. 0.51142 Z = (0.5114 – 0.5) / sqrt(0.5x0.5/100,000)
D. 7.2
= 7.2
In Weed control Study 2, the crop scientist subsequently
planted 10 experimental plots and obtained a sample mean
of 20.9 kg/acre with a sample standard deviation of 4.1
kg/acre. Use these data to test the hypothesis at the 5% level of
significance. (Note that the results of earlier studies should not
be included in this subsequent sample of 10 experimental
plots.)
17. In Gender-Specific Birth-Rate Study 2, the observed Pvalue is
A. P > 0.10
12. Using Weed control Study 2 Data, calculate the
standard error of the mean
A. −1.62
B. 0.41
C. 1.30
SE = s / sqrt(n) = 4.1 / sqrt(10) = 1.297
D. 1.37
E. 1.62
Test3-13f-A-KEY.docx
Fall 2013
B. 0.05 < P < 0.10
C. 0.01 < P < 0.05
D. P < 0.01
2
11/22/2013
Stat 3615
Test 3, Version A KEY
21. If the correct, true standard deviation is  = 10 kg (Curve
B), then the smallest effect  that one could detect with a
power of at least 80% is approximately
A. 0.2
Q22
B. 0.8
A food scientist is planning a study of the effect, , in
kilograms, of an experimental food supplement on the mean
weight gain (kg) of juvenile male apes over a six week period.
She plans to test the null hypothesis H0:  = 0 versus the
alternative hypothesis HA:  > 0.
Only twelve (12) male apes are available for study. The
plot below shows the power for samples of size n = 12 with
three different standard deviations:  = 5, 10, and 20 kg.
σ=5
Fall 2013
C. 4.2
A. n↓β↑
B. α↓β↑
D. 7.2
E. 15.0
22. If the correct, true standard deviation were  = 20 kg,
and if the goal were to detect an effect of  = 4.0 kg, then
σ = 10
σ = 20
A. the study could be performed with fewer apes.
B. the study could be performed with a smaller Type 1
error rate.
C. the study is not worth performing.
Miscellaneous Significance/Hypothesis-Testing Questions
23. If the P-value is P = 0.04, is the result significant at the α
= 0.05 level? At the α = 0.01 level?
A. It is not significant at the 0.05 level, and it is not
significant at the 0.01 level.
B. It is significant at the 0.05 level, but not at the 0.01
level.
C. It is not significant at the 0.05 level, but it is
significant at the 0.01 level.
D. It is significant both at the 0.05 level, and at the 0.01
level.
24. If the null hypothesis is rejected at the 0.05 level of
significance, would it be rejected at the 0.01 level?
A. Yes, it would be rejected at the 0.01 level.
B. No, it would not be rejected at the 0.01 level.
C. Not enough information is given.
25. If the null hypothesis is rejected at the 0.05 level of
significance, would it be rejected at the 0.10 level?
A. Yes, it would be rejected at the 0.10 level.
B. No, it would not be rejected at the 0.10 level.
C. Not enough information is given
18. The vertical axis of the power plot represents
A. The probability of rejecting the null hypothesis
B. The probability of rejecting the alternative
hypothesis
C. The probability of NOT rejecting the null
hypothesis
D. The probability of NOT rejecting the alternative
hypothesis
19. Given that the three curves are for the standard deviations
 = 5, 10, and 20 kg., label the curves with their standard
deviations on the graph, and indicate here which curve is
for the standard deviation of  = 5.
A. A
B. B
C. C
20. If the correct, true standard deviation of weight gain is
 = 10 kg (Curve B), what is the Type 2 error rate of the
test against the alternative that the effect is  = 4.0 kg?
A. 0.01
B. 0.05
C. 0.10
D. 0.4
E. 0.6
Test3-13f-A-KEY.docx
When you are done, carefully check that the answers circled
on these pages are the ones marked in the opscan, make sure
you have your name, VT Student ID, and Form (A or B)
marked on the opscan. Keep these questions to study for the
final. Put your opscan in the hands of an instructor, then leave
quietly without disturbing the other students.
