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Instructions. Please print your name at the top of each page now.
This is a closed-book, closed-note, 50 minute test. You may use a calculator but you may not
use scratch paper. Show all work on these pages or on the backs. Please place all books and
notes on the floor under your seat, out of sight. Pledge below and print your name and Stat
ID at the top of each page.
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Test Scoring:
Multiple choice questions (Circle the one best answer.)
Each numerical box (Use 3 significant digits)
Stating conclusions verbally using no symbols
2 points
3 points
4 points
Table Z. The Standard Normal CDF (partial listing).
z
-3.090
-3.000
-2.500
-2.326
-2.000
F(z)
.001
.0013
.0062
.01
.0228
z
-1.960
-1.645
-1.500
-1.282
-1.000
F(z)
.025
.05
.0668
.10
.1587
z
1.000
1.282
1.500
1.645
1.960
F(z)
.8413
.90
.9332
.95
.975
z
2.000
2.326
2.500
3.000
3.090
F(z)
.9772
.99
.9938
.9987
.999
1. The life span in years of a patient born with a certain genetic defect is normally distributed
with a (population) mean of 26 years and a (population) standard deviation of 4 years.
a) Compute the probability that such a patient has a life span of more than 32 years.
b) Compute the expected mean life span for a random sample of 4 patients. I.e., if you took a
random sample of 4 patients, what would you expect the value of the sample mean to be.
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Stat 5605, Biometry 1, Test 2, Fall 2008
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Name ____________________________________________ Stat ID__________
c) Recall that the standard deviation of the sample mean is called the standard error of the
mean. Compute the standard error the mean life span for a random sample of 4 patients.
d) Compute the probability that the mean life span for a random sample of 4 patients is
greater than 32 years.
e) (Circle the one best answer.) As the sample size increases, the standard error of the
mean
i) increases
ii) remains constant
iii) decreases
f) (Circle the one best answer.) As the sample size increases, the probability that the sample
mean life span is greater than 32 years
i) increases
ii) remains constant
iii) decreases
2. Which of the following formulas is used to estimate the population mean when the
population standard deviation is known?
a) X  z1  
n
2
b) X  tn1,1  s
n
2
c)
 n  1 s 2 ,  n  1 s 2
2
n 1,1
d) p  z1 
2

2
2
n 1,

2
p 1  p  n
e) None of the above.
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Stat 5605, Biometry 1, Test 2, Fall 2008
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Name ____________________________________________ Stat ID__________
3. A nutritionist wants to learn about the mean blood glucose level (mg/dl) of diabetic
teanagers.
a) The parameter of interest is
i) The mean blood glucose level (mg/dl) for all teenage diabetics.
ii) The blood glucose level (mg/dl) for the i-th randomly sampled teenage diabetic.
iii) The standard deviation of blood glucose level (mg/dl) for all teenage diabetics.
iv) The number of teenage diabetics in a large school district.
b) What sample size would be required to estimate the parameter of interest to within 5
mg/dl of the true value with 95% confidence, based on a pilot study that yielded the
following data? Follow the steps, put the numerical answers in the box, and show your
work on the right. (Hint: The purpose of the pilot study is to provide an estimate of
dispersion to be used for determining sample size.)
Pilot Study Sample: 138, 184, 144, 167, 158
i) The sample mean of the pilot study is
ii) The sample standard deviation of the pilot study is
iii) Temporarily assuming that the standard deviation found in the pilot study is the
population standard deviation, compute the sample size required to estimate the
parameter of interest to within 5.0 mg/dl of the true value with 95% confidence.
(Hint: Use the margin-of-error formula with z as though the standard deviation of the
pilot study were the population standard deviation. This will give you a “close
enough” estimate of the sample size required.) Show your work.
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Stat 5605, Biometry 1, Test 2, Fall 2008
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Name ____________________________________________ Stat ID__________
c) Subsequent to the pilot study, the investigator observed 64 randomly-selected teenage
diabetics, and obtained a (sample) mean of 163 mg/dl with a (sample) standard deviation
of 19.2 mg/dl. Use these data to estimate the parameter of interest with 95% confidence.
