King M.R., Mody N.A. Numerical and Statistical Methods for Bioengineering: Applications in MATLAB
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σ
2
c
1
c
2
¼ Ec
1
β
1
ðÞc
2
β
2
ðÞðÞ:
Since c
i
¼
P
m
j¼1
K
ij
y
j
and β
i
¼
P
m
j¼1
K
ij
μ
y
j
, where K ¼ A
T
A
1
A
T
and μ
y
i
¼ Ey
i
ðÞ,
σ
2
c
1
c
2
¼ E
X
m
j¼1
K
1j
y
j
μ
y
j
!
X
m
k¼1
K
2k
y
k
μ
y
k
! !
¼ E
X
m
j¼1
K
1j
K
2j
y
j
μ
y
j
2
þ
X
m
j¼1
X
m
k¼1;k6¼j
K
1j
K
2k
y
j
μ
y
j
y
k
μ
y
k
!
:
Box 3.9B Kidney functioning in human leptin metabolism
In Box 3.9A we calculated s
2
y
¼ 0:9039. In this example, K ¼ A
T
A
1
A
T
is a 2 × 16 matrix. Squaring
each of the elements in K (in MATLAB you can perform element-wise operations on a matrix such as
squaring by using the dot operator) we get
K:^2 ¼
:2106 :0570 :0266 :0186 :0172 :0035 :0046 :0053 :0082 :0099 :0109 :0150 :0154 :0161 :0166 :0175
:0071 :0002 :0001 :0004 :0004 :0066 :0071 :0073 :0084 :0089 :0092 :0103 :0105 :0106 :0108 :0110
;
s
2
c
1
¼
X
m
j¼1
K
2
1j
s
2
y
¼ s
2
y
X
m
j¼1
K
2
1j
¼ 0:9039 0:4529ðÞ¼0:4094;
s
2
c
2
¼ s
2
y
X
m
j¼1
K
2
2j
¼ 0:9039 0:1089ðÞ¼0:0985:
Although we have extracted the variance associated with the coefficients, we are not done yet. We wish
to determine the variances for K
m
and R
max
, which are different from that of c
1
¼ K
m
=R
max
and
c
2
¼ 1=R
max
.
First we calculate the mean values of the model parameters of interest. If we divide c
1
by c
2
we recover
K
m
. Obtaining R
max
involves a simple inversion of c
2
. So,
K
m
¼ 10:87; R
max
¼ 1:732:
We use Equation (3.43) to estimate the variances associated with the two parameters above. We assume
that the model parameters also exhibit normality in their distribution curves.
Let’s find the variance for R
max
first. Using Equation (3.43) and
z ¼ R
max
¼ 1=c
2
, we obtain
s
2
R
max
R
2
max
¼
s
2
c
2
c
2
2
or s
2
R
max
¼ 1:732
2
0:0985
0:5774
2
¼ 0:8863:
The 95% confidence interval is given by
R
max
t
0:975; f¼14
s
R
max
:
Since t
0:025; f¼14
¼ 2:145,
1:732 2:145ð0:941Þ or 0:286; 3:751
is the 95% confidence interval for R
max
. Since negative values of R
max
are not possible, we replace the
lower limit with zero. We can say with 95% confidence that the true value of R
max
lies within this range.
Next, we determine the variance associated with K
m
using Equation (3.43). Note that Equation (3.43)
has a covariance term for which we do not yet have an estimate. Once we investigate the method
(discussed next) to calculate the covariance between coefficients of a model, we will revisit this problem
(see Box 3.9C).
197
3.7 Linear regression error
Since
E
X
m
j¼1
X
m
k¼1;k6¼j
K
1j
K
2k
y
j
μ
y
j
y
k
μ
y
k
!
¼
X
m
j¼1
X
m
k¼1;k6¼j
K
1j
K
2k
Ey
j
μ
y
j
y
k
μ
y
k
¼ 0;
we have
σ
2
c
1
c
2
¼ E
X
m
j¼1
K
1j
K
2j
y
j
μ
y
j
2
!
