King M.R., Mody N.A. Numerical and Statistical Methods for Bioengineering: Applications in MATLAB

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sample design. In an independent sample design, the effects of the uncontrolled

confounding factors are mixed with the treatment effects and potentially corrupt

the test result.

Even though a paired experimental design generates two data sets, the data points

in each set pair up in a regular way, and therefore the data sets are not independent.

A hypothesis test meant for testing two independent samples, such as the two-sample

z test or two-sample t test, should not be used to analyze paired data. Instead, special

hypothesis tests that cater to paired data sets should be used. If the measured

variable is cardinal and the difference in measurements, d, is normally distributed,

you can use a one-sample t test, called a paired sample t test, to assess statistical

signiﬁcance of the difference in treatment(s) . Note that the individual d ’s must be

independent of each other.

In a paired sample t test, if x

and x

are two paired data points from a data set

containing n pairs, then the difference between the two paired data points is calcu-

lated as d

¼ x

 x

. The sample mean difference is given by

d ¼

i¼1

 x

ðÞ¼







The expected value of the sample mean difference is





¼ μ

¼ E







ðÞ¼μ

 μ

Thus, the population mean difference between the paired data points, μ

, is equal to

the difference of the tw o population means, μ

 μ

. If the two population means are

the same, then μ

¼ 0. If μ

, then μ

0. The only estimate available of the

standard deviation of the population distribut ion of differences between the data

pairs is s

, the standard deviat ion of the sample differences.

The relevant hypotheses for the paired sample t test are:

: “The treatment has no effect, μ

¼ 0.”

: “The treatment has some effect, μ

6¼ 0.”

We have assumed in the hypotheses stated above that the null hypothesis is a

statement of no effect. This does not have to be the case. We can also write the null

hypothesis as

: μ

¼ μ

;

where μ

is the exp ected population mean difference under the null hypothesis. The

alternate hypothesis becomes



d 6¼ μ

This is an example in which the null hypothesis does not necessarily represent the

status quo. A non-zero mean difference μ

is expected under the null hypothesis. If d is

normally distributed within the population, and if the null hypothesis is true, then the

sampling distribution of ð



d  μ

Þ=ðs

ﬃﬃﬃ

Þ is the t distribution with n – 1 degrees of

freedom. Since we wish to consider the null hypothesis of status quo, we set μ

¼ 0.

The test statistic for the paired sample t test is

t ¼



ﬃﬃﬃ

: (4:13)

247

4.6 The t test

The t distribution approximates the z distribution when n > 30.

Box 4.5 Narrowing of the optic canal in fibrous dysplasia

Fibrous dysplasia of the bone is a rare condition that results from replacement of normal bone tissue

with fibrous tissue. The fibrous growth causes expansion and weakening of the bone(s). This disease is

caused by a somatic cell mutation of a particular gene and is therefore acquired after birth. This

condition may be present in only one bone (monostotic) or in a large number of bones (polyostotic).

In the latter case, expansion of the bone within the anterior base of the cranium is often observed, which

leads to compression of the optic nerve. A study (Lee et al., 2002) was performed to determine if

compression of the optic nerve is symptomatic of fibrous dysplasia, and if it correlates with vision loss.

The areas of the right and left optic canals of individuals belonging to two groups, (1) patients with

fibrous dysplasia and (2) age- and gender-matched controls, were measured using computed tomo-

graphic (CT) imaging. The results are presented in Table 4.4.

The non-directional null hypothesis to be tested is

: μ

rightðÞ¼0; H

: μ

leftðÞ¼0:

The mean difference in the areas of the right and left optic canals between group 1 and matched controls

in group 2 were found to be

right canal:



d ¼2:37  4:8; n ¼ 32; f ¼ 31;

left canal:



d ¼1:83  4:9; n ¼ 35; f ¼ 34:

We test the hypothesis for the right optic canal. The t statistic is

t ¼

2:37

4:8=

ﬃﬃﬃﬃﬃ

¼2:793:

The corresponding p value is obtained using the MATLAB tcdf function. The statement

p = 2*(1-tcdf(2.793,31))

gives us p = 0.009. Since 0.009 < 0.05, we reject H

for the right eye. Thus, the areas of the right optic

nerve canals of patients with craniofacial fibrous dysplasia are narrower than the dimensions observed in

healthy individuals.

