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Statistics For Analytical Chemistry by using MINITAB programby using MINITAB program

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By Mohamed Salama

I t d tiIntroductionGraph in MINITAB:p

A pictorial gallery from which to choose a graph type.Flexibility in customizing graphs from sub setting of Flexibility in customizing graphs, from sub setting of data to specifying titles and footnotes.Ability to change most graph elements, such as fonts, Ability to change most graph elements, such as fonts, symbols, line, placement of tick marks and data display after the graph is creates.Ability to automatically update graphs.

IntroductionIntroductionEnhancements to Specific Graphs:

Multiple levels of categorical variables.Contour plots use color ramps and label contour lines.p pUse summarized data in making a bar chart.Fit regression lines and distributions to selected ggraphs.You can use Empirical CDF (cumulative distribution

)function) graphs to evaluate the fit of a distribution to your data or to compare different sample distributions.

IntroductionIntroductionData, Limits and Details:

A worksheet can contain up to 4000 columns, 1000 constants and up to 10,000,000 rows depending on p , , p ghow much memory your computer has.Three stored constants have default values (you can change them if you wish):

K998 = ∗ (missing), Κ999 = 2.71828 (e) and K1000 = 3 14159 (pi)3.14159 (pi).

IntroductionIntroductionReport Pad:pIn report pad you can,Store MINITAB results and graphs in a single Store MINITAB results and graphs in a single document.Add comments and headings.gRearrange your output.Change font sizes.Print entire output from an analysis.Create Web‐ready reports.

C ti d t t i MINITABCreating a data set using MINITABTo creating a data set you must know the variables and o c eat g a data set you ust o t e a ab es a drecords.

The variables are stored in columnsThe variables are stored in columns.

The individual records go in rows.

Variable names are inserted at the top of column in the row with no numerical designation.

C ti tt d d t lCreating a patterned data columnIf there is more than level for the variable.Make sequence for each level and replications.Know the first value, last value and the increment units.Know the levels for labeling, replicates in each level

d li t f h l l (th h l )and replicates for each level (the whole sequence).You can make all this by manual data entry.

D i l i d i fData manipulation and creation of new variablesnew variables

Data Coding:Variable coding is done by creating a new variable have a name with categories corresponding to the l llevels.Could coding a numerical variable into a categorical variable (Numeric to Text)variable (Numeric to Text).

Data manipulation and creation of new variablesnew variables

Transformation of variables:Arithmetic operation such as addition, multiplication, division, exponentiation, log‐transformation,……..

h f dLogarithmic transformation: sometimes used in statistical analysis for normalizing data or for stabilizing variances.Square root transformation.Others, there are many transformation available under

( )the Functions box (antilog, arccosine, cosine, etc.)

G hi l di l f d tGraphical display of dataHistograms.gPie Chart.Scatter Plots.

All this kind of graphics are often used for data visualization.

Some graphs may help in assessing the shape of distribution of data, whereas other types my help in summarizing data at hand or in describing summarizing data at hand or in describing relationships between variables of interest.

Producing a report outputThere are many different ways to produce a report e e a e a y d e e t ays to p oduce a epo tusing MINITAB project.

Use Microsoft Word, by copy and paste the data which will be reported.will be reported.

Use MINITAB Word.

C l iConclusionMinitab Statistical Software:Minitab Statistical Software:

Easy to use.

State of the Art, Graphs and Graph Editing.

Regression Analysis.g y

Statistical Process Control.

Measurement Systems Analysis.

Reliability/Survival Analysis.

ConclusionMinitab Statistical Software (continue):Minitab Statistical Software (continue):

Multivariate Analysis.

Nonparametric.

Simulations and Distributions.

Data and File Management.

General statistics.

Analysis of variance.

C l iConclusionMinitab Statistical Software (continue):tab Stat st ca So t a e (co t ue):

Quality Tools.

Design of experiments.

Power and sample size.

Time Series and Forecasting.

T blTables.

Macros and Customizability.

Statistics IntroductionStatistics IntroductionStatistics is the science ofStatistics is the science of…….

Collecting, Describing and Interpreting data

To make……

Predictions and Decision.

