Statistics For Analytical Chemistry 3nt16

Training Programme “Statistics for Analytical Chemistry”

PREPARED BY

MOHAMED SALAMA

Training Programme “Statistics for Analytical Chemistry”

PART I

INTRODUCTION

What is Statistics? Statistics is the science of  collecting,  describing, and  interpreting data to make  predictions, and  decisions

A statistical Analysis …includes  describing the problem  gathering data  summarizing data  Analyzing the data and communicating meaningful conclusions

Statistics in chemical analysis Lab Chemists are concerned with the chemical analysis processes that quantify analytes in different matrices. All these processes are subject to • systematic variation (e.g. Instrument effects, matrix effect) • random variation (e.g. measurement errors) Statistics is a tool to help us understand the effects of random variation

Areas of statistics DESCRIPTIVE STATISTICS Methods for summarizing and describing a given set of data. E.g. using

 tables  graphs  numerical summaries

Areas of statistics II INFERENTIAL STATISTICS Methods for drawing conclusions about a population based on a given data set from that population

Population & Sample POPULATION The collection, or set, of ALL individuals, items, or events of interest.

SAMPLE The subset of the population for which we have data available.

Example Suppose we are investigating the daily water consumption of students at Cairo University 50 students are selected at random and interviewed.  The POPULATION is the set of all students.  The SAMPLE is the 50 students randomly selected.

Population & Sample

Population Use parameters to summarize features

Sample Use statistics to summarize features

Inference on the population from the sample

Probability and statistics PROBABILITY:  Usually assumes knowledge of the population, i.e  distributions are known.  The theory behind statistics. PRINCIPLE OF INFERENTIAL STATISTICS: Only have a sample Use this to infer details about the population.

Variable A characteristic of interest about each individual element of a population or sample. In the water consumption study, a example of a variable is the consumption per week in units e.g. litres. Other examples for a group of students could be:  age, weight, height  etc

Parameter  A piece of numerical information about a population.  We use statistical inference to estimate the parameter from the given sample. Example: Suppose we want to find the average height of a lab staff. The POPULATION is the set of all lab staff. The average height is a PARAMETER of the population. The QC section is an example of a SAMPLE.

Statistic A statistic is number calculated from a sample. Typical examples we will learn about: • the average (or mean) of a sample of measurements • the standard deviation

Types of data Qualitative (non numerical data)  NOMINAL – Names that describe categories, but no order. Example: eye colour, gender  ORDINAL – Ordered categories. Example: year groups (1st year, 2nd year, 3rd year etc.)

Types of data II Quantitative (numerical data)  DISCRETE – number of possible values can be counted E.g. number of children in a family, marks out of 50 in an exam.  CONTINUOUS – possible values are in a continuum E.g. measurements in height, weight and volume

Data collection and sampling techniques Surveys and experiments should use methods to select a sample that is “representative” of the population of interest. Possible methods to obtain “representative” samples  deterministic sampling.  random sampling.

Data collection and sampling techniques Deterministic Sample Elements are selected on the basis some algorithmic approach, for example, selecting every 10th member of the population. Danger of biased results. Random Sample Each member of the population is selected with a certain probability.

So that ! Statistics is a valuable tool in all sciences. However, making proper use of statistics to • select data samples • analyse these samples • draw conclusions is not a simple task.

Why Statistics? 1. Existence of errors

  

Gross Random Systematic

2. Small number of measurements

 

Sample Population

Definitions (ISO 5725-1:1994) 1. Accuracy 2. Trueness 3. Bias 4. Laboratory bias (Total bias) 5. Bias of measurement method 6. Laboratory component of bias 7. Precision 8. Repeatability (Conditions & Limit) 9. Reproducibility (Conditions & Limit)

Definitions 1. Accuracy The closeness of agreement between a test result and the accepted reference value 2. Trueness The closeness of agreement between the average value obtained from large series of test results and an accepted reverence value

Definitions 3. Bias The difference between the expectation of the test results and an accepted reference value 4. Laboratory bias (Total bias) The difference between the expectation of the test results from a particular lab and an accepted reference value

Definitions 5. Bias of the measurement method The difference between the expectation of test results obtained from all laboratories using that method and an accepted reference value 6. Laboratory component of bias The difference between the lab bias and the bias of the measurement method 7. Precision The closeness of agreement between independent test results obtained under stipulated conditions

Definitions

Definitions

value of param eter

value of param eter

(a) N o bias, high precision

(b) N o bias, low precision

value of param eter

value of param eter

(c) B iased, high precision

(d) B iased, low precision

F igure 7.4.1 B ias and precision. From C hance Encounters by C .J. W ild and G .A .F. Seb er, © John W iley & S ons, 2000.