Now we test the hypothesis for the left optic canal. The t statistic is

t ¼

1:83

4:9=

ﬃﬃﬃﬃﬃ

¼ 2:178:

The corresponding p value is 0.036. Since 0.036 < 0.05, we reject H

for the left eye and draw the same

conclusion as that for the right eye.

What if we had treated the two groups as independent samples and performed a two-sample z test

(or two-sample t test) to evaluate the hypothesis? What p value would you get? Calculate this. You will

Table 4.4. Measurements of left and right optic canal areas in the study population

(Lee et al. 2002)

Group 1 Group 2

Sample size, n 32 35

Area (mm

) right 9.6 ± 3.8 12.0 ± 2.9

left 9.9 ± 3.6 11.9 ± 2.7

248

Hypothesis testing

One shortcoming of the paired design is that it assumes that a treatment or

condition will have the same effect

on all individuals or experimental units in

the population. If the treatment has different effects (e.g. large and positive on

some, large and negative on others) on different segments of the population, a paired

analysis will not reveal this and may in fact show negligible or small treatment effects

when actually the opposite may be true. This limitation exists in independent-sample

designs as well. Therefore all data points should be carefully inspected for such trends.

Using MATLAB

The ttest function can be used to perform a paired sample analysis. The syntax for

the ttest for this purpose is

[h, p] = ttest(x, y, alpha, tail)

where x and y are the vectors storing data values from two matched samples. Each

data point of x is paired with y, and therefore the two vectors must have equal

lengths.

4.6.2 Independent two-sample t test

When we wish to make inferences regarding the difference of two population means,

 μ

, and the random samples that are drawn from their respective populations

are independent or not related to one another, we use the independent or unpaired

samples t test. When the sampl e sizes are small, it is imperative that the populations

in question have a normal distribution in order to use the t test. If the sample sizes are

large, the requirement for normality is relaxed. For large n, the sampling distribution







becomes increasingly normal. The shape of the t curve approaches the

standard normal distribution and the z test may be used instead.

As discussed in Section 4.5.2, when the normality conditions mentioned above

are met, the difference of two sample means







is distributed as

N μ

 μ

;

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

þ σ



. The best estimate we have for the standard error

uses the sample standard deviation, i.e.







find that the two-sample test yields a similar result. Why do you think this is so? Compare the SE for

both the paired test and the unpaired test. You will notice that there is little difference between the two.

This means that precision was not gained by pairing. Matching on the basis of age and gender may not

be important, as possibly these factors may not be biasing the measurements. Perhaps height, weight,

and head circumference have more influence on the variability in optic canal cross-sectional area.

Pairing of data improves the power of the test when the SE of paired data is smaller than the SE of

unpaired data.

By treating paired data as independent samples, essentially we unpair the data, and thereby lose the

precision that is gained when related measurements are paired. In a paired design the variation is

minimized between the paired observations since the paired measurements have certain key identical

features. Sometimes, a well-designed paired analysis analyzed using an independent two-sample test

will yield a non-significant result even when a paired sample test demonstrates otherwise.

249

4.6 The t test

The method used to calculate the test statistic hinges on whether the populations are

homoscedastic (population variances are equal). The following two situations are

possible.