Includes……

Describing the problem gathering data summarizing Describing the problem, gathering data, summarizing data, analyzing data and communicating meaningful conclusions.

Statistics IntroductionStatistics IntroductionLab chemist are concerned with the chemical analysis ab c e st a e co ce ed t t e c e ca a a ys sprocesses that quantify analytes in different matrices.

All these processes are subject to:All these processes are subject to:

Systematic variation. (e.g. Instrument effects, matrix effect).effect).

Random variation. (e.g. measurement errors)

Statistics is a tool to help us understand the effects of random variation.

Statistics IntroductionStatistics IntroductionProbability:

Usually assumes knowledge of the population.

Distributions are knownDistributions are known.

The theory behind statistics.

Principle of inferential statistics:

Only have a sample.y p

Use this to infer details about the population.

Statistics IntroductionStatistics IntroductionDefinition according to ISO 5725‐1:1994gAccuracy.Trueness.Bias.Laboratory bias (total bias).Bias of measurement method.Laboratory component o bias.P i iPrecisionRepeatability (conditions & limit).Reproducibility (conditions & limit)Reproducibility (conditions & limit).

Statistics IntroductionStatistics Introduction

Accuracy

Trueness PrecisionTrueness Precision

Bias Repeatability Reproducibilityp y p yconditions r conditions R

Withi l b ithi l bWithin lab within labvariation +

Between labvariation

Statistics IntroductionStatistics IntroductionAccuracy:yThe closeness of agreement between a test result and the accepted reference value.

Trueness:The closeness of agreement between the average value obtained from large series of test result and an obtained from large series of test result and an accepted reference value.

Bias:Bias:The difference between the expectation of the test result and an accepted reference value. (sys. error)

St ti ti I t d tiStatistics IntroductionLaboratory Bias (Total Bias):abo ato y as ( ota as):The difference between the expectation of the test results from a particular lab and an accepted reference lvalue.

Bias of the measurement method:Th d f b t th t ti f t t lt The deference between the expectation of test result obtained from all laboratories using that method and an accepted reference value.p

Laboratory component of bias:The difference between the lab bias and the bias of the measurement method.

St ti ti I t d tiStatistics IntroductionPrecision:ec s o :The closeness of agreement between independent test results obtained under stipulated conditions. (r. error)

Statistics IntroductionStatistics IntroductionRepeatability:yPrecision under repeatability conditions.Conditions where independent test results are obtained with the same operator on identical test 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.q p

Reproducibility:Precision under reproducibility conditions.p yCondition where test results are obtained with the same method on identical test items in different labs i h diff i diff iwith different operators using different equipment.

E l blExample problemsExample problem I:

A reference material known to contain 1.00% by weight of a particular component is studied by four analysts (A–D) each analyst performing 5 replicate (A D), each analyst performing 5 replicate measurements. The results are as follows:

A 1.03 1.05 1.03 1.07 1.07A 1.03 1.05 1.03 1.07 1.07

B 0.98 0.88 1.09 1.06 0.94

C 1.06 1.11 1.13 1.04 1.26

D 0 98 1 03 1 02 0 99 1 03

Comment on the random and systematic errors in the results A‐D

D 0.98 1.03 1.02 0.99 1.03

results A D

S l tiSolutionInter the data in the MINITAB

Bias:From mean menu select Stat > Quality Tools > GageStudy > Gage Linearity and Bias StudyStudy > Gage Linearity and Bias Study.From the appeared window choose:

Part Numbers: from variables A, B, C or D.Reference Value: from the reference column.Measurement Data: the same chooses from Part NumberO ti t l t th th d f ti ti t bilit Option: to select the methods of estimating repeatability standard deviation from:

Sample Range.Sample Standard Deviation.

S l tiSolutionPrecision:From mean menu select Stat > Basic Statistics > Display Descriptive Statistics.F h d i d hFrom the appeared window choose:

Variables: all variables A, B, C and D.Statistics: to select the parameter need to be calculated:Statistics: to select the parameter need to be calculated:

Mean.Standard Deviation. (Precision)MinimumMinimum.Maximum.Range.