¼
X
m
j¼1
K
1j
K
2j
Ey
j
μ
y
j
2
;
σ
2
c
1
c
2
¼
X
m
j¼1
K
1j
K
2j
σ
2
y
: (3:50)
Using Equation (3.50) to proceed with the calculation of s
2
^
y
0
, we obtain
s
2
^
y
0
¼ f
1
x
0
ðÞðÞ
2
s
2
c
1
þ f
2
x
0
ðÞðÞ
2
s
2
c
2
þ 2f
1
x
0
ðÞf
2
x
0
ðÞ
X
m
j¼1
K
1j
K
2j
s
2
y
: (3:51)
Equation (3.51) calculates the variance of the predicted mean value of y at x
0
.
The foregoing developments in regression analysis can be generalized as shown
below using concepts from linear algebra. We begin with m observations of y
i
associated with different values of x
i
(1 ≤ i ≤ m). We wish to describe the dependency
of y on x with a predictive model that consists of n parameters and n modeling
functions:
y
i
X
n
j¼1
c
j
f
j
x
i
ðÞ;
where c
j
are the parameters and f
j
are the modeling functions. As described earlier,
the above fitting function can be rewritten as a matrix equation, y = Ac, where
A
ij
¼ f
j
x
i
ðÞ. The solution to this linear system is given by the normal equations, i.e.
c ¼ A
T
A
1
A
T
y,orc ¼ Ky, where K ¼ A
T
A
1
A
T
. The normal equations mini-
mize k r k
2
2
, where r is defined as the residual: r = y – Ac.
Recall that the covariance of two variables y and z is defined as
σ
2
yz
¼ Eððy μ
y
Þðz μ
z
ÞÞ, and the variance of variable y is defined as
σ
2
y
¼ Eððy μ
y
Þ
2
Þ. Accordingly, the matrix of covariance of the regression parame-
ters is obtained as
Σ
2
c
E c μ
c
ðÞc μ
c
ðÞ
T
where μ
c
¼ E cðÞ¼β, and Σ
2
c
is called the covariance matrix of the model parameters.
Note that the vector–vector multiplication νν
T
produces a matrix (a dyad), while the
vector–vector multiplication ν
T
ν is called the dot product and produces a scalar. See
Section 2.1 for a review on important concepts in linear algebra. Substituting into
the above equation c ¼ Ky and μ
c
¼ Kμ
y
, we get
Σ
2
c
¼ KE y μ
y
y μ
y
T
K
T
:
198
Probability and statistics
Note that ðKðy μ
y
ÞÞ
T
¼ðy μ
y
Þ
T
K
T
(since ABðÞ
T
¼ B
T
A
T
).
Now, Eððy μ
y
Þðy μ
y
Þ
T
Þ is the covariance matrix of y, Σ
2
y
. Since the individual
y
i
observations are independent from each other, the covariances are all identically
zero, and Σ
2
y
is a diagonal matrix (all elements except those along the diagonal are
zero). The elements along the diagonal are the variances associated with each y
i
observation. If all of the observations y have the same variance, then Σ
2
y
¼ Iσ
2
y
, where
σ
2
y
can be estimated from Equation (3.48). Therefore,
Σ
2
c
¼ σ
2
y
KK
T
: (3:52)
We can use the fitted model to predict y for k different values of x. For this we create
a vector x
ðmÞ
that contains the values at which we wish to estimate y . The model
predictions
^
y
ðmÞ
are given by
^
y
ðmÞ
¼ A
ðmÞ
c;
where A
ðmÞ
ij
¼ f
j
x
ðmÞ
i
and A
ðmÞ
is a k × n matr ix. We want to determine the error in
the prediction of
^
y
ðmÞ
. We construct the matrix of covariance for
^
y
ðmÞ
as follows:
Σ
2
^
y
ðmÞ
¼ E
^
y
ðmÞ
μ
ðmÞ
y
^
y
ðmÞ
μ
ðmÞ
y
T
:
Substituting
^
y
ðmÞ
¼ A
ðmÞ
c, we obtain
Σ
2
^
y
ðmÞ
¼ A
ðmÞ
E c μ
c
ðÞc μ
c
ðÞ
T
A
mðÞT
¼ A
ðmÞ
Σ
2
c
A
mðÞT
:
Therefore
Σ
2
^
y
ðmÞ
¼ σ
2
y
A
ðmÞ
KK
T
A
mðÞT
: (3:53)
The diagonal elements of Σ
2
^
y
ðmÞ
are what we seek, since these elem ents provide the
uncertainty (variance) in the model predictions. Note the distinction between σ
2
y
and
the diagonal elements of Σ
2
^
y
ðmÞ
. The form er is a measure of the observed variability of
y about its true mean μ
y
, while the latter measures the error in the prediction of μ
y
yielded by the fitted model and will, in general, be a function of the independent
variable.