Population variances are equal

The best estimate of the population variance is obtained by pooling the two estimates

of the variance. We calculate the pooled sample variance, which is a weighted average

of the sample variances. The weights are the degrees of freedom. This means that if the

samples are of unequal size, then the larger sample provides a more accurate estimate

of the sample variance and will be given more weight over the other sample variance

based on a smaller sample size. The pooled sample variance is given by

pooled

 1ðÞs

þ n

 1ðÞs

þ n

 2

; (4:14)

pooled

is associated with n

þ n

 2 degrees of freedom, which is the sum of the

degrees of freedom for sample 1, n

– 1, and sample 2, n

– 1. The larger number of

degrees of freedom implies greater precision in the statistical analysis. The standard

error (SE) based on the pooled sample variance is given by

SE ¼ s

pooled

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

The sample estimate of the difference of two population means is







, which when

standardized follows the t distribution with n

þ n

 2 degrees of freedom. The test

statistic for the unpaired sample t test for equal population variances is

t ¼







ðÞμ

 μ

ðÞ

pooled

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

If the null hypothesis is H

: μ

 μ

¼ 0, then

t ¼







ðÞ

pooled

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

(4:15)

is distributed under the null hypothesis according to the t distribution with

þ n

 2 degrees of freedom.

The population variances are not equal

If the sample variances differ by a factor of 4, or there is reason to believe that the

population variances are not equ al, then the best estimate of the standard error of

the difference of two sample means is

SE ¼

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

The t statistic is calculated as

250

Hypothesis testing

t ¼







ðÞ

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

; (4:16)

but does not exactly follow a t distribution when H

is true. The best approximate t

distribution followed by the t statistic calculated in Equation (4.16) corresponds to f

degrees of freedom calculated using Satterthwaite’s formula:

f ¼



 1ðÞ



 1ðÞ



: (4:17)

If both sample sizes are the same, then the t statistic calculation is identi cal for both

the equal variance and unequal variance case. Show that when n

¼ n

; SE

6¼σ

¼σ

for the two-sample t test.

Using MATLAB

The two-sample t test can be performed for both equal and unequal sample variances

using the Statistics Toolbox’s function ttest2. One form of syntax for ttest2 is

[h, p] = ttest2(x, y, alpha, tail, vartype)

where x and y are vectors containing the data values of the two independent samples

and do not need to have equal lengths; alpha and tail have their usual meanings;

vartype speciﬁes whether the variances are equal, and can be assigned one of two

values, either “equal” or “unequal.” If “equal” is speciﬁed, then MATLAB calcu-

lates the pooled sample variance and uses Equation (4.15) to calculate the t statistic.

If “unequal” is speciﬁed, MATLAB calculates the t statistic using Equation (4.16).

The degrees of freedom for unequal variances are calculated using Satterthwaite’s

approximation (Equation (4.17)); h and p have their usual meanings.

4.7 Hypothesis testing for population proportions

Sometimes, an engineer/scientist needs to estimate the fraction or proportion of a

population that exhibits a particular characteristic. On the basis of a single charac-

teristic, the entire population can be divided into two parts:

(1) one that has the characteristic of interest, and

(2) the other that is devoid of that characteristic.

Any member chosen from the population will either possess the characteristic

(“successful outcome”) or not possess it (“failed outcome”). The event – the

presence of absence of a characteristic – constitutes a random variable that can

assume only two non-numeric values. A nominal variable of this type is called a

dichotomous variable because it divides the population into two groups or catego-

ries. We are now in a position to link the concept of population proportion to the

theory of the Bernoulli trial and the binomial distribution that was developed in

Section 3.4.1 .