Graphs: you can select which graph describe the dataGraphs: you can select which graph describe the data.

E iExercisesExercise I:

A standard sample of pooled human blood serum contains 42.0g of albumin per litter.Five laboratories (A‐E) each do six determinations of the Five laboratories (A E) each do six determinations of the albumin concentrations (on the same day), with the following results (g/l):A 42.5 41.6 42.1 41.9 41.1 42.2

B 39.8 43.6 42.1 40.1 43.9 41.9

C 43.5 42.8 43.8 43.1 42.7 43.3

C h d d i i h

D 35.0 43.0 37.1 40.5 36.8 42.2

E 42.2 41.6 42.0 41.8 42.6 39.0

Comment on the random and systematic errors in the results A‐D

F d t l f St ti tiFundamentals of StatisticsMeasures of Central Tendency:easu es o Ce t a e de cy:

Average (M):

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 order sample).p )

Mode:which is defined as the most frequent value in a frequency distribution.

F d t l f St ti tiFundamentals of StatisticsMeasures of Dispersion:easu es o spe s o :

Range (R):

Standard Deviation (S):Standard Deviation (S):

Coefficient of Variation (CV):

Fundamentals of StatisticsFundamentals of StatisticsMeasures of Dispersion:

Variance (S2):

Pooled Standard Deviation (Sp):Pooled Standard Deviation (Sp):

E l blExample problemsExample problem II:

Ten measurements of the ratio of two peak areas in LC experiment gave the following values:

0.921 0.2898 0.2923 0.302 0.3

0.296 0.2947 0.2986 0.29 0.288

Calculate the average, standard deviation and relative standard deviation.

S l tiSolutionFrom mean menu select Stat > Basic Statistics >

lDisplay Descriptive Statistics.From the appeared window choose:

V i bl i lVariables: ratio column.Statistics: to select the parameter need to be calculated:

Mean.Standard Deviation.Variance.Coefficient of variation.N total.

Graphs: you can select which graph describe the data.

E iExercisesExercise II:

To investigate the reproducibility of a method for the determination of selenium in food, nine measurements were made on a single batch of brown rice with the were made on a single batch of brown rice with the listed below results:

No. 1 2 3 4 5 6 7 8 9

Se (ug/g)

0.07 0.07 0.08 0.07 0.07 0.08 0.08 0.09 0.08

Calculate the average, median, mode, range, SD, and RSD

E i Cl i l A l iErrors in Classical AnalysisDistribution of Errors:st but o o o s:

Continuous random variable: values from interval numbers, absence of gaps.Continuous probability distribution: distribution of continuous random variable.Most important continuous probability distribution: the Most important continuous probability distribution: the normal distribution.

Normal Distribution: Bell shapedf(X)

pmean, median and mode are equalrandom variable has infinity range

X

μg

Mean Median Modeμ

Normal DistributionNormal Distribution

( )211 X μ( )

( )2

2

12

Xf X e

μσ

πσ2− −

=

( ) : density of random variable 3 14159 2 71828

f X X3.14159; 2.71828

: population meaneπ

μ= =

( )

p p: population standard deviation

l f d i blX X

μσ

( ): value of random variable X X−∞ < < ∞

N l Di t ib tiNormal Distribution

shiftsthec r ealongthea is increasesthespreadandflattensthec r e

(a) Changing (b) Increasing

shifts the curve alongtheaxis

=61 =1 = 6

increases the spreadandflattensthecurve

2=61 =

2= 12

160 180 200140 160 180 200 140

2=174 2=1701=1= 160


(c) Probabilities and numbers of standard deviations

Shaded area = 0.683 Shaded area = 0.954 Shaded area = 0.997

68% chance of fallingbetween and

− +

+ 95% chance of fallingbetween and

+2

+2

3+

99.7% chance of fallingbetween and 3+

−2 −3

32between and + between and +2 between and 3+−3− −2

N l Di ib iNormal Distribution


Probability is the area under ( ) ?P c X d≤ ≤ =the curve!

f(X)

( )

f( )

Xc d

X


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

E l blExample problemsExample problem III:Part 1:

An analytical chemist wishing to evaluate a new method i t li i i ti ti i hi h h k carries out preliminary investigation in which he makes

six replicates determination of the Dimethoate content of a standard solution which is known to have Dimethoate content of 60 ppb. Each determination requires the preparation of 10ml sample and the determination using GC‐NPD are:determination using GC NPD are:

Can the analyst draw any conclusions from this limited

58.2 61 56.6 61.5 53.8 56.9

amount of data?