The x
ðmÞ
vector should not contain any x values that correspond to the data set
that was used to determine the model parameters. Can you explain why this is so?
Replace A
(m)
with A in Equation (3.53) and simplify the right-hand side. What do
you get?
3.8 End of Chapter 3: key points to consider
(1) Statistical information can be broadly classified into two types: descriptive statistics,
such as mean, median, and standard deviation, and inferential statistics, such as
confidence intervals . In order to draw valid statistical inferences from the data, it is
critical that the sample is chosen using methods of random sampling.
199
3.8 Key points
Box 3.9C Kidney functioning in human leptin metabolism
Let’s complete the task we began in Box 3.9B of determining the variance of K
m
¼
z ¼ c
1
=c
2
.
We have
s
2
y
¼ 0:9039; c
1
¼ 6:279; c
2
¼ 0:5774; s
2
c
1
¼ 0:4094; s
2
c
2
¼ 0:0985;
s
2
c
1
c
2
¼ s
2
y
X
m
j¼1
K
1j
K
2j
¼ 0:9039 0:1450ðÞ¼0:1311:
From Equation (3.43),
s
2
z
z
2
¼
s
2
c
1
c
2
1
þ
s
2
c
2
c
2
2
2
s
2
c
1
c
2
c
1
c
2
"#
¼
0:4094
6:279
2
þ
0:0985
0:5774
2
2
0:1311
6:279ð0:5774Þ
¼ 0:3782:
So,
s
2
z
¼ 0:3782 10:87
2
¼ 44:68:
The 95% confidence interval is given by
K
m
t
0:975; f¼14
s
K
m
:
Since t
0:975; f¼14
¼ 2:145,
10:87 2:145ð6:684Þ or 3:467; 25:208
is the 95% confidence interval for K
m
. However, K
m
cannot be less than zero. We modify the confidence
interval as (0, 25.2). The interval is large due to the large uncertainty (s
K
m
) in the value of K
m
.
Next, we predict kidney leptin uptake at a plasma leptin concentration S of 15 ng/ml. Our
mathematical model is 1=
^
R ¼ 6:279ð1=SÞþ0:5774; from which we obtain 1=
^
R ¼ 0:996. Here,
f
1
SðÞ¼1=S and f
2
SðÞ¼1. The variance associated with this estimate of the mean value at
1=S ¼ 1=15 is calculated using Equation (3.51):
s
2
^
y
0
¼
1
15
2
0:4094 þ 0:0985 þ
2
15
0:1311ðÞ¼0:0828:
Note that the variance s
2
^
y
0
associated with the prediction of the mean
^
y
0
is much less than the variance s
2
y
associated with an individual observation of y
i
about its mean value
^
y
i
. Why is this so?
We also wish to estimate the error in the model predictions over the interval S ∈[0.75, 35.71] or 1/S
∈[0.028 1.333]. Let
x ¼
1
S
and x
mðÞ
¼
0:03
0:1
0:25
0:5
0:75
1:0
1:3
2
6
6
6
6
6
6
6
6
4
3
7
7
7
7
7
7
7
7
5
:
Then,
A
ðmÞ
¼
x
ðmÞ
1
1
x
ðmÞ
2
1
:
:
:
x
ðmÞ
7
1
2
6
6
6
6
6
6
6
4
3
7
7
7
7
7
7
7
5
:
200
Probability and statistics
(2) The mean
x ¼
P
n
i¼1
x
i
n
of a sample drawn by random sampling is an unbiased estimator of the population
mean μ. The variance
s
2
¼
P
n
i¼1
ðx
i
xÞ
2
n 1
of a sample drawn by random sampling is an unbiased estimator of the population
variance σ
2
.