3
11/22/2013
Critical Values of Student’s t Distribution
Degrees of Freedom
Percentile rank (probability of a lesser value):
Confidence level (central area):
P-value for two-sided alternative:
P-value for one-sided alternative:
1
Example: t with 4 deg of freedom
2
3
4
5
6
7
8
9
10
11
12
0
13
-5 -4 -3 -2 -1 0 1 2 3 4 5
14
t with 4 df
15
The vertical lines reference the critical values of
16
the t statistic for the purpose of statistical
17
inference.
18
For confidence interval estimation, the critical
values are the t-multipliers of standard error that
19
yield the margin of error in the formula (MOE) =
20
(t)(s.e.). For a 90% confidence interval with 4
21
degrees of freedom, the t-multiplier is t = 2.132,
22
as indicated by the black vertical lines in the
graph.
23
For hypothesis testing, the critical values are the t24
values that correspond to the commonly used
25
significance levels of  = 0.10, 0.05, and 0.01. For
26
testing H0:  ≤ 150 vs. HA:  > 150, an observation
of t = 2.2 with 4 degrees of freedom indicates that
27
the P-value is 0.025 < P < 0.05, because t = 2.2 is
28
between the critical values of t = 2.132 and
29
t = 2.776.
30
40
50
60
70
0
80
2
2.2
2.4
2.6
2.8
3
90
t with 4 df
100
1000
Standard Normal =
Infinite
P-value for one-sided alternative:
P-value for two-sided alternative:
Confidence level (central area):
Percentile rank (probability of a lesser value):
0.90
0.80
0.20
0.10
3.078
1.886
1.638
1.533
1.476
1.440
1.415
1.397
1.383
1.372
1.363
1.356
1.350
1.345
1.341
1.337
1.333
1.330
1.328
1.325
1.323
1.321
1.319
1.318
1.316
1.315
1.314
1.313
1.311
1.310
1.303
1.299
1.296
1.294
1.292
1.291
1.290
1.282
1.282
0.10
0.20
0.80
0.90
0.95
0.90
0.10
0.05
6.314
2.920
2.353
2.132
2.015
1.943
1.895
1.860
1.833
1.812
1.796
1.782
1.771
1.761
1.753
1.746
1.740
1.734
1.729
1.725
1.721
1.717
1.714
1.711
1.708
1.706
1.703
1.701
1.699
1.697
1.684
1.676
1.671
1.667
1.664
1.662
1.660
1.646
1.645
0.05
0.10
0.90
0.95
0.975
0.95
0.05
0.025
12.706
4.303
3.182
2.776
2.571
2.447
2.365
2.306
2.262
2.228
2.201
2.179
2.160
2.145
2.131
2.120
2.110
2.101
2.093
2.086
2.080
2.074
2.069
2.064
2.060
2.056
2.052
2.048
2.045
2.042
2.021
2.009
2.000
1.994
1.990
1.987
1.984
1.962
1.960
0.025
0.05
0.95
0.975
0.99
0.98
0.02
0.01
31.821
6.965
4.541
3.747
3.365
3.143
2.998
2.896
2.821
2.764
2.718
2.681
2.650
2.624
2.602
2.583
2.567
2.552
2.539
2.528
2.518
2.508
2.500
2.492
2.485
2.479
2.473
2.467
2.462
2.457
2.423
2.403
2.390
2.381
2.374
2.368
2.364
2.330
2.326
0.01
0.02
0.98
0.99
0.995
0.99
0.01
0.005
63.657
9.925
5.841
4.604
4.032
3.707
3.499
3.355
3.250
3.169
3.106
3.055
3.012
2.977
2.947
2.921
2.898
2.878
2.861
2.845
2.831
2.819
2.807
2.797
2.787
2.779
2.771
2.763
2.756
2.750
2.704
2.678
2.660
2.648
2.639
2.632
2.626
2.581
2.576
0.005
0.01
0.99
0.995