S l tiSolutionPart I:From mean menu select Stat > Basic Statistics > Display Descriptive Statistics.F h d i d hFrom the appeared window choose:

Variables: Determination column.Statistics: to select the parameter need to be calculated:Statistics: to select the parameter need to be calculated:

Mean.Standard Deviation.VarianceVariance.Coefficient of variation.N total.

Graphs: you can select which graph describe the dataGraphs: you can select which graph describe the data.


In addition to the six determination we have just id d th l ti l h i t h considered, the analytical chemist has many

determinations of the Dimethoate content of his standard solution by his old method, this particular solution has been used for QC purposes over a period of weeks.

Q 1: Estimate the % determinations which would be Q 1: Estimate the % determinations which would be greater than 65.0.

Q 2: Estimate which value is likely to be exceeded by the highest 10% determinations.


The sixty most recent determinations are:

61.0 65.4 60.0 59.2 57.0 62.5 57.7 56.2 62.9 62.5

56.5 60.2 58.2 56.5 64.7 54.5 60.5 59.5 61.6 60.8

58.7 54.4 62.2 59.0 60.3 60.8 59.5 60.0 61.8 63.8

64.5 66.3 61.1 59.7 57.4 61.2 60.9 58.2 63.0 59.5

56.0 59.4 60.2 62.9 60.5 60.8 61.5 58.5 58.9 60.5

61.2 57.8 63.4 58.9 61.5 62.3 59.8 61.7 64.0 62.7

S l tiSolutionPart II:From mean menu select Stat > Tables > Tally Individual Variables.F h d i d hFrom the appeared window choose:

Variables: R. Dete. column.Display: to select the parameter need to be calculated:Display: to select the parameter need to be calculated:

Counts.Percents.


6.2 5 0.1210

XZ μσ− −

= = =10σ

Normal Distribution Standardized Normal Distribution

10σ = 1Zσ =

5μ =6.2 X Z0Zμ =

0.12

C fid I t lConfidence IntervalsThe problem:

How large are the error bounds when we use data from a sample to estimate parameter of the underlying populationpopulation.Compute confidence intervals for µ.when σ2 is known.when σ2 is unknown.

Suppose an estimate for the mean ( ) is given, and we want to describe the precision of the estimate.

We do this by giving a range of likely values for the t S h i ll d C fid I t lparameter. Such a range is called Confidence Interval.

El t f CI E ti tiElements of CI EstimationA probability that the population parameter falls

h h h lsomewhere within the Interval.

SampleConfidence Interval Sample Statistic

Confidence Limit (Lower)

Confidence Limit (Upper)(Lower) (Upper)

Confidence limits for Mean

Parameter = Statistic ± Its Error ±= Xμ Error

= Error = X−μμ−X

XX

XZσσ

μ=

−=

Error

xZ σ=

ZX ±

Error

XZX σμ ±=

C fid I t lConfidence Intervalsσ

σ_nZXZX X

σσ •±=•±

σx_

90% Samplesxx σμσμ 645.1645.1 +−

X

90% Samples

95% Samplesxx σμσμ 96.196.1 +−

95 p

xx σμσμ 58.258.2 +−99% Samples

F t ff t I t l idthFactor affect Interval width

D V i i Data Variation measured by σ Intervals Extend from

X Z X ZSample Size X ‐ Zσ to X + Z σxx

n/XX σ=σ

Level of Confidence

n/XX σσ

Level of Confidence(1 ‐ α)

C fid I t lConfidence Intervals

Concluding Remark:

ll h f dAs smaller we choose α as more confident we get that the interval contains the parameter µ But at the same time the parameter µ. But at the same time the confidence interval gets wider and is therefore less precise.

