Supplemental Material for

The Use of Statistics in the High School Chemistry Laboratory

By: Paul S. Matsumoto1 and Chris Kaegi2

Galileo Academy of Science and Technology; Department of Science1 or

Mathematics2; San Francisco, CA 94109

Given the capability of websites (1), graphing calculators, and Microsoft Excel to

do the various statistical calculations, the reader should strive to gain a conceptual

understanding of statistics, rather than a capability of doing the calculations. Probability

is the basis of statistics; these topics will be presented using calculus (2) or algebra (3, 4).

The purpose of this supplement is to provide the reader with a background in

statistics. In addition, selected experimental data from my Honors, Advanced Placement,

and regular Chemistry classes will be analyzed using various statistical tests.

The presentation of statistics in this supplement will be somewhat mathematically

rigorous. The rationale for this level of rigor is analogous to the rationale that a high

school Chemistry teacher should have more than a first-year college Chemistry course.

That is, in both situations, a greater depth of understanding is more likely to produce a

good lesson for the students.

1

Random variables (2) may be either discrete (e.g. # heads in flipping a coin 100

times) or continuous (e.g. a person’s mass). The probability function involves a discrete

random variable and uses algebra, while the probability density function (pdf) involves a

continuous random variable and uses calculus. These functions generate the probability

for a specific value of the random variable.

Mean, standard deviation, and standard error of the mean. In an introductory

statistics laboratory exercise, students roll a die many times and record the value on the

face of a die, then analyze their results. The value on the face of a die is a discrete

random variable and its probability function is the uniform probability function (2)

[1] f(x) =1k

where k = number of faces of the die = 6; x = number of dots on the face of the die, and

f(x) = 0 for x ≠ 1, 2, 3, 4, 5, 6.

In this laboratory exercise, students determine the mean and standard deviation

(SD) of their sample. The mean of the sample is

[2] sample mean =

Σ xi

n

while the SD of the sample is

[3] sample SD = √ Σ (mean−xi)2

n−1

where n = sample size. The mean of a population containing discrete random variables is

[4] Population mean = ∑ x f ( x )

where in the case of rolling a die, the theoretical mean is

2

[5] mean = ∑x=1

6

x f (x ) = (1+2+3+4+5+6)

16 = 3.5.

The variance of a population containing discrete random variables is

[6] Population variance = ∑ (mean−x i)2 f ( x )

and [7] variance = SD2 [or SD=√ var iance ]

in the case of rolling a dice, the theoretical SD is

[8] SD = √∑i=1

6

(mean− xi )2 f ( x )

= √[ (3 .5-1 )2+(3 .5-2 )2+(3.5-3 )2+(3.5-4 )2+(3 .5-5 )2+(3 .5-6 )2 ] 16

= 1.71.

In addition, the standard error of the mean (SEM) is

[9] SEM = population SD

√n

where the SEM is a measure of the uncertainty in the value of the mean (2, 3). Equations

2, 3, 7, & 9 are valid for any random variable, including continuous random variables (2,

3). Notice that

[10]limn→∞

SEM=0

that is, as the sample size increases, the SEM converges to zero. The SEM has a higher

frequency of use than the SD, since the SEM is less than the SD. Nonetheless, the SD is

preferred as it provides a better indication of the variation in the data than the SEM (3).

Figure 1 shows the results for this laboratory exercise from a student group.

Notice that the experimental values of the mean, SD, and SEM converges to its

theoretical values. The chi-square “Goodness of fit” test (2, 3, 8) could be used to

3

examine if the data was described by the uniform probability function.

1 2 3 4 5 60

2

4

6

8

10

12

14

value on the face of the die

freq

uenc

yA.

0 10 20 30 40 50 600

1

2

3

4

sample size

valu

e

B.

SD

SEM

mean

4

Figure 1. A. The histogram of the number of dots on the face of a die. B. The sample

mean, sample SD, and sample SEM as a function of sample size are shown.

The Q-test is used to identify an outlier – either the smallest or largest data in the

data set (5). The data is deleted if the Q-statistic

[11] Q = value of the suspect data - value of nearest data > Qc.

value of largest data - value of smallest data

Table 1 contains values of Qc. The Q-test is used to detect an outlier and should be used

to delete only a single entry.

Table 1. Q-test values for 90% confidence.