(3) A Bernoulli trial can have only two outcomes – success or failure – which are
mutually exclusive. A binomial process involves several independent Bernou lli trials,
where the probability of success remains the same in each trial. x su ccesses out of
n trials is a binomial variable. The binomial probability distribution is given by
C
n
x
p
x
ð1 pÞ
nx
for 0 ≤ x ≤ n, where x is an integ er.
(4) The Poisson distribution describes the probability of events occurring in a finite
amount of time or space. If x is a discrete rand om variable that takes on positive
integral values and γ is the mean or exp ected value of x, then the Poisson distribution
is given by γ
x
e
γ
=x!.
(5) A density curve is a probability distribution plot for a continuous rando m variable x
that has an area under the curve equal to 1.
The variance (error) associated with each model prediction at x
ðmÞ
i
is calculated using Equation (3.53).
The diagonal of Σ
2
^
y
ðmÞ
gives us the variance vector s
2
y
ðmÞ
, i.e. the variance (error) associated with each
prediction contained in the vector
^
y
ðmÞ
¼ A
ðmÞ
c. Upon taking the square root of each element of s
2
^
y
ðmÞ
,
we obtain the standard deviations associated with the model predictions
^
y
ðmÞ
. Figure 3.18 shows a plot
of the predicted
^
y
ðmÞ
values and the curves depicting the 68% (1σ) confidence intervals (magnitude of
error or uncertainty) for the model predictions.
Upon careful examination you will note that the 1σ curves are not parallel to the regression curve. The
error curves lie closest to the regression curve at the mean value of x =1/S of the original data set. In
general, the uncer tainty increases in the model’s predictions as one moves away from the mean and
towards the limits of the measured range.
Figure 3.18
The 68% confidence intervals that depict the error in model predictions.
0 0.5 1 1.5
0
2
4
6
8
10
1/R (min/nmol)
1/S (ml/n
g
)
Model predictions
1σ error curves
201
3.8 Key points
(6) The normal density function describ es a unimodal bell-shaped curve symmetric about
the peak or mean value and is given by
fxðÞ¼
1
ffiffiffiffiffi
2π
p
σ
e
ðxμÞ
2
2σ
2
;
where μ is the mean and σ is the standard deviation of the normal distribution N(μ, σ).
(7) The standard error of the mean or SEM is the standard deviation of the sampling
distribution of a sample mean, i.e. the probability distribution of all possible values of
sample means obtained from a single population, where each sample is of size n;
SEM = s
x
¼ σ=
ffiffiffi
n
p
.
(8) The z statistic foll ows the standard normal distribution N(0, 1), where z ¼ðx μÞ=σ,
and x is a normally distributed random variable. Note that 95% of all observations
of x are located within 1.96 standard deviations from the mean μ of the distribution.
(9)
x 1:96 σ=
ffiffiffi
n
p
is called the 95% confidence interval for the population mean μ
because it estimates, with 95% confidence, the range of values which contain the
population mean.
(10) The Student’s t distribution is used to estimate the width of the confidence interval
when the population standard deviation is not known and the sample size n is small.
The shape of the t curve is a function of the degrees of freedom available in
calculating the sample standard deviation.
(11) The central-limit theorem states that if a sample is drawn from a non-normally
distributed population with mean μ and variance σ
2
, and the sample size n is large,
the distribution of the sample means will be approximately normal with mean μ and
variance σ
2
/n.
(12) When combining measur ements associated with error, one must also combine the
errors (variances) in accordance with prescribed methods.
(13) It is not enough merely to fit a linear least-squares curve to a set of data thereby
estimating the linear model parameters. One should also report the confidence one
has in the values of the model coefficients and model predictions in terms of standard
deviation (err or) or confidence intervals.
3.9 Problems
Problems that require adva nced critical an alysis are labeled **.
3.1. A study on trends in cigarette smoking among US adolescents has a sample size of
16 000 students from grades 8, 10, and 12. It was found that in the year 2005, 25% of
12th-grade boys and 22% of 12th-grade girls were current smokers, i.e. they had
smoked at least one cigarette in the past 30 days (Nelson et al., 2008).
(a) If a class contains seven girls and eight boys, what is the probability that all
students in the class are smokers?
(b) If a class contains seven girls and eight boys, what is the probability that none of
the students in the class are smokers?