CI ( k H dl T )CI (σ known – Hardly True)Assumptions:Assumptions:

Population standard deviation is known.Population is normally distributed.p yIf not normal, use large samples.

Confidence interval estimate:

CI (σ unknown)CI (σ unknown)Assumptions:Assumptions:

Population standard deviation is unknown.Sample size must be large enough for central limit p g gtheorem or population must be normally distributed.

Use Student’s t distribution.

Confidence interval estimate:

E l blExample problemsExample problem IV:

The DDT content of a fish samples in ppb was determined using GC‐ECD.The following values were obtained:The following values were obtained:

102 97 99 98 101 106

What are the 95% and 99% confidence limits for the DDT concentration?

S l tiSolutionStandardized normal distribution:From mean menu select Stat > Basic Statistics > 1‐sample Z (test and confidence interval).F h d i d hFrom the appeared window choose:

Samples in columns: data column.Standard deviation: but the value of SD previous Standard deviation: but the value of SD previous calculated.Options: to choose the confidence level.pGraphs: you can select which graph describe the data.

S l tiSolutionT distribution:From mean menu select Stat > Basic Statistics > 1‐sample T (test and confidence interval).F h d i d hFrom the appeared window choose:

Samples in columns: data column.Options: to choose the confidence levelOptions: to choose the confidence level.Graphs: you can select which graph describe the data.

E l blExample problemsExample problem V:

The absorbance scale of a spectrometer is tested at a particular wavelength with a standard solution which has an absorbance given as 0 47has an absorbance given as 0.47.Ten measurements of the absorbance with the spectrometer give average absorbance 0.461 and standard deviation 0.003.

Find the 95% confidence interval for the mean absorbance Find the 95% confidence interval for the mean absorbance as measured by the spectrometer and hence decide whether a systematic error is present.

S l tiSolutionFrom mean menu select Stat > Basic Statistics > 1‐

l ( d d l)sample Z (test and confidence interval).From the appeared window choose:

S l i l d lSamples in columns: data column.Standard deviation: but the value of SD previous calculated.calculated.Options: to choose the confidence level.Graphs: you can select which graph describe the data.

E iExercisesExercise III:

A 0.1M solution of acid was used to titrate 10ml of 0.1M solution of alkali and the following volumes of acid were record:record:

9.88 10.18 10.23 10.39 10.25

Calculate the 95% confidence limits of the mean and use them to decide if there is any evidence of systematic errorerror.

St d t’ t T tStudent’s t TestWhen solving probability problems for the sample e so g p obab ty p ob e s o t e sa p emean, one of the steps was to convert the sample mean values to z‐scores using the following formula:

where and

What happens if we do not know the population standard deviation σ?, if we substitute the population standard deviation σ with the sample standard standard deviation σ with the sample standard deviation s can we use the standard normal table?

Answer: No

Student’s t TestStudent s t TestThis question was addressed in 1908 when W.S. Gossets quest o as add essed 908 e .S. Gossetfound that if we replace σ with the sample standard deviation s the distribution becomes a t‐distribution. IfIf:

then T has a t distribution with n 1 degrees of then T has a t–distribution with n‐1 degrees of freedom. the t‐distribution is similar to 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 α and degrees of freedomα and degrees of freedom.

Student’s t Test

One sample t‐test:

When using t‐test TAKE CARE !!! Is it :il d il done‐tailed or two‐tailed

Student’s t Test

Independent sample t‐test (equal variances):

D f f d ( ) Degrees of freedom = (n1 + n2) ‐ 2

Student’s t Test

Independent sample t‐test (unequal variances):

Student’s t TestIndependent sample t test (unequal ariances)Independent sample t‐test (unequal variances)In such complicated case degrees of freedom is calculated from:calculated from:

St d t’ t T tStudent’s t TestPaired sample t‐test:Paired sample t‐test:

The sign of the difference is very important.

E l blExample problemsExample problem VI:

In a method for determination of mercury by cold‐vapor atomic absorption the following values were obtained for a standard reference material containing 38 9% for a standard reference material containing 38.9% mercury:

38 9% 37 4% 37 1%

Is there any evidence of systematic error?