251

4.7 Hypothesis testing for population proportions

Box 4.6 Flavone-8-acetic acid (FAA) reduces thrombosis

Flavone-8-acetic acid (FAA) is a non-cytotoxic drug that has antitumor activity in mice. It has also been

found that FAA inhibits human platelet aggregation by blocking platelet surface receptor GPIbα binding

to von Willebrand factor, a large plasma protein. An in vivo study using a porcine animal model was

conducted to determine the effect of FAA on platelet aggregation (thrombosis) (Mruk et al., 2000). Pigs

were randomly assigned either to a treatment group that received an FAA bolus followed by continuous

infusion or to a control group that was administered placebo (saline solution). Balloon angioplasty was

performed for the carotid artery of all 20 pigs. Platelet deposition occurred in deeply injured arterial

regions that underwent dilatation. The mean deposited platelet concentration for the treated group was

(13 ± 3) × 10

cells/cm

, and for the control group was (165 ± 51) × 10

cells/cm

. The individual

values are listed in the following:

treated group: 5.5, 6.5, 8, 8, 9.5, 11.5, 14.5, 15, 15, and 33 × 10

/cm

, n

= 10,

control group: 10, 17, 60, 85, 90, 170, 240, 250, and 450 × 10

/cm

, n

=9.

To reduce the large variability in platelet deposition amounts, the data are transformed by taking the

natural logarithm. We want to determine if FAA has an effect on platelet deposition. The variances of the

log-transformed data of both groups differ by a factor of 6. We perform the t test for unequal variances

using the ttest2 function:

x1 = [5.5 6.5 8 8 9.5 11.5 14.5 15 15 33]*1e6;

x2 = [10 17 60 85 90 170 240 250 450]*1e6;

[h, p] = ttest2(log(x1), log(x2), 0.05, ‘both’, ‘unequal’)

8.5285e-004

Thus, the log of the amount of platelet deposition (thrombosis) in the porcine carotid artery during

FAA administration is significantly different from that observed in the absence of FAA dosing.

Fibrinogen deposition for both groups was also measured, and was (40 ± 8) × 10

molecules/cm

for the treated group and (140 ± 69) × 10

molecules/cm

for the control group.

The individual values are listed in the following:

treated group: 14, 16, 19, 22, 37, 39, 41, 50, 65, and 96 × 10

/cm

, n

= 10,

control group: 19, 20, 37, 38, 70, 120, 125, 200, and 700 × 10

/cm

, n

=9.

To reduce the large variability in fibrinogen deposition amounts, the data are transformed by taking

the natural logarithm:

x1 = [14 16 19 22 37 39 41 50 65 95]*1e12;

x2 = [19 20 37 38 70 120 125 200 700]*1e12;

[h, p] = ttest2(log(x1), log(x2), 0.05, ‘both’, ‘unequal’)

0.0883

Since p > 0.05, we conclude that the log of the amount of fibrinogen deposition in the porcine

carotid artery during FAA administration is not significantly different from that observed in the absence

of FAA dosing. These results indicate that FAA interferes with platelet binding to the injured vessel wall

but does not affect coagulation.

252

Hypothesis testing

In this section, we take the concept of a Bernoulli trial further. The “experiment” –

referred to in Chapter 3 – that yields one of two mutually exclusive and exhau stive

outcomes is deﬁned here as the process of sampling from a large population. In a

Bernoulli trial, the members of a population are solely described by a dichotomous

nominal variable, which is called a Bernoulli random variable x

(i refers to the

Bernoulli trial number). A dichotomous variable can have only two values that

represent two mutually exclusive and exhaustive outcomes such as

(1) age > 20 or age ≤ 20,

(2) sleeping or awake, and

(3) pH of solution < 7 or pH ≥ 7.

Box 4.2B Treatment of skeletal muscle injury using angiotensin II receptor blocker

The mean areas of fibrosis within the injured zone observed in the three groups after 3 weeks were

control group: 34% ± 17%, n

=4;

low-dose ARB group: 16% ± 9%, n

=8;

high-dose ARB group: 7% ± 4%, n

=8.

We wish to compare the extent of fibrous tissue development in the control group with that observed in

both the low-dose group and the high-dose group. We use the two-sample t test for dissimilar variances

to test the null hypothesis of no difference.