Definitions 8. Repeatability Precision under repeatability conditions 9. Repeatability conditions Conditions where independent test results are obtained with the same operator on identical test items in the same lab by the same operator using the same equipment within short intervals of time.

Definitions 10. Reproducibility Precision under reproducibility conditions 11. Reproducibility conditions Conditions where test results are obtained with the same method on identical test items in different laboratories with different operators using different equipment

Statistical Model 1. Basic Model (ISO 5725-1:1994) Test Result (Y) = m + B + e



M is the general mean (expectation);



B is the laboratory component of bias under repeatability conditions (between-lab variation - Var (B) = σL2);



e is the random error occurring in every measurement under repeatability conditions (within-lab variation - Var (e) = σr2).

Statistical Model 5. Basic Statistical Model & Precision Repeatability Standard Deviation = σr = √ Var (e) Reproducibility Standard Deviation = σR = √ σL2 + σr2

i.e. Repeatability variance is estimated directly from Random error while reproducibility variance depends on the sum of both within- & between- lab variance

Conclusions Accuracy Trueness

Bias

Precision

Under Repeatability Conditions r

Within-lab variation

Under Reproducibility Conditions R

Within-lab variation

+

Between-lab variation

Example Problems

Exercises

Training Programme “Statistics for Analytical Chemistry”

PART II

FUNDAMENTALS OF STATISTICS

Statistical Fundamentals 1. Measures of Central Tendency 1.1 Average n

 xi X=

i=1

n

Statistical Fundamentals 1. Measures of Central Tendency 1.2 Median (Md) which is simply the middle value of the sample when the measurements are arranged in numerical order. (If n is even, then the median is the average of the two middle values of the ordered sample)

Statistical Fundamentals 1. Measures of Central Tendency 1.3 Mode The mode is defined as the most frequent value in a frequency distribution.

Statistical Fundamentals 2. Measures of Dispersion 2.1 Range (R) 2.2 Standard Deviation (s)

2.3 Coefficient of Variation (CV)

Statistical Fundamentals 2. Measures of Dispersion 2.4 Variance (s2) 2.5 Pooled Standard Deviation (sp)

Example Problems

Exercises

Training Programme “Statistics for Analytical Chemistry”

PART III

ERRORS IN CLASSICAL ANALYSIS

Training Programme “Statistics for Analytical Chemistry”

1. DISTRIBUTION OF ERRORS

Normal Distribution Continuous random variable Values from interval of numbers Absence of gaps

Continuous probability distribution Distribution of continuous random variable

Most important continuous probability distribution The normal distribution

Normal Distribution “Bell shaped” Symmetrical Mean, median and mode are equal Interquartile range equals 1.33  Random variable has infinite range

f(X)

 Mean Median Mode

X

Normal Distribution Is symmetric about the mean  Mean = Median 

Figure 6.2.2

50% 50% Mean

Normal Distribution f X  

1



e

1 2

2 X     

2 2 f  X  : density of random variable X

  3.14159; e  2.71828  : population mean  : population standard deviation X : value of random variable    X   

Normal Distribution (a) Changing

(b) Increasing

shifts the curve alongthe axis

increases the spreadandflattens the curve 1 =6

1=

2 =6 2 =12

140

160 1 =160

180 2=174

200

140

160

180 1 = 2=170

200

Normal Distribution (c) Probabilities and numbers of standard deviations Shaded area = 0.683





68% chance of falling between  and 

Shaded area = 0.954





95% chance of falling between   and 

Shaded area = 0.997





99.7% chance of falling between   and 

Normal Distribution

Normal Distribution Probability is the area under the curve!

P c  X  d   ?

f(X)

c

d

X

Normal Distribution

An infinite number of normal distributions means an infinite number of tables to look up!