N (sample size) Qc

345678910

0.940.760.640.560.510.470.440.41

For example, is the value of 25 an outlier in the following data set ?

10, 11, 13, 14, 25.

In this example,

Q= 25−14 25 −10 =

0.73 > Qc = 0.64

thus, the value of 25 is an outlier and may be deleted in subsequent data analysis.

Regression analysis is used in “curve fitting” (2, 5, 6), a method to determine the

parameters in an equation that describes the data. The method is applicable to nonlinear

equations, but for simplicity, we shall demonstrate its use with the equation

5

[12] y = mx.

The basis of regression analysis using the method of least-squares is to minimize

the square of the difference between the actual experimental value and its predicted value

based on the best-fit line. That is, our goal concerning equation 12 is to find the value of

m, where

[13] R = Σ (experimental value - predicted value)2

= Σ (yi - m xi)2

= Σ (yi2 – 2 m xi yi + m2 xi

2)

is a minimum, which would occur at

[14] ∂R∂m = - 2 Σ xi yi + 2 m Σ xi

2 = 0

since∂2R

∂m2(m) > 0

thus

[15] m =

Σ xi yi

Σ xi2

where there is a single minimum. In cases with multiple minima, chose the absolute

minimum. When the equation has more parameters, simply evaluate the derivatives with

respect to those parameters, then solve the resulting system of equations. Both graphing

calculators and MS Excel have build-in capabilities to evaluate a number of functions,

thus the students may not need to calculate the values of the parameters in the function –

they only need to understand its basis. The use of a weighted least-squares regression

analysis (4, 7) is beyond the scope of this supplement.

Table 2 provides an example of using regression analysis that involves the

calibration curve in a Beer’s Law laboratory exercise.

6

Table 2. Calibration curve for Cu2+ at λ = 660 nm.

[Cu2+] (mM) Absorbance250 1.308125 0.68662.5 0.35931.25 0.186

The concentration of an unknown solution of Cu2+ can be calculated using the Beer-

Lambert Law

[16] Absorbance = [molar absorptivity * path length] * concentration

= constant * concentration.

The value of the constant, based on the data in table 2 and equations 15 and 16 is

[17] constant =

∑ concentration∗absorbance

∑ concentration2

=

0. 25∗1. 308 + 0. 125∗0 .686 + 0 . 0625∗0 .359 + 0 .03125∗0 . 1860. 252 + 0 . 1252 + 0 . 06252 + 0. 031252

= 5.313 M-1

thus equation 16 becomes

Absorbance = 5.313 M-1 [Cu2+].

For example, if an unknown Cu2+ solution has an absorbance = 1.0, then

[Cu2+] =

absorbancecons tan t = 1 . 0

5 . 313 M−1=0. 190 M

p-value A statistical test determines the p-value (P), the probability of error (2, 3,

8), which decides if there is a “statistically significant difference”. By convention, there

is no difference when P ≥ 0.05, while there is a difference when P < 0.05.

7

There are two types of errors (see table 3) in a statistical test. The subsequent p-

values in this supplement refers to the type I error. The evaluation of the type II error is

beyond the scope of this supplement.

Table 3. Types of errors in a statistical test.

accept the null hypothesis reject the null hypothesis

null hypothesis is valid Correct decision Type I error

null hypothesis is invalid Type II error Correct decision

The p-value depends on three factors. First, the value of the test statistic, which is

a continuous random variable that provides an index of the “difference” in the data that is

been compared in the statistical test. Second, P is equal to the definite integral of the pdf,

thus P depends on the nature of the pdf. Third, the limits of integration depend upon the

null (Ho) and alternative (H1) hypothesis.

For the Student-t pdf [equation 22 & figure 2a], there are two types of p-values

corresponding to a type I error, which depends on Ho and H1 in the statistical test, and are

known as a 1-sided or 2-sided p-value. The 1-sided (or 1-tail) p-value refers to the case,

where

[18] Ho: X1 ≥ X2 H1: X1 < X2 and P = ∫−∞

− x

f ( x ) dx.

In this situation, X1 is the mean of sample 1, X2 is the mean of sample 2, x is the test

statistic and f(x) is its pdf. Another 1-sided p-value refers to the case, where

8

[19] Ho: X1 ≤ X2 H1: X1 > X2 and P = ∫x

∞

f ( x ) dx.