38.9% 37.4% 37.1%

S l tiSolutionFrom mean menu select Stat > Basic Statistics >

h lGraphical Summary.From the appeared window choose:

V i bl d lVariables: data column.Confidence level: to choose the confidence level.

From mean menu select Stat > Basic Statistics > 1From mean menu select Stat > Basic Statistics > 1‐sample T (test and confidence interval).From the appeared window choose:From the appeared window choose:

Samples in columns: data column.Options: to choose the confidence level.Perform hypothesis test: record the value of Mean

E l blExample problemsExample problem VII:

In a comparison of LC and GC methods for the determination of Primicarb in vegetables the following results (ug/g) were obtained:results (ug/g) were obtained:

Mean SD

HPLC method 28.00 0.30

For each method 10 determinations where made.

HPLC method 28.00 0.30

GC‐NPD method 26.25 0.23

Assuming that the two samples have standard deviation which are not significantly different, could you decide that both method give results having means which that both method give results having means which differ significantly?

S l tiSolutionFrom mean menu select Stat > Basic Statistics > 2‐

l ( d d l)sample T (test and confidence interval).From the appeared window choose:

S i d DSummarized Data:First:

Sample size, Mean and Standard Deviation.p ,Second:

Sample size, Mean and Standard Deviation.

A l iAssume equal variances.Options: to choose the confidence level.Graphs: you can select which graph describe the data.p y g p

E l blExample problemsExample problem VIII:

The pesticide levels (ppm) in four thoroughly homogenized apple samples are determined once by each (a) a chromatographic method and (b) an each (a) a chromatographic method and (b) an immunoassay method. The results are:

Apple Pesticide Level

Chrom. Immuno.

1 7.1 7.6

2 6.1 6.8

Is there any evidence that the two analytical methods give

3 5.0 4.8

4 6.0 5.7

Is there any evidence that the two analytical methods give significantly different results?

S l tiSolutionFrom mean menu select Stat > Basic Statistics > Paired ( d d l)T (test and confidence interval).

From the appeared window choose:S l i C lSamples in Columns:

First Sample:The column of variables need to paired evaluates.p

Second Sample:The column of variables need to paired evaluates.

O ti t h th fid l lOptions: to choose the confidence level.Graphs: you can select which graph describe the data.

E l blExample problemsExample problem IX:

It is suspected that an acid‐base titrimetric method has a significant indicator error and thus tends to give results with a positive systematic error (i e positive bias)results with a positive systematic error (i.e. positive bias)To test this an exactly 0.1M solution of acid is used to titrate 25.0ml of an exactly 0.1M solution of alkali with

( )the following results (ml):

25.06 25.18 24.87 25.51 25.34 25.41

Test for positive bias in these results.

S l tiSolutionUsing 1‐Sample Z:From mean menu select Stat > Basic Statistics > 1‐sample Z (test and confidence interval).F h d i d hFrom the appeared window choose:

Samples in columns: data column.Standard deviation: but the value of SD previous Standard deviation: but the value of SD previous calculated.Options: to choose the confidence level and data one tail por two tail.Graphs: you can select which graph describe the data.

S l tiSolutionUsing 1‐Sample T:From mean menu select Stat > Basic Statistics > 1‐sample T (test and confidence interval).F h d i d hFrom the appeared window choose:

Samples in columns: data column.Options: to choose the confidence level and data one tail Options: to choose the confidence level and data one tail or two tail.Perform hypothesis test: record the value of Mean.

S l tiSolutionUsing Gage Linearity and Bias Study:From mean menu select Stat > Quality Tools > GageStudy > Gage Linearity and Bias Study.F h d i d hFrom the appeared window choose:

Part Numbers: from variables A, B, C or D.Reference Value: from the reference columnReference Value: from the reference column.Measurement Data: the same chooses from Part NumberOption: to select the methods of estimating repeatability Option: to select the methods of estimating repeatability standard deviation from:

Sample Range.S l S d d D i iSample Standard Deviation.

f T tf ‐ Testf – Test is used to compare the standard deviation of f – Test is used to compare the standard deviation of two samples and to make a test to determine whether the population from which they come have equal variances.