We first compare the control group with the low-dose group. The t statistic is calculated as

t ¼

16  34ðÞ

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

¼1:983:

The degrees of freedom associated with this test statistic are

f ¼



81ðÞ



41ðÞ



¼ 3:9:

The p value corresponding to this test statistic is 0.1202. If the null hypothesis is true, then we will

observe a difference of this magnitude between the control group and the low-dose treated group about

12.02% of the time. Since p > 0.05, we fail to reject the null hypothesis. The data do not provide

evidence that fibrosis is reduced by administering low doses of ARB for 3 weeks. Next, we compare the

control group with the high-dose group. The t statistic is calculated as

t ¼

7  34ðÞ

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

¼3:133:

The degrees of freedom associated with this test statistic are

f ¼



81ðÞ



41ðÞ



¼ 3:167:

The p value corresponding to t

f¼3:167

¼3:133 is 0.0483. Since p < 0.05, the evidence reveals a

significant difference between the extent of fibrosis occurring in the untreated group and that in the high-

dose treated group.

253

4.7 Hypothesis testing for population proportions

When a member is selected from a population, it is classiﬁed under only one of the

two possible values of the Bernoulli variable. One of the two mutually exclusive and

exhaustive outcomes is termed as a “success” and x

takes on the value of 1 when a

success occurs. The probability with which a success occurs in each Bernoulli trial is

equal to p. If the selec ted member does not possess the characteristic of interest, then

the outcome is a “fai lure” and the Bernoulli variable takes on the value of 0. A failure

occurs with probability 1 – p.

The expected value of the Bernoulli variable x

is the mean of all values of the

variable that turn up when the trial is repeated inﬁnitely. The probability of obtain-

ing a success (or failure) depends on the proportion of success (or failure) within the

population, and this determines the frequency with which a success (or failure)

occurs when the experiment is performed many times. The expected value of x

calculated using Equation (3.14):

ðÞ¼μ

ðÞ¼1  p þ 0  1 pðÞ¼p: (4:18)

The variance associated with the Bernoulli variable x

is calculated using Equations

(3.16) and (3.17) to yield

 μ

ðÞ



¼ σ

¼ Ex



 E μ



ðÞp

¼ 1

 p þ 0

 1 pðÞp

¼ p 1 pðÞ:

(4:19)

As discussed in Chapter 3, the binomial distribution is a discrete probability distri-

bution of the number of successes obtained when a Bernoulli trial is repeated n times.

A detailed description of the properties of this distribution is presented in

Section 3.4.1. The binomial variable x is a discrete random variable that assumes

integral values from 0 to n and represents the number of successes obtained in a

binomial process. The binomial variable x is simply the sum of n Bernoulli variables

produced during the binomial experiment,

x ¼

i¼1

(4:20)

The expected values of the mean and variance of the binomial variable x are

calculated below using Equations (3.14), (3.16), and (4.20):

ExðÞ¼μ ¼ E

i¼1

ðÞ¼np (4:21)

and

Ex μðÞ



¼ σ

¼ E

i¼1

ðÞnp

¼ E

i¼1

 μ

ðÞ

¼ E

i¼1

 μ

ðÞ

i¼1

j¼1;i6¼j

 μ

ðÞx

 μ



i¼1

 μ

ðÞ



i¼1

j¼1;i6¼j

 μ

ðÞx

 μ



254

Hypothesis testing

The expectation for x

 μ

ðÞx

 μ



can be easily calculated for a binomial process,

since the individual trials are independent of each other:

 μ

ðÞx

 μ



 μ

ðÞx

 μ



ðÞPx



(4:22)

The double summation in Equation (4.22) occurs over all possible values of x

and x

Since x

and x

both take on only two values, the double summation consists of only

four terms. Substituting the values of the variables and their respective pro babilities,

we can show that Ex

 μ

ðÞx

 μ



¼ 0. Try to obtain this result yourself.