The 2-sided (or 2-tail) p-value refers to the case, where

Ho: X1 = X2 H1: X1 ≠ X2

which could be interpreted as:

Ho: X1 = X2 H1: X1 < X2 or X1 > X2

[20] and P =∫−∞

− x

f ( x ) dx + ∫x

∞

f ( x ) dx.

Notice that if the pdf is symmetric about x = 0, ∫−∞

− x

f ( x ) dx = ∫x

∞

f ( x ) dx, and P (2-tail)

= 2 P (1-tail) for the same value of the test statistic, e.g. see tables 4 and 5.

To further clarify the difference between the 1- and 2- tail p-values, examine

figure 2a, which shows an example of the pdf for the Student-t distribution. While

changes in the number of degrees of freedom (df; refers to the sample size) changes the

pdf, its shape remains the same. The area under the curve represents the probability. For

P < 0.05, the 1-tail case [equation 19] is represented by the area under the curve in figure

2a from t = 1.8 to infinity, while the 2-tail case [equation 20] is represented by the sum of

the area under the curve in figure 2a from t = 2.2 to infinity and t = - infinity to -2.2 .

That is, for a given P (i.e. area under the curve), the value of the test statistic in the 1-tail

case is less than the 2-tail case, thus it is easier to detect a statistically significant

difference using the 1-tail p-value than the 2-tail p-value.

9

-6 -4 -2 0 2 4 60

0.1

0.2

0.3

0.4

0.5

t

prob

abili

ty

1-tail

2-tail 2-tail

A.

0 1 2 3 4 5 60

0.2

0.4

0.6

F

prob

abili

ty

B.

Figure 2. An example of the probability density function for the Student-t

distribution (A; df = 10) and F-distribution (B; df1 = df2 = 6).

t-test. There are two types of t-tests (2, 3, 8). The 2-sample t-test (also known as

an independent sample t-test or unpaired-sample t-test) compares the means of two

independent samples, while the 1-sample t-test (also known as a correlated sample t-test

or paired-sample t-test) usually determines if the difference between a paired set of data

or the mean of a single sample is zero. Data from independent samples must be analyzed

using the 2-sample t-test, while data from a paired sample should be analyzed by the 1-

10

sample t-test, but may be analyzed by the 2-sample t-test. The 1-sample t-test is more

likely to detect a difference than the 2-sample t-test.

The test statistic in the 2-sample t-test is

[21a]t = difference between the means

var iance in data

=

X1 − X2

S p√ 1n1+ 1

n2

where Sp2 = (n1 – 1) SD1

2 + (n 2-1) SD22

n1 + n2 - 2

and n1 and n2 are the sample size of group 1 and 2, respectively. If n1 = n2 = n, then

[21b]t =

(X 1 − X2) √n

√ SD12 + SD 2

2

.

The pdf of the test statistic is the Student-t distribution (shown in figure 2a)

[22] f (t, ν) =

Γ( ν +12 )

Γ ( ν2 ) √πν

(1+ t 2

ν)− ν+ 1

2

for -∞ < t < ∞ with ν degrees of freedom, where ν = n1 + n2 - 2

where the gamma function is

[23]

Γ ( x ) = ∫0

∞

y x−1 e− y dy = (x-1) ! (for x = a positive integer)

and

[24] Γ (m + ½) = Γ (2m+1) 2 -2m √π (for m = integer).Γ(m+1)

The probability of making a type I error, the probability of rejecting the null hypothesis

when it is true is

11

P(1−tail )=∫−∞

−t

f ( t , ν ) dt=∫t

∞

f ( t , ν ) dt

where ∫0

∞

f ( t , ν ) dt=∫0

t

f ( t , ν ) dt+∫t

∞

f ( t , ν ) dt = 0 .5 (see figure 2a to visualize

this relationship) thus,

[25]P(1−tail ) =∫

t

∞

f ( t , ν ) dt

= 0 .5 −∫0

t

f ( t , ν ) dt

=0. 5−Γ ( ν+1

2 )

Γ ( ν2 ) √πν∫0

t

(1+ t 2

ν)− ν +1

2 dt

and

P(2−tail )=1−2 Γ ( ν+1

2 )

Γ ( ν2 ) √πν∫0

t

(1+ t 2

ν)− ν +1

2 dt.

Notice that both equations 23 and 24 were used to evaluate equation 25.