Normal Distribution Cumulative Standardized Normal Distribution Table (Portion)

Z  0 Z

.00

.01

Z 1

.02 .5478

0.0 .5000 .5040 .5080

Shaded Area Exaggerated

0.1 .5398 .5438 .5478 0.2 .5793 .5832 .5871 Probabilities

0.3 .6179 .6217 .6255

0 Z = 0.12

Only One Table is Needed

Normal Distribution Z

X 



6.2  5   0.12 10 Standardized Normal Distribution

Normal Distribution

  10

 5

Z 1

6.2

X

Z  0

0.12

Z

Example Problems

Example Problems

Exercises

Training Programme “Statistics for Analytical Chemistry”

2. CONFIDENCE INTERVALS

Confidence Intervals  The problem  How large are the error bounds when we use data from a sample to estimate parameters of the underlying population?  Compute confidence intervals for  when  2 is known  when  2 is unknown



Confidence Intervals Suppose an estimate e.g. an estimate for the mean x is given

We want to describe the precision of the estimate

We do this by giving a range of likely values for the parameter. Such a range is called confidence interval.

Confidence Intervals Population Mean, , is unknown Sample

Random Sample Mean X = 50

I am 95% confident that  is between 40 & 60.

Confidence Intervals Provides 

Range of Values

Based on Observations from 1 Sample

Gives Information about Unknown Population Parameter Stated

Closeness

in of Probability

Never 100% Sure

to

Elements of CI Estimation A Probability That the Population Parameter Falls Somewhere Within the Interval. Confidence Interval

Confidence Limit (Lower)

Sample Statistic

Confidence Limit (Upper)

Confidence Limits for Mean Parameter = Statistic ± Its Error

  X  Error X   = Error =   X

Z 

X  



X



Error



Error  Z 

  X  Z X

X

x

Confidence Intervals

 X  Z  X  X  Z  n

x_ _ X

  1.645 x   1.645 x 90% Samples

  1.96 x

  1.96 x

95% Samples

  2.58 x

  2.58 x 99% Samples

Intervals & Level of Confidence 



Probability that the unknown population parameter is in the confidence interval in 100 trials. Denoted (1 - ) % = level of

confidence

90%, 95%, 99% 

Is Probability That the Parameter Is Not Within the Interval in 100 trials (NOT THIS TRIAL ALONE!)

e.g.

Intervals & Level of Confidence Sampling Distribution of the Mean /2 Intervals Extend from

_

x

1 -

X  

X  Z X to

X  Z X

/2

_ X (1 - ) % of Intervals Contain . % Do Not.

Confidence Intervals

Factors Affect Interval Width 

Data Variation



measured by 

Intervals Extend from



Sample Size

X - Z



 XofConfidence X / n Level (1 - )

x

to X + Z 

x

Confidence Intervals CONCLUDING REMARK As smaller we choose  as ‘more confident’ we get that the interval contains the parameter  . But at the same time the confidence interval gets wider and is therefore less precise.

CI (σ Known – Hardly True)  Assumptions   

Population Standard Deviation Is Known Population Is Normally Distributed If Not Normal, use large samples

 Confidence

Interval Estimate

X  Z / 2 

 n

   X  Z / 2 

 n

CI (σ Unknown) 

Assumptions  

Population Standard Deviation Is Unknown Sample size must be large enough for central limit theorem or Population Must Be Normally Distributed



Use Student’s t Distribution



Confidence Interval Estimate

S X  t / 2,n 1  n



S X  t / 2,n1  n

Student’s t Distribution Standard Normal Bell-Shaped Symmetric ‘Fatter’ Tails

t (df = 13) t (df = 5)

Z

0

t

Degrees of Freedom (df) 



Number of Observations that are free to vary after sample Mean has been calculated Example 

Mean of 3 Numbers Is 2 X1 = 1 (or Any Number) X2 = 2 (or Any Number) X3 = 3 (Cannot Vary) Mean = 2

degrees of freedom = n -1 = 3 -1 =2

Student’s t Table /2 Upper Tail Area df

.25

.10

.05

Assume: n = 3 =n-1=2

df

 = .10 /2 =.05

1 1.000 3.078 6.314

2 0.817 1.886 2.920

.05

3 0.765 1.638 2.353

t Values

0

2.920

t

Example Problems

Example Problems

Exercises

Training Programme “Statistics for Analytical Chemistry”

PART IV

SIGNIFICANCE TESTS

Training Programme “Statistics for Analytical Chemistry”

1. T-TEST

Student’s t Test When solving probability problems for the sample mean, one of the steps was to convert the the sample mean values to z-scores using the following formula:

z

x  x

x

where

x  

and

x 

 n

What happens if we do not know the population standard deviation  ? If we substitute the population standard deviation  with the sample standard deviation, s, can we use the standard normal table? Answer: no.