Put the 95% lower and upper confidence limits in the following two boxes, and show
your computations on the right. You need not formally show the 6-step method here.
(Note: “Subsequent to the pilot study” connotes that we are no longer assuming that the
standard deviation found in the pilot study is the population standard deviation. Indeed,
we make no further use of the data of the pilot study.)
d) State the conclusion verbally, using no symbols.
4. Using the data of the the previous question, i.e., the sample of 64 teenage diabetics with mean
glucose level of 163 mg/dl and a standard deviation of 19.2 mg/dl, estimate the standard
deviation of of blood glucose (mg/dl) for all teenage diabetics with 95% confidence. Do so by
completing the following steps.
a) The parameter of interest is
i) The mean blood glucose level (mg/dl) for allteenage diabetics.
ii) The blood glucose level (mg/dl) for the i-th randomly sampled teenage diabetic.
iii) The standard deviation of blood glucose level (mg/dl) for all teenage diabetics.
iv) The number of teenage diabetics in a large school district.
582788529
Stat 5605, Biometry 1, Test 2, Fall 2008
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Name ____________________________________________ Stat ID__________
b) The required assumption(s) is (are) that
i)
mean blood gucose level for all diabetic teens = 163 mg/dl.
ii)
the standard deviation of glucose level for all diabetic teens is 19.2 mg/dl.
iii)
blood glucose levels are normally distributed.
iv)
(i) and (ii)
v)
(i) and (iii)
vi)
(ii) and (iii)
vii) (i), (ii), and (iii).
c) The confidence level, i.e., confidence coefficient called for in this study is the number
d) The sample size is
e) Within this table of Chi-Squared critical values, circle the two critical values, i.e.,
percentiles, required.
degrees of
freedom
60
61
62
63
64
65
66
67
68
69
70
Lower–tail probability
0.01
0.025
0.05
37.5
40.5
43.2
38.3
41.3
44.0
39.1
42.1
44.9
39.9
43.0
45.7
40.6
43.8
46.6
41.4
44.6
47.4
42.2
45.4
48.3
43.0
46.3
49.2
43.8
47.1
50.0
44.6
47.9
50.9
45.4
48.8
51.7
Upper-tail probability
0.95
0.975
0.99
79.1
83.3
88.4
80.2
84.5
89.6
85.7
81.4
90.8
82.5
86.8
92.0
83.7
88.0
93.2
84.8
89.2
94.4
86.0
90.3
95.6
87.1
91.5
96.8
88.3
92.7
98.0
89.4
93.9
99.2
90.5
95.0
100.4
f) The lower confidence limit is
g)
The upper confidence limit is
582788529
Stat 5605, Biometry 1, Test 2, Fall 2008
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Name ____________________________________________ Stat ID__________
h) State the conclusion verbally, using no symbols.
5. New Study. The objective of a study is to estimate the proportion of teanagers who are
diabetic. This requires an expensive blood test, so the sample size is of great concern.
a) What sample size would be required to estimate the proportion with 90% confidence, a
margin of error of 4% (i.e., 0.04)? Show your work.
b) Calculate the sample size required to estimate the proportion with the same 90%
confidence, the same margin of error of 4% (i.e., 0.04), but assuming that the actual
population proportion is no larger than 25% (i.e., 0.25). Show your work.
c) How would using a smaller confidence level, say 80% confidence, affect the sample size
required to achieve the same precision?
i) A smaller confidence level would decrease the sample size required.
ii) A smaller confidence level would not affect the sample size required.
iii) A smaller confidence level would increase the sample size required.
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Name ____________________________________________ Stat ID__________
Miscellaneous Questions
MOE
50
6. The figure to the right is a graph of the
margin of error (MOE) of a confidence
40
interval for the mean as a function of the
sample size, n, for a fixed sample standard
30
deviation. The two curves are for two
A
different confidence levels, 90% and 99%.
Which of the curves is for the 90%
20
confidence level?
B
10
a) Curve A is for 90% confidence.
b) Curve B is for 90% confidence.