Table 4 provides an example of an experiment using the 2-sample t-test. The

composition of pre- and post- 1982 pennies are different, thus their densities may differ.

As P < 0.05 (for either the 1-tail or 2-tail p-value), there is a statistically significant

difference between the (average) density of pre- versus post- 1982 pennies.

12

Table 4. The density of pre- and post- 1982 pennies. The 2-sample t-test was

used to determine the p-value: P (1-tail) = 0.000383; P (2-tail) = 0.000765.

Density (g / mL) of pre-1982 pennies

Density (g / mL) of post-1982 pennies

9.329.699.258.86

6.597.577.56.29

X ± SD 9.28 ± 0.34 6.99 ± 0.65

The test statistic [equation 21b] using the data in table 4 is

t =

(9 .28−6 . 99)√4

√0. 342+0. 652 =6 . 24

and the p-value [equation 25] was determined by using numerical methods (6) to evaluate

the definite integral. A description of numerical methods is beyond the scope of this

supplement.

The test statistic in the 1-sample t-test is

[26] t =

differenceSEM

= difference √nSD

or sample mean √nsample SD

which is also described by the Student-t distribution and used to determine its p-value

[equations 22 - 25].

The 1-sample t-test was used to analyze the data in table 5. In this experiment,

the experimentally determined value of the change in enthalpy to dissolve solid sodium

hydroxide in water was compared to its theoretical value. As P < 0.05 (for either the 1-

tail or 2-tail case), there is a statistically significant difference between the experimental

and theoretical value.

13

Table 5. The experimental determination of the change in enthalpy to dissolve solid

sodium hydroxide in water. The 1-sample t-test was used to determine the p-value: P(1-

tail) = 0.0058; P(2-tail) = 0.012.

Experimental value Theoretical value difference- 37.3 kJ / mole- 39.5- 38.8

- 44.5 kJ / mol 7.2 kJ / mole5.05.7

X ± SD - 38.5 ± 1.1 5.97 ± 1.12

The test statistic [equation 26] using the data in table 5 is

t = 5. 97 √3

1. 12=9 . 23

and the p-value [equation 25] was determined by using numerical methods (6) to evaluate

the definite integral.

In the 2-sample t-test [equation 21] or 1-sample t-test [equation 26], the value of

the test statistic may increase due to an increase in the difference between the means, an

increase in the sample size, or a decrease in the SD. An increase in the value of the test

statistic produces a decrease in P [equation 25 & see figure 2a], thus an increase in the

difference, an increase in the sample size, or a decrease in the SD raises the likelihood of

detecting a statistically significant effect using either the 1- or 2- sample t-test. As there

is limited control in determining the difference or SD in the data, the method under the

investigator’s control to detect a statistically significant effect is to increase the sample

size.

F-test was used to compare the variance of two groups of data (2, 3, 8). The test

statistic is

[27]F =

SD12

SD2

2

14

where SD12>SD 2

2 with n1 and n2 degrees of freedom, respectively and its pdf, the F-

distribution, is

[28] g(F, n1, n2) =

Γ(n1+n2

2 )

Γ ( n1

2 ) Γ ( n2

2 )(n1

n2

)n1

2 Fn1

2 −1(1+ n1

n2

F )−

n1+ n2

2

for F > 0;

otherwise, g (F) = 0. The pdf of the F-statistic is shown in figure 2b. The P-value is

[29]

P=Γ (

n 1+n 2

2 )

Γ ( n1

2 ) Γ ( n 2

2 )(

n1

n2

)n1

2∫F

∞

Fn1

2 −1(1+ n1

n2

F )−

n1+ n2

2 dF

=1 −

Γ (n1+ n2

2 )

Γ ( n1

2 ) Γ ( n2

2 )(

n1

n2

)n1

2∫0

F

Fn1

2−1(1+

n1

n2

F )−

n1+n2

2 dF(see figure

2b to visualize this relationship)

where equations 23 and 24 were used to evaluate the gamma functions. Notice that as the

difference between the SD increases, the value of F increases, thus the value of P

decreases, which raises the likelihood that the SD are different.