Student’s t Test This question was addressed in 1908 when W.S. Gosset found that if we replace  with the sample standard deviation s, the distribution becomes a t-distribution. If

x T s/ n then T has a t-distribution with n-1 degrees of freedom. The t-distribution is similar to the z-curve in that it is bell shaped, but the shape of the t-distribution changes with the degrees of freedom. We will use the T-tables to get the critical t-values at different levels of  and degrees of freedom.

Student’s t Test 1.One-sample t-test

x T  s/ n

When using t-test TAKE CARE !!! Is it one-tailed or two-tailed test?

Student’s t Test 2. Independent sample t-test (Equal variances)

Degrees of freedom= (n1+n2) -2

Student’s t Test 3. Independent sample t-test (Unequal variances)

Student’s t Test 3. Independent sample t-test (Unequal variances) In such complicated case Degrees of freedom is calculated from:

Degrees of freedom= (n1+n2) -2

Student’s t Test 4. Paired sample t-test

The sign of the difference is very important

Training Programme “Statistics for Analytical Chemistry”

2. F-TEST

f -Test f - Test is used to compare the standard deviation of two

samples, and to make a test to determine whether the populations from which they come have equal variances

Training Programme “Statistics for Analytical Chemistry”

3. HYPOTHESIS TESTING

Hypothesis Testing A

hypothesis is a claim (assumption) about the population parameter 



Examples of parameters are population mean or proportion The parameter must be identified before analysis

I claim the mean GPA of this class is   3.5!

Hypothesis Testing Assume the population mean age is 50. ( H 0 :   50 )

Identify the Population

Is X  20 likely if    ? No, not likely! REJECT Null Hypothesis

Take a Sample

 X  20 

Hypothesis Testing Sampling Distribution of X It is unlikely that we would get a sample mean of this value ...

... Therefore, we reject the null hypothesis that m = 50. ... if in fact this were the population mean.

20

 = 50 If H0 is true

X

Hypothesis Testing H0: Innocent Jury Trial

Hypothesis Test

The Truth

The Truth

Verdict

Innocent Guilty Decision H0 True H0 False

Innocent

Do Not Reject H0

Guilty

Correct

Error

Error

Correct Reject H0

1-

Type II Error (  )

Type I Error ( )

Power (1 -  )

Hypothesis Testing If you reduce the probability of one error, the other one increases so that everything else is unchanged.

 

Hypothesis Testing 

True value of population parameter 





Increases when



Increases when  increases



Increases when n decreases

 decreases

Population standard deviation 



Increases when the difference between hypothesized parameter and its true value decrease

Significance level 





Sample size 

 

   n

Hypothesis Testing Convert sample statistic (e.g.: X ) to test statistic (e.g.: Z, t or F –statistic)  Obtain critical value(s) for a specified from a table or computer 



If the test statistic falls in the critical region, reject H0  Otherwise do not reject H0 

Hypothesis Testing H0: 0 H1:  > 0

H0: 0 H1:  < 0

Reject H0

Reject H0



 0

Z Must Be Significantly Below 0 to reject H0

Z

0

Small values of Z don’t contradict H0 Don’t Reject H0 !

Z

Example Problems

Example Problems

Example Problems

Example Problems

Example Problems

Exercises

Training Programme “Statistics for Analytical Chemistry”

4. ANALYSIS OF VARIANCE (ANOVA)

Analysis of Variance (ANOVA) The statistical tests described previously are used in the comparison of two sets of data, or to compare a single sample of measurements with a standard or reference value. Frequently, however, it is necessary to compare three or more sets of data, and in that case we can make a use of a very powerful statistical method with a great range of applications – Analysis of Variance (ANOVA).

Analysis of Variance (ANOVA)

If there is only one source of variation apart from this measurement area, a one-way ANOVA calculation is appropriate, if there are two sources of variation we use two-way ANOVA calculations, and so on.

Analysis of Variance (ANOVA) Example: A sample of fruit is analysed for its pesticide content by a liquid chromatographic procedure, but four different extraction procedures A-D (solvent extraction with different solvents, solid-phase extraction etc) are used, the concentration in each case being measured three times. The results (mg kg-1) are indicated in the shown table, Is there any evidence that the four different sample preparation methods yield different results?