0
7. Improving the protocol of an experiment
0
10
20
30
40
50
such that the underlying variability is
Sample Size, n
decreased, would result in a 95%
confidence interval estimate that is
a) more accurate.
b) less accurate.
c) more precise.
d) less precise.
8. In confidence interval estimation, one would obtain a narrower, more precise confidence
interval by using
a) A smaller confidence level.
b) A larger sample size.
c) Less underlying variability.
d) All of the above.
e) None of the above.
9. An invistigator reported that she is 95% confident that mean height is between 24.0 and 30.0
cm. This means that
a) the sample mean is
b) the margin of error is
cm., and
cm.
10. Moe and Curly estimated the same parameter from the same population using the same
sample. The only difference in their methodology was that Moe estimated with 90%
confidence and Curly estimated with 95% confidence. Since they used the same data,
a) Moe was more confident in accuracy and Moe was more precise.
b) Moe was more confident in accuracy and Curly was more precise.
c) Curly was more confident in accuracy and Moe was more precise.
d) Curly was more confident in accuracy and Curly was more precise.
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Stat 5605, Biometry 1, Test 2, Fall 2008
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T-Table: 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
0
11
-5 -4 -3 -2 -1 0 1 2 3 4 5
12
t with 4 df
13
The vertical lines reference the critical
14
values of the t statistic for the purpose of
15
statistical inference.
16
For confidence interval estimation, the
17
critical values are the t-multipliers of
standard error that yield the margin of
18
error in the formula (MOE) = (t)(s.e.).
19
For a 90% confidence interval with 4
20
degrees of freedom, the t-multiplier is
21
t = 2.132, as indicated by the black
vertical lines in the graph.
22
23
For hypothesis testing, the critical values
are the t-values that correspond to the
24
commonly used significance levels of 
25
= 0.10, 0.05, and 0.01. For testing H0: 
26
≤ 150 vs. HA:  > 150, an observation of
27
t = 2.2 with 4 degrees of freedom
indicates that the P-value is
28
0.025 < P < 0.05, because t = 2.2 is
29
between the critical values of t = 2.132
30
and t = 2.776.
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):
582788529
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
Stat 5605, Biometry 1, Test 2, Fall 2008
0.95
0.975 0.99
0.995
0.90
0.95
0.98
0.99
0.10
0.05
0.02
0.01
0.05
0.025 0.01
0.005
6.314 12.706 31.821 63.657
2.920 4.303 6.965 9.925
2.353 3.182 4.541 5.841
2.132 2.776 3.747 4.604
2.015 2.571 3.365 4.032
1.943 2.447 3.143 3.707
1.895 2.365 2.998 3.499
1.860 2.306 2.896 3.355
1.833 2.262 2.821 3.250
1.812 2.228 2.764 3.169
1.796 2.201 2.718 3.106
1.782 2.179 2.681 3.055
1.771 2.160 2.650 3.012
1.761 2.145 2.624 2.977
1.753 2.131 2.602 2.947
1.746 2.120 2.583 2.921
1.740 2.110 2.567 2.898
1.734 2.101 2.552 2.878
1.729 2.093 2.539 2.861
1.725 2.086 2.528 2.845
1.721 2.080 2.518 2.831
1.717 2.074 2.508 2.819
1.714 2.069 2.500 2.807
1.711 2.064 2.492 2.797
1.708 2.060 2.485 2.787
1.706 2.056 2.479 2.779
1.703 2.052 2.473 2.771
1.701 2.048 2.467 2.763
1.699 2.045 2.462 2.756
1.697 2.042 2.457 2.750
1.684 2.021 2.423 2.704
1.676 2.009 2.403 2.678
1.671 2.000 2.390 2.660
1.667 1.994 2.381 2.648
1.664 1.990 2.374 2.639
1.662 1.987 2.368 2.632
1.660 1.984 2.364 2.626
1.646 1.962 2.330 2.581
1.645 1.960 2.326 2.576
0.05
0.025 0.01
0.005
0.10
0.05
0.02
0.01
0.90
0.95
0.98
0.99
0.95
0.975 0.99
0.995
8