Analysis of variance (ANOVA) refers to a group of statistical tests that differ in

the number of factors that may affect the outcome of the experiment (2, 3, 4, 8). One-

factor ANOVA (1-ANOVA) is used to examine one factor that may affect the outcome of

an experiment, while two-factor ANOVA (2-ANOVA) examines two factors that may

affect the outcome of an experiment. A factor has many levels, e.g. the concentration of

a drug is a factor, while the different concentrations of the drug refer to the levels of this

factor. Higher-order ANOVA (examines three or more factors that may affect the

15

outcome of an experiment) or repeated measures ANOVA (equivalent to multiple

correlated sample t-tests) are beyond the scope of this supplement.

1-ANOVA was used to compare the mean of three or more levels (or

experimental conditions) in a pair-wise manner. While multiple independent sample t-

tests could achieve the same result, this method increases the probability of error (3, 4).

1-ANOVA achieves these multiple pair-wise comparisons without incurring an increase

in P. 1-ANOVA detects, but does not identify, the presence of pair(s) of data that are

different. The specific pair(s) of data that are different are identified by a post-ANOVA

multiple comparison test (3, 4, 8), e.g. the Tukey test.

In 1-ANOVA, the test statistic is

[30]F=MS(Tr )

MSE

where the MSE is “mean sum error” and MS(Tr) is “mean sum treatment”. The basis of

1-ANOVA is as follows. The determination of the MSE and MS(Tr) are different

methods to estimate the variance. If there was a treatment effect, then the mean among

the various experimental conditions are not from the same population (i.e. they are

different), thus using the SEM to estimate the population variance (basis of MS(Tr) as an

estimate of the variance) is invalid. An assumption in 1-ANOVA is that all experimental

conditions, irrespective of a treatment effect, have the same variance, thus the MSE,

which is the average variance of all groups, is a valid estimate of the variance. The F-test

was used to compare these two estimates of the variance. If the MSE and MS(Tr) are the

same, then there was no treatment effect. If the MSE and MS(Tr) are different, then there

was a treatment effect.

The MSE is

16

[31a]MSE=

∑i=1

n1

( x1 i−X1 )2 +∑

i=1

n2

( x2 i−X 2 )2+ . . . +∑

i=1

nk

(xki−X k )2

nt − k

where X k is the mean of the kth group, xki is the ith sample in the kth group, nk = number

of observations in the kth group, k = number of experimental conditions, and nt is the total

number of observations = n1 + n2 + … + nk. Equation 31a is simplified by using

equations 3 & 7 to obtain

[31b]MSE=

(n1−1 ) var iance1 + (n2−1 ) var iance2 + . . .+ (nk−1) var iancek

nt − k .

For n1 = n2 = … = nk = n, equation 31b becomes

[31c]MSE=

(n−1 ) [var iance1 + var iance2 +.. .+ var iancek ]

kn − k

=var iance1 + var iance2 +.. .+ var iancek

k .

Depending if the groups have the same sample size, the MSE is the average variance

[equation 31c] or the weighted average variance [equation 31b] based on the variance of

each group.

The MS(Tr) is

[32a]MS (Tr )=

∑i=1

k

n i(X i−XT )2

k − 1

where XT = mean of all observations. For n1 = n2 = ... = nk = n, equation 32a becomes

[32b]MS (Tr )=n [∑i=1

k( Xi−X T )

2

k − 1 ].

17

The SEM is the SD of a sample of means (SDmeans) from the same population and using

equation 3,

[32c]SEM=SDmeans=√∑

i=1

k( Xi−X T )

2

k − 1 .

Squaring equation 9 and solving for SD2

[32d] population SD2 = n SEM2

then substituting equation 32c into equation 32d, followed by using equation 7

[32e] variance = n [∑i=1

k(X i−XT )

2

k − 1 ]

refers to the variance of all observations. Substituting equation 32b into equation 32e,

[32f] MS(Tr) = variance = n SEM2 = n [∑i=1

k(X i−XT )

2

k − 1 ]which shows that the MS(Tr) is an estimate of the variance based on the SEM. In

contrast, when the sample size is different, the relationship between the MS(Tr) and SEM

is less clear.

Table 6 contains an example of an experiment, where 1-ANOVA and Tukey’s test

were used in its analysis. This experiment compares the melting points of various

organic compounds. As P < 0.05, there was at least one statistically significant difference

among a pair of samples. The subsequent use of the Tukey test identifies the differences

to be between samples B versus C and between samples A versus C.