Analysis of Variance (ANOVA) Results 1 2 3 Average

A 10.5 11.5 10.7 10.9

B 9.9 10.8 10.8 10.5

Overall Average = 9.9

C 9.9 9.1 8.9 9.3

D 9.2 8.5 9.0 8.9

Analysis of Variance (ANOVA) Solution: The principle underlying a one-way ANOVA calculation is that two separate estimates of So2 are made, one from the within-sample variation, the other from the between-sample variation. If H0 is correct, then these two estimates of So2, which can be compared using the F-test, should not be significantly different.

Analysis of Variance (ANOVA) Solution: In practice simple spreadsheets and other statistical software provide the results very rapidly, but here we demonstrate the underlying principles by going through the calculations step by step. At each stage we use the expression ∑ (xi -x)2/(n – 1) to calculate the variances.

Analysis of Variance (ANOVA) Solution: 1. Within-sample variation: This is readily found by calculating the variance of the three measurements for each of the four methods A-D, and averaging them. The variance for sample A is given by {(10.5 – 10.9)2 + (11.5 – 10.9)2 + (10.7 – 10.9)2}/2 = (0.16 + 0.36 + 0.04)/2 = 0.28.

Analysis of Variance (ANOVA) Solution: 1. Within-sample variation: Similarly the variances for B, C and D are 0.27, 0.28 and 0.13. The average of these four variances, 0.24, is the within-sample estimate of So2 . This estimate has 8 degrees of freedom, i.e. 2 from each of the four samples.

Analysis of Variance (ANOVA) Solution: 2. Between-sample variation: If the variance of the four average values for the samples A-D is determined, it is found to be {(10.9 – 9.9)2 + (10.5 – 9.9)2 + (9.3 – 9.9)2 + (8.9 – 9.9)2}/3 = 2.72/3 = 0.907. However this variance is determined from numbers which are themselves the means of three replicates, so it estimates So2/n rather than So2.

Analysis of Variance (ANOVA) Solution: 2. Between-sample variation: In the present experiment n = 3, so the betweensample estimate of So2 is 0.907  3 = 2.72. This estimate has three degrees of freedom, being derived from 4 sample means.

Analysis of Variance (ANOVA) Solution: 3. F-Test: The two estimates of So2 can be compared using the F-test. ing that F > 1, the experimental value of F is 2.72/0.24 = 11.33. When evaluating this result using statistical tables it is very important to note that the F-test in ANOVA is used as a one-tailed (one-sided) test. This is because the only realistic possibility is that the betweensample variation might be significantly greater than the within sample variation – the opposite is not feasible.

Analysis of Variance (ANOVA) Solution: 3. F-Test: At the P = 0.05 level, the critical one-tailed value of F3,8 is 4.07. The experimental value is much larger than this, so the null hypothesis can be rejected: it is very likely that there are significant differences between the four mean values.

Analysis of Variance (ANOVA) EXCEL PRINTOUT FOR ANOVA EXAMPLE

Training Programme “Statistics for Analytical Chemistry”

5. THE CHI-SQUARED TEST

The CHI-Squared Test  The significance tests so far described in this course, in

general, assume that the data analyzed: 1. Be continuous, interval data comprising a whole population or sampled randomly from a population. 2. Have a normal distribution. 3. Sample size should not differ hugely between the groups. In Contrast, CHI-Squared Test is concerned with frequency i.e. the number of times a given event occurs.

The CHI-Squared Test Example: The numbers of glassware breakages reported by four laboratory workers over a given period are given below. Is there any evidence that the workers differ in their reliability? Number of breakages: 24, 17, 11,9 Solution: Ho : The same number of breakages by each worker (no difference

in reliability assuming that the workers use the lab for an equal length of time)

HA : Different number of breakages by each worker

The CHI-Squared Test The null hypothesis implies that since the total number of breakages is 61, the expected number of breakages per worker is 61 / 4 = 15.25. Obviously it is not possible in practice to have a non integral number of breakages (this number is only a mathematical concept). The question to be answered is whether the difference between the observed and expected frequencies is so large that the null hypothesis should be rejected. The calculation of chi-squared, χ2, the quantity used for test for a significant difference, is shown in the following table:

The CHI-Squared Test Observed Frequency (O)

Expected Frequency (E)

O-E

(O-E)2/E

24

15.25

8.75

5.020

17

15.25

1.75

0.201

11

15.25

- 4.25

1.184

9

15.25

- 6.25

2.561 χ2= ∑ (O-E)2/E = 8.966

The CHI-Squared Test If χ2, (propability, df) exceeds a certain critical value the null hypothesis is rejected.