E l blExample problemsExample problem X:

A proposed method for the determination of TCDD in water was compared with the standard method. The following results were obtained for a river water sample:following results were obtained for a river water sample:

Mean (ug/l) SD (ug/l)

Standard method 72 0 3 31Standard method 72.0 3.31

Proposed method 72.0 1.51

For each method 8 determinations were mad. Is the precision of the proposed method is significantly greater than that of the standard method?than that of the standard method?

S l tiSolutionFrom mean menu select Stat > Basic Statistics > 2‐Variances.From the appeared window choose:

S i d DSummarized Data:First:

Sample size and Variance.pSecond:

Sample size and Variance.

St th d t ld b l l t d d d d i Storage: the data could be calculated and recorded in MINITAB worksheet.Options: to choose the confidence level.p

H th i T tiHypothesis TestingA hypothesis is a claim (assumption) about the

lpopulation parameter.Examples of parameters are

l ti I claim the mean GPA ofpopulation mean or proportion.The parameter must be

I claim the mean GPA of this class is 3.5!μ =

The parameter must be identified before analysis.

H th i T tiHypothesis Testing

A thAssume thepopulation

mean age is 50.Identify the Population

g 5( )0 : 50H μ =

Take a SampleN t lik l !

X 20 likely if Is ?μ= = 50

REJECT

No, not likely!

( )Null Hypothesis ( )20X =

H th i T tiHypothesis TestingSampling Distribution of XSampling Distribution of

It is unlikely that ld t

... Therefore, j t th

X

we would get a sample mean of this value

we reject the null hypothesis that m = 50this value ... that m = 50.

... if in fact this wereth l ti

= 50μ20 X

the population mean.

= 50μ20

If H0 is true

X

H th i T tiHypothesis TestingH0: Innocent

The Truth The Truth

Jury Trial Hypothesis Test

The Truth The Truth

Verdict Innocent Guilty Decision H0 True H0 False

Do NotInnocent Correct Error

Do NotRejectH0

1 ‐ α Type IIError (β )

T IGuilty Error Correct Reject

H0

Type IError(α )

Power(1 ‐ β )

H th i T tiHypothesis TestingIf you reduce the probability of oneIf you reduce the probability of one error, the other one increases so that everything else is unchanged.

β

α

H th i T tiHypothesis TestingTrue value of population parameterβ Increases when the difference between hypothesized parameter and its true value decrease.

Significance levelβ increases when α decreases α

β

Population standard deviationβ increases when σ increases

β σβ c eases e σ c eases

Sample size ββ increases when n decreases n

E iExercisesExercise IV:

A new operator is given a sample containing a known concentration 50ug/kg of profenofos and instructed to make eight determinationmake eight determination.He obtained the following results:

49.4 49.8 50.8 49.3 51.3 50.0 50.8 51.8

a) Is there evidence that operator is biased?b) Calculate a 95% confidence interval and hence state

49.4 49.8 50.8 49.3 51.3 50.0 50.8 51.8

) C %the maximum possible bias for operator.

c) What sample size is needed to estimate operator bias i hi 0 3 /kto within ± 0.3ug/kg.

A l i f V i (ANOVA)Analysis of Variance (ANOVA)The statistical tests described previously are used in h f f dthe comparison of two sets of data, or to compare a single sample of measurements with a standard or reference value frequentlyreference value frequently.

However, it is necessary to compare three or more sets 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

lapplications.

A l i f V i (ANOVA)Analysis of Variance (ANOVA)

A l i f V i (ANOVA)Analysis of Variance (ANOVA)

If there is only one source of variation apart from this measurement area, a one‐way ANOVA calculation is appropriateappropriate.

If there are two sources of variation we use two way If there are two sources of variation we use two‐way ANOVA calculation.

And so on…..

E l blExample problemsExample problem XI:

A sample of fruit is analyzed for its pesticide content by a liquid chromatographic procedure, but four different extraction procedures A – D are used the concentration extraction procedures A D are used, the concentration in each case being measured three times.The results are indicated in the shown table:Results A B C D

1 10.5 9.9 9.9 9.2

2 11 5 10 8 9 1 8 5

Is there any evidence that the four different sample

2 11.5 10.8 9.1 8.5

3 10.7 10.8 8.9 9.0

Is there any evidence that the four different sample preparation methods yield different results?