Substituting Equation (4.19) into the above equation, we get

Ex μðÞ



¼ σ

i¼1

pð1 pÞ¼np 1 pðÞ: (4:23)

Equations (4.21) and (4.23) are identical to the derived mean and variance of a

binomial distribution given by Equations (3.15) and (3.18).

A sample derived from a large population can be used to estimate the population

proportion of a characteristic. A sample of size n is obtained through a bino mial

sampling process that consists of n Bernoulli trials. The sample proportion

p is

calculated a s

p ¼

i¼1

;

where x

is a Bernoulli variable and x is a binomial variable, and the hat above p

signiﬁes a statistic calculated from observed data;

p can be deﬁned as a sample mean

of the Bernoulli varia ble x

. Different samples of the same size will yield different

values of

p. The sampling distribution of

p is the relative frequency distribution of all

possible values of

p obtained from samples of size n, and is discrete in nature. Each

sample of size n can be viewed as a single binomial experiment consisting of n

Bernoulli trials. Since the sample proportion is simply the binomial variable divided

by the sample size n, the sampling distribution of a sample propo rtion is given by the

binomial probability distribution, where the x-axis binomial variable of the binomial

distribution curve is divided by n.

The expected value of the sample proportion is the mean of the sampling distri-

bution of

p, and is equal to the population proportion p:

pðÞ¼E

i¼1

EðxÞ

¼ p: (4:24)

The variance associated with the sampling distribution of

p is given by

p  pðÞ



¼ σ

¼ E

 p





¼ E

x  npðÞ



Substituting Equation (4.23) into the above expression, we get

p  pðÞ



¼ σ

pð1 pÞ

: (4:25)

Figure 4.7 demonst rates the effect of sample size n (the number of independent

Bernoulli trials) on the shape of the sampling distribution of

p. In the ﬁgure, the

population proportion p is equal to 0.7. If n is small, then the sampling distribution

255

4.7 Hypothesis testing for population proportions

of the sample proportion is considerably discrete. The sample will usually not provide

an accurate estimate of the true population proportion, and the sampling distribution

p will be relatively spread out over a larger range of

p values. When n is large, the

sample proportion is a better approximation of the true population proportion and

the sampling distribution of

p is more compact and spread over a narrower range.

In fact, if n is large, such that n

5andn 1

pðÞ

p will have a sampling

distribution that can be approximated by a continuous distribution – the normal

curve. This is a consequence of the central-limit theorem, which states that the

sampling distribution of the sample mean when n is large tends to a normal distribu-

tion. This theorem applies even if the random variable is not a continuous variable.

Since

p is essentially the sample mean of a set of 1’s and 0’s, i.e. Bernoulli variables,

its sampling distribution at large n is dictated by the central-limit theorem. This

theorem also applies to the sum

i¼1

of all sample values x

.Whenn is large, the

sampling distribution of the sum of n values of a sample will be normal. Thus, the

binomial distribution of x (x ¼ n

p), which is the sampling distribution of the sum of

n Bernoulli variables, will be approximately normal for large n.

When the conditions of n

5 and n 1

pðÞ

5 are satisﬁed,

p will be distributed

approximately as

p  Np;

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

pð1 pÞ

When the above conditions are satisﬁed, a conﬁdence interval for p can be obtained

using the normal probability distribution. We can also perform hypothesis testing

on population proportions using methods similar to those discussed in Sections 4.5

and 4.6.

4.7.1 Hypothesis testing for a single population proportion

When n is large, a hypothesis for a single population proportion can be tested using

the one-sample z test. A non-directional null hypothesis assumes that the population

Figure 4.7

The discrete sampling distribution of the sample proportion contracts into an approximately normal distribution as

sample size n increases (p = 0.7).

0 0.5 1

0.05

0.1

0.15

0.2

0.25

0.3

Probability

n = 10

0 0.5 1

0.05

0.1

0.15

0.2

n = 30

0 0.5 1

0.05

0.1

0.15

ˆp

= 60

(a) (b) (c)

256

Hypothesis testing