3.2. A cone and plate viscometer is used to shear a solution of suspended platelets. If the
concentration of platelets in the solution is 250 000 platelets/μl, how many unique
two-body collisions can occ ur per microliter?
3.3. Obesity is a rapidly growing epidemic in the USA. On average, an adult needs a daily
intake of 2000 calories. However, there has been an increase in daily calorie intake of
200–300 calories over the past 40 years. A fast-food meal usually has high energy
content, and habitual intake of fast-food can lead to weight gain and obesity. In fact,
202
Probability and statistics
in 2007, 74% of all food purchased from a US restaurant was fast-food. In 2007,
most fast-food chains did not provide calorie or nutritional information, with the
exception of Subway, which posted calorie information at the point of purchase. A
study (Bassett et al., 2008) surveyed the eating habits of numerous cu stomers of New
York City fast-food chains to assess the benefits of providing calorie information to
patrons. A summary of their results is provided in Tables 3.7 and 3.8.
From Tables 3.7 and 3.8, the following is evident.
*
On average, the calorie consumption of Subway customers who saw the calorie
information reduced.
*
The calorie purchasing habits of Subway customers who did not view the calorie
information and of those who viewed the calorie information but disregarded it
are similar. This indicates that it is not only the calorie-conscious population that
view the calorie informat ion.
Determine the following.
(a) What is the probability that three people walking into a fast-food restaurant
categorized as “All except Subway” will all purchase a very high calorie content
meal (≥1250 calories)? Calculate this probability for Subway customers for
comparison.
(b) What is the probability that exactly five out of ten people walking into a fast-food
restaurant categorized as “All except Subway” will purchase a normal calorific
meal of <1000 calories? What is this probability for Subway customers?
(c) Using the bar function in MATL AB, plot a binomial distribution of the
probabilities that i out of ten people (0 ≤ i ≤ 10) entering a fast-food chain
Table 3.8.
Subway customers
Purchases
<1000 calories (%)
Purchases 1000–1250
calories (%)
Purchases ≥1250
calories (%)
Customers did not see
calorie info.
77.0 12.7 10.3
Customers saw calorie info. 82.6 10.0 7.4
Calorie info. had effect on
purchase
88.0 8.0 4.0
Calorie info. had no effect on
purchase
79.8 11.1 9.1
Table 3.7.
Fast-food chain
Purchases <1000
calories (%)
Purchases 1000–1250
calories (%)
Purchases ≥1250
calories (%)
All
a
except
Subway
66.5 19 14.5
Subway 78.7 11.9 9.4
a
Includes Burger King, McDonald’s, Wendy’s, Kentucky Fried Chicken, Popeye’s, Domino’s, Papa
John’s, Pizza Hut, Au Bon Pain, and Taco Bell.
203
3.9 Problems
categorized as “All except Subway” will purchase a normal calorific meal of
<1000 calori es. Repeat for Subway customers.
(d) Wh at is the probability that three Subway patrons do not see calorie informa-
tion and yet all three purchase a very high calorie content meal (≥1250
calories)?
(e) What is the probability that three Subway patrons see calorie information and
yet all three purchase a very high calorie content meal (≥1250 calories)?
(f) **If there are ten people who enter a Subway chain, and five individuals see the
calorie information and five do not, what is the probability that exactly five
people will purchase a normal calorific meal of <1000 calories? Compare your
answer with that available from the binomial distribution plotted in (c). Why is
the probability calculated in (f) less than that in (c)?
(g) **If all ten people who purchase their meal from a Subway restaurant admit to
viewing the calorie information before making their purchase, and if three out of
the ten people report that the information did not influence their purchase, then
determine the probability that exactly three of the ten customers purchased a
meal with calorie content greater than 1000. Compare your answ er with that
available from the binomial distribution plotted in (c). How is your answer
different, and why is that so?