18

Table 6. The melting points of 3 samples. 1-ANOVA based p-value = 0.00015 and

Tukey’s test: P > 0.05 for A versus B, while P < 0.01 for A versus C and B versus C.

Sample A Sample B Sample C

Experimental data (melting point; ºC)

808083

817277

555852

X ± SD 81 ± 1.7 77 ± 4.5 55 ± 3

Using equation 31c

MSE= (1.72+4 . 52+32 )

3=10 .71

while using equation 2

mean = 80 + 80 + 83 + 81 + 72 + 77 + 55 + 58 + 529 =71

and using equation [32f],

MS(Tr) = n [∑i=1

k(X i−XT )

2

k − 1 ]=3 ∗ (81−71)2+ (77−71)2+ (55−71 )2

2=588

.

The test statistic [equation 30] is

F = var iance based on the SEMaverage var iance of groups=

MS (Tr )MSE = 588

10 . 71=54 .9

which was used to calculate the p-value [equation 29] by using numerical methods (6) to

evaluate the definite integral.

The Tukey test has a q-statistic (4, 8)

19

q=X big − X small

√ MSE2 ( 1

nbig+ 1

nsmall)

with k and nT-k degrees of freedom, where nT= total number of observations. Unlike the

preceding test statistics, we do not know the pdf of the q-statistic, but values of the q-

statistic for a given P and degrees of freedom are found in tables (4) or using a web-based

calculator (8). A significant difference between a pair of means occurs when (8)

[33]X big − X small ≥ q √ MSE

2( 1

nbig

+ 1nsmall

).

For example, to compare sample A versus B in table 6, using q (k=3, nT-k= 6; P = 0.05)

and equation 33

81 – 77 ≥ 4.34 √ 10 . 712 ( 1

3 +13 ) ?

4 ≥ 8.2

is false, thus there is no significant difference between samples A and B. On the other

hand, to compare samples A versus C using q (k=3, nT-k= 6, P = 0.01) and equation 33

81 – 55 ≥ 6.32 √ 10 . 712 ( 1

3+13 ) ?

26 ≥ 11.9

is true, thus there is a significant difference between samples A and C.

2-ANOVA (2, 4, 8) is similar to 1-ANOVA, but examines the effect of two, rather

than one factor that may affect the outcome of an experiment. In addition, 2-ANOVA

identifies any interaction (4) between the two factors, which means that the effects of the

two factors are not independent, i.e. the effect of a factor depends on the value of the

other factor. A description of multiple comparison tests associated with 2-ANOVA is

beyond the scope of this supplement.

20

The data in table 7 examines the effect of two factors, the [HCl] and the duration

of HCl treatment, on the determination of the % zinc in post-1982 pennies using 2-

ANOVA. The basis of this experiment is that these pennies contain zinc and copper,

where the selective oxidation of zinc by HCl, allows the determination of the % zinc in

these pennies. The statistical analysis shows that there was no interaction between the

effects of [HCl] and the duration of HCl treatment, thereby simplifying our conclusions.

Furthermore, the statistical analysis shows that a higher [HCl] increases the oxidation of

zinc by HCl, but the duration of the HCl treatment did not affect the results.

Table 7. The effects of the concentration of HCl and the duration of the HCl treatment on

the experimentally determined % zinc in post-1982 pennies. 2-ANOVA was used in data

analysis. P(column) = 0.0002; P(row) = 0.38; P (interaction) = 0.60

[HCl] = 3.0 M [HCl] = 6.0 M X ± SD (row)1-day HCl treatment

[X ± SD; n = 5]

4523304118

[31 ± 12 %]

7473827454

[71 ± 10 %]

51 ± 23 %

2-day HCl treatment

[X ± SD; n = 6]

427616541551

[42 ± 24 %]

845454956494

[74 ± 19 %]

58 ± 26 %

X ± SD (column) 37 ± 19 % 73 ± 15 %

The calculation of the test statistic in 2-ANOVA is similar to 1-ANOVA, where

F (row )= MS(Tr : row )MSE and F (column )=MS(Tr : column )

MSE

21

and numerical methods (6) were used to evaluate the definite integral of equation 29 to

obtain the p-value. The subsequent calculations in this section will use the data in table

7. Using equation 2,

Mean = 45 + 23 + 30 + . .. (add all data in table 7 )

22 =55 .14 .