S l tiSolutionFrom mean menu select Stat > ANOVA > One –Way

l ( k d)Analysis of Variance (unstacked).From the appeared window choose:

R (i l )Responses (in separate column):All variables A, B, C and D.

Confidence level.Confidence level.

Note: If the p‐value is below a specified significance youf p p f g f ycan declare the statistic to be statistically significantand reject the test's null hypothesis.

2 S d t tχ2‐Squared testThe significance tests so far described in this course, in

l h h d l dgeneral, assume that the data analyzed:Be continuous, interval data comparison a whole population or sampled randomly from a populationpopulation or sampled randomly from a population.Have a normal distribution.Sample size should not differ hugely between the p g ygroups.

χ2= ∑ (O‐E)2/EIn contrast, chi‐squared test is concerned with frequency .

i e the number of times a given event occursi.e. the number of times a given event occurs.

E l blExample problemsExample problem XII:

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

24 27 11 9

Is there any evidence that the workers differ in their reliability?

Solution:Ho: the same number of breakages by each worker.Ha: different number of breakages by each workers.

S l tiSolutionFrom mean menu select Stat > Tables > Chi‐Square

d ( bl )goodness – of – Fit test (one variable).From the appeared window choose:

R (i l )Responses (in separate column):All variables A, B, C and D.

Confidence level.Confidence level.

Note: If the p‐value is below a specified significance youf p p f g f ycan declare the statistic to be statistically significantand reject the test's null hypothesis.

C l i f Si ifi T tConclusion from Significance Testas we have just explained that a significance test at for

l P l l i l % i k th t ll example P = o.o5 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 yp yp (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 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 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 h h ihypothesis.

C l i f Si ifi T tConclusion from Significance TestConsider the situation where a certain chemical

d f h h bproduct is meant to contain 3% of phosphorus by weight.

I i d h hi i h i dIt is suspected that this proportion has increased.

To test such increase the composition is analyzed by a t d d th d ith k t d d d i ti f standard method with known standard deviation of 0.036%.

Suppose 4 measurements are made and a significance Suppose 4 measurements are made and a significance test is performed at level of P = 0.05. a one‐tailed test is required, as we are interested only in an increase.q , y

C l i f Si ifi T tConclusion from Significance TestHo, µ = 3.0%

h l d l h hThe solid line shows the sampling distribution of the mean if null hypothesis is true. ypThis sampling distribution has mean 3.0 and Standard deviation = 0 036/√4 % If the deviation = 0.036/√4 %. If the sample mean lies above the indicated critical value Xc the

ll h th i i j t d null hypothesis is rejected, thus the shaded region with area 0.05 represent the probability of a type 1 error

C l i f Si ifi T tConclusion from Significance TestThe only way in which both

b d d berrors can be reduced is by increasing the sample size. The effect of increasing n to 9 for 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 decrease in both types of error for a given value of Xc.

C l i f Si ifi T tConclusion from Significance Testthe probability that a false null hypothesis is rejected k h f h h h fis 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 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 useful to compare the powers of the tests to decide which is more appropriatethe tests to decide which is more appropriate.

E l blExample problemsExample problem XIII:

The nitrate level (mg/l) in a sample of river water was measured four times, with the following results:

Can the last value be rejected as an outlier?

0.404 0.400 0.398 0.379

Can the last value be rejected as an outlier?

S l tiSolutionFrom mean menu select Stat > Basic Statistics > Display p yDescriptive Statistics.From the appeared window choose:

Variables: Determination columnVariables: Determination column.Statistics: to select the parameter need to be calculated:Graphs: you can select which graph describe the data.

F l t St t B i St ti ti l T From mean menu select Stat > Basic Statistics > 1‐sample T (test and confidence interval).From the appeared window choose:

Samples in columns: data column.Options: to choose the confidence level.Perform hypothesis test: record the value of Meanf yp

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