3.4. An experiment W yields n outcomes. If S is the sample space then S =(ξ
1
, ξ
2
, ξ
3
, ...,
ξ
n
), where ξ
i
is any of the n outcomes. An event is a subset of any of the n outcomes
produced when W is performed, e.g. E
1
:(ξ
1
, ξ
2
)orE
2
:(ξ
1
, ξ
2
, ξ
3
). An event is said to
occur when any of the outcomes defined within the set occurs. For example, if a six-
sided die is thrown and the event E is defined as (1, 2, 3), then E occurs if either a 1, 2,
or 3 shows up. An event can contain none of the outcomes (the impossible event),
can contain all of the outcomes (the certain event), or be any combination of the n
outcomes. Using the binomial theorem,
x þ aðÞ
n
¼
X
n
k¼0
C
n
k
x
k
a
nk
and your knowledge of combinato rics, determine the total number of events (sub-
sets) that can be generated from W.
3.5. Reproduce Figure 3.4 by simulating the binomial experiment using a computer. Use
the MATLAB uniform random number generator rand to generate a random
number x between 0 and 1. If x ≤ p, then the outcome of a Bernoulli trial is a success.
If x > p, then the outcome is a failure. Repeating the Bernoulli trial n times generates
a Bernoulli process characterized by n and p. Perform the Bernoulli process 100, 500,
and 1000 times and keep track of the number of successes obtained for each run. For
each case
(a) plot the relative frequency or probability distribution using the bar function;
(b) calculate the mean and variance and compare your result with that provided by
Equations (3.15) and (3.18).
3.6. Show that the mean and variance for a Poisson probability distribution are both
equal to the Poisson parameter γ.
3.7. Dizziness is a side-effect of a common type II diabetic drug experienced by 1.7 people
in a sample of 100 on average. Using a Poisson model for the probability distribu-
tion, determine the probability that
(a) no person out of 100 patients treated with this drug will experience dizziness;
(b) exactly one person out of 100 patients treated with this drug will experience
dizziness;
204
Probability and statistics
(c) exactly two people out of 100 patients treated with this drug will experience
dizziness;
(d) at least two people out of 100 patients treated with this drug will experience
dizziness.
3.8. Prolonged exposure to silica dust generated during crushing, blasting, drilling, and
grinding of silica-containing materials such as granite, sandstone, coal, and some
metallic ores causes silicosis. This disease is caused by the deposition of silica
particles in the alveoli sacs and ducts of the lungs, which are unable to remove the
silica particles by coughing or expelling of mucous. The deposited silica particles
obstruct gaseous exchange between the lung and the blood stream. This sparks an
inflammatory response which recruits large numbers of leukocytes (neutrophils) to
the affected region. Roursgaard et al.(2008) studied quartz-induced lung inflam-
mation in mice. The increase in the number of neutrophils is associated with higher
levels of MIP-2 (anti-macrophage inflammatory protein 2). Bronc hoalveolar fluid
in the lungs of 23 mice exposed to 50 μg quartz was found to have a mean MIP-2
expression level of 11.4 pg/ml of bronchoalveolar fluid with a standard deviation of
5.75.
(a) What would you predict is the standard error of the mean of the MIP-2
expression levels?
(b) The expression levels may not be normally distributed, but since n =23 we
assume that the central-limit theorem holds. Construct a 95% confidence inter-
val for the population mean.
3.9. Clinical studies and animal model research have demonstrated a link between
consumption of black tea and a reduced risk for cancer. Theaflavins are polyphe-
nols – biologically active comp onents – abundantly present in black tea that have
both anti-oxidative and pro-oxidative properties. Their pro-oxidant behavior is
responsible for generating cytotoxic ROS (reactive oxygen species) within the cells.
Normal cells have counteracting mechanisms to scavenge or reduce ROS with anti-
oxidant enzymes. Cancer cells, on the other hand, are more sensitive to oxidative
stress in their cells. Theaflavins were recently shown to be more toxic to tongue
carcinoma cells than to normal gingival human fibroblasts, resulting in pronounced
growth inhibition and apoptosis in cancer cells. These results suggest the useful
chemopreventive properties of black tea for oral carcinoma. Exposure of human
fibroblasts to 400 mM TF-2A (theaflavin 3-gallate) for 24 hours resulted in a mean
viability of 82% with a standard error of the mean 3.33 (n = 4). Construct the 95%
confidence interval for the population mean viability of normal fibroblasts.
Exposure of carcinoma cells to 300 mM TF-2A (theaflavin 3-gallate) for 24 hours
resulted in a mean viability of 62% with a standard error of the mean of 2.67 (n = 4).