χ2calculated = 8.966 χ2critical = 7.81 χ2calculated > χ2critical The null hypothesis is rejected at the 5 % significance level i.e. there is evidence that the workers do differ in their reliability

df

χ2, (P = 0.05)

1

3.84

2

5.99

3

7.81

4

9.49

5

11.07

6

12.59

7

14.07

8

15.51

9

16.92

10

18.31

Training Programme “Statistics for Analytical Chemistry”

6. CONCLUSIONS FROM SIGNIFICANCE TESTS

Significance Tests - Conclusions What are the conclusions which may be drawn from significance test? As we have just explained that a significance test at , for example, P = 0.05 level involves a 5 % risk that a null hypothesis will be rejected even though it is true. This type of error is known as a Type 1 error : the risk of such an error can be reduced by altering the significance level of the test to P = 0.01 or even P = 0.001. This, however, is the not only possible type of error: it is also possible to retain a null hypothesis even when it is false. This is called Type 2 error. In order to calculate the probability of type 2 error it is necessary to postulate an alternative to the null hypothesis known as alternative hypothesis.

Significance Tests - Conclusions

Significance Tests - Conclusions Consider the situation where a certain chemical product is meant to contain 3% of phosphorus by weight. It is suspected that this proportion has increased. To test such increase the composition is analyzed by a standard method with known standard deviation of 0.03%. Suppose 4 measurements are made and a significance test is performed at the level of P = 0.05. A one-tailed test is required, as we are interested only in an increase.

Significance Tests - Conclusions Ho , µ = 3.0 % The solid line shows the sampling distribution of the mean if the null hypothesis is true. This sampling distribution has mean 3.0 and s.e.m = 0.03 / √4 % . If the sample mean lies above the indicated critical value , Xc , the null Hypothesis is rejected. Thus the shaded region, with area 0.05, represent the probability of a Type 1 error.

Significance Tests - Conclusions H1 , µ = 3.05 % The broken line shows the sampling distribution of the mean if the alternative hypothesis is true. Even if this is the case, The null hypothesis will be retained if the Sample mean lies below Xc. The probability of this type 2 error is Represented by the hatched area

Significance Tests - Conclusions The diagram makes clear the interdependence of the two types of error. If, for example, the significance level is changed to P = 0.01 in order to reduce a risk of type 1 error , Xc will Be increased and the risk of a type 2 Error is also increased. Conversely, A decrease in the risk of type 2 error Can only be achieved at the expense of an increase in the probability of Type 1 error.

Significance Tests - Conclusions The only way in which both errors can be reduced is by increasing the sample size. The effect of increasing n to 9, for example, is illustrated in the shown figure : the resultant decrease in the standard error of the mean produces a decrease in both types of error for a given value of Xc .

Significance Tests - Conclusions The probability that a false null hypothesis is rejected is known as the POWER of the test. that is, the power of a test is (1 – the probability of a type 2 error). In the studied example the POWER is a function of the mean specified in the alternative hypothesis, and depends on the sample size , the significance level of the test, and whether the test is one- or two- tailed. IF TWO OR MORE TEST ARE AVAILABLE TO TEST THE SAME HYPOTHESIS, IT MAY BE USEFULL TO COMPARE THE POWERS OF THE TESTS TO DECIDE WHICH IS MORE APPROPRIATE

Training Programme “Statistics for Analytical Chemistry”

PART V

OUTLIERS

Outliers – Dixon’s Test (|Suspect Value – nearest value|) Q = (|largest value – smallest value|)

Example The nitrate level (mg l-1) in a sample of river water was measured four times, with the following results: 0.404, 0.400, 0.398, 0.379. Can the last value be rejected as an outlier?

Outliers – Dixon’s Test Solution: Using the mentioned equation Q is given by (0.398-0.379)/(0.404 – 0.379) = 0.019/0.025 = 0.76. The critical value of Q (n = 4, P = 0.05, 2-tailed) is 0.831. Since the experimental value does not exceed the critical Q value , the null hypothesis is retained, i.e. the measurement 0.379 cannot be rejected. n

4

5

6

7

8

9

10

Q (P = 0.05)

0.831

0.717

0.621

0.570

0.524

0.492

0.464