The MS(Tr: row) and MS(Tr: column) terms were calculated in a similar manner as in

the 1-ANOVA case [equation 32a],

MS (Tr )=∑i=1

k

ni(X i−XT )2

k − 1

where k, ni, and Xi correspond to the values in the columns or rows (we did not use

equation 32f, since the number of samples in each experimental condition is different).

The estimates of the population variances are

MS (Tr : rows )=∑i=1

k

ni( Xi−X T )2

k − 1=

10 (51. 4 − 55.14 )2 + 12 (58 . 25 − 55 .14 )2

2 − 1=255 .9

and

MS (Tr : columns)=∑i=1

k

ni( Xi−X T )2

k − 1=

11 (37 .36 − 55 .14 )2 + 11 (72. 91 − 55 .14 )2

2 − 1=6951.

The calculation of the MSE is similar to the 1-ANOVA case, where we shall use equation

31b (we did not use equation 31c, since the number of samples in each experimental

condition was different)

MSE=(n1−1 ) var iance1 + (n2−1 ) var iance2 + . . .+ (nk−1) var iancek

nt − k

=

4 (122) + 4 (102) + 5 (24 2) + 5 (192)

22 − 4=314 .5

.

Substituting these values into the test statistic

22

F (row )= MS(Tr : row )MSE = 255. 9

314 . 5=0 . 81

and

F (column )=MS(Tr : column )MSE = 6951

314.5=22 .1

which were used to calculate the p-value. The calculation of F (interaction) is beyond

the scope of this supplement.

The statistical tests in this supplement are parametric tests, where the data can be

described by the normal pdf. Nonparametric tests (2, 3, 8) do not have this requirement

and have less assumptions on the nature of the data than parametric tests. Nonparametric

tests were not used in my Chemistry classes; because, I did not wish to overwhelm my

students with statistics. The reasons for using parametric rather than nonparametric

statistical tests were due to the robustness and greater power of the parametric tests (3, 4).

Further arguments concerning the use of a parametric versus nonparametric test or a

description of nonparametric tests are beyond the scope of this supplement. Ultimately, it

is a matter of opinion in the use of a parametric versus nonparametric test.

Table 8 is a summary of the situations to use the various statistical tests described

in this supplement.

23

Table 8. Comparison of various statistical tests.

24

Statistical test Compare SD

Compare mean

# groups Ho & H1

F-test √ 2 Ho: SD1=SD2

H1: SD1≠SD2

1-tail, 1-sample t-test √ 1 or 2 Ho: ∆ x ≤ 0H1: ∆ x > 0

orHo: ∆ x ≥ 0H1: ∆ x < 0

2-tail, 1-sample t-test √ 1 or 2 Ho: ∆ x = 0H1: ∆ x ≠ 0

1-tail, 2-sample t-test √ 2 Ho: X1 ≤ X2

H1: X1 > X2

or

Ho: X1 ≥ X2

H1: X1 < X2

2-tail, 2-sample t-test √ 2 Ho: X1 = X2

H1: X1 ≠ X2

1-anova √ 3 or more;1 factor Ho: X1 = X2

H1: X1 ≠ X2

2-anova √ 3 or more;2 factors Ho: X1 = X2

H1: X1 ≠ X2

References

1. http://faculty.vassar.edu/lowry/VassarStats.html ;

http://members.aol.com/johnp71/javastat.html (accessed August 2004)

2. Freund, J.E.; Walpole, R.E. Mathematical Statistics, 4th ed.; Prentice Hall, NJ

1987

3. Gantz, S.A. Primer of Biostatistics, 4th ed. McGraw-Hill, NJ 1997.

4. Kleinbaum, D.J.; Kupper, L.L.; Muller, K.E. Applied Regression Analysis and

Other Multivariable Methods. 2nd ed.; PWS-Kent; MA. 1988.

5. Shoemaker, D.P.; Garland, C.W.; Steinfeld, J.I. Experiments in Physical

Chemistry. 3rd ed. McGraw Hill, NY 1974.

6. Cheney, W.; Kincaid, D. Numerical Mathematics and Computing, 2nd ed. Brooks

/ Cole Publ., CA 1985.

7. deLevie, R. J. Chem. Educ. 1986, 63, 10 – 15.

8. http://faculty.vassar.edu/lowry/webtext.html (accessed August 2004)

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