Construct the 95% confidence interval for the population mean viability of carci-
noma cells.
3.10. A 20.00 ml sample of a 0.1250 molarity CuSO
4
solution is diluted to 500.0 ml. If the
error in measurement of the molarity is ±0.0002, that of the 20.00 ml pipet is
±0.03 ml, and that of the volume of the 500 ml flask is ±0.15 ml, determine how
the molarity of the resulting solution should be reported. That is, what is the
molarity of the resulting solution, and what error is there in this measurement?
Recall that the molarity M of a solution is given by M = n/V, where n is the number
of moles of solute and V is the volume of the solution in liters. Thus, in our case,
M
init
V
init
= M
final
V
final
,orM
final
= M
init
V
init
/V
final
.
3.11. The pH of a solution is found to be 4.50 ± 0.02. What is the error in the concen-
tration of H
+
ions?
205
3.9 Problems
3.12. In quantitative models of human thermal regulation, the convective heat transfer
coefficient (h) combines with the conductive heat transfer coefficient through cloth-
ing (h
c
) to yield an overal l heat transfer coefficient H in the following way:
H ¼
1
1=h þ 1=h
c
:
If the convective heat transfer coefficient is measured as h = 43.0 ± 5.2 W/(m
2
K),
and the conductive heat transfer coeffici ent through clothing is measured as h
c
=
26.1 ± 1.8 W/(m
2
K), calculate the overall heat transfer coefficient H and report the
error in that estimate.
3.13. Many collections of biomo lecular fibers are modele d as semi-flexibl e polyme r
solutions. A n important length scale in such systems is the m esh size ξ,which
represents the typical spacing between polymers in solution. The following
equation relates the mesh size to the fiber radius a and the fiber volume
fraction :
ξ ¼
a
ffiffiffi
p
:
In the adventitia of the thoracic artery of the rat, the collagen fibrils have a diameter
of 55 ± 10(SD) nm and a volume fraction of 0.0777 ± 0.0141(SD). Using the error
propagation formulas, calculate the mean mesh size in the adventitia of rat thoracic
artery, along with its 95% confidence interval.
3.14. A sample of acute myeloid leukemia (AML) cells are run through a Coulter Z2 Cell
Counter to determine the distribution of cell sizes. This instrument determines the
volume of each cell in the sample quite accurately, and reports a mean cell volume of
623.6 ± 289.1(SD) μm
3
. Recognizing that the spherical cell diameter (d) is simply
related to the volume (V) by the relation
d ¼
6V
π
1=3
;
use the error propagatio n formulas to calculate the mean cell diameter and standard
deviation in diameter for this sample of AML cells.
3.15. A linear regression model is
^
y ¼ c
1
f
1
xðÞþc
2
f
2
xðÞ, where f
1
xðÞ¼x; f
2
xðÞ¼1, and
you have two data points (x
1
, y
1
)and(x
2
, y
2
). The predicted mean value of y at x
0
is
^
y
0
¼ c
1
x
0
þ c
2
, where c
1
and c
2
have already been determined. The variance asso-
ciated with y is σ
2
. You know from Equation (3.51) that the variance associated with
^
y
0
is given by
s
2
^
y
0
¼ f
1
x
0
ðÞðÞ
2
s
2
c
1
þ f
2
x
0
ðÞðÞ
2
s
2
c
2
þ 2f
1
x
0
ðÞf
2
x
0
ðÞ
X
m
j¼1
K
1j
K
2j
s
2
y
;
K ¼ A
0
AðÞ
1
A
0
, where A ¼
x
1
1
x
2
1
:
(a) Find K in terms of the given constants and data points.
(b) Using Equation (3.49) and your result from (a), show that s
2
c
1
¼ σ
2
=
P
x
i
xðÞ
2
:
(c) Using Equation (3.49) and your result from (a), show that s
2
c
2
¼ σ
2
P
x
2
i
=m
P
x
i
xðÞ
2
:
(d) Using Equation (3.50) and your result from (a), show that s
2
c
1
c
2
¼σ
2
x=
P
x
i
xðÞ
2
:
Note that the summation term indicates summation of all x from i =1tom, where
m = 2 here.
206
Probability and statistics