Part 2: statistics Exam Contents: A. Basic concepts of descriptive statistics. B. Statistics of grouped variables

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1 Part : statistics Exam 4.10 Cotets: A. Basic cocepts of descriptive statistics * variable types * measures of cetral tedecy * measures of variatio B. Statistics of grouped variables C. Estimatio of parameters * cofidece itervals D. Two variable statistics * regressio aalysis * correlatio E. Hypothesis tests oe-sample t-test Two sample t-test Chi square test 1

2 BASIC CONCEPTS 1. Populatio ( also called sample space) Populatio icludes all objects of iterest, which we wish to research. Examples: All customers of a supermarket, all citizes of Rovaiemi, all studets of RAMK. Sample Sample is a portio of populatio, which is researched. I most cases it is ot possible to do research to the whole populatio because it would be too expesive or because it would be almost impossible to reach all members of populatio. That is why i statistical research the coclusios are made based o iformatio, which is obtaied from samples, which are usually radomly chose subsets of populatio. 3. Variables The properties of objects of the populatio, which we wat to research, are called VARIABLES The type of the variable determies, which kids of statistical parameters ca be calculated from measured variable values 4. Discrete ad cotiuous variables Discrete variables: ** set of possible values is eumerable, fiite Cotiuous variables: ** set of possible values is cotiuous, ifiite -umber of childre i a family -geder (male of female) - grade i statistics -mothly salary -legth 5. Statistical parameters Parameters are umbers which characterize the distributio of values of variables Examples: mea, media, quartiles, percetiles, variace, stadard deviatio NOTATION: Populatio parameters are expressed with Greek letters, Sample parameters are expressed with Lati letters x, s

3 Levels of measuremet Statistical variables ca be divided ito four categories, which are called the levels of measuremets level of measuremet descriptio examples omial level variable or class variable ordial level variable iterval level variable ratio level variable variable values are just ames or labels, which defie classes values are labels, but a order ca be added to the values -a umerical variable, where it is meaigful to compare differeces, but ot ratios - absolute zero does ot exist a umerical variable where also ratios are comparable -geder (male, female) -marital status (married, sigle, divorced, widow) -color (red,blue, ) -quality class of potatoes -a players locatio o the rakig list - Celsius temperature -Kelvi temperature -mothly salary - age i years 8. Measures of cetral tedecy Parameter ame ad symbol Mea Media ME Mode Md Formula x i Media = a variable value which separate the higher half from the lower half of ordered data If is odd, Media = the value i the middle of ordered data list If is eve, Media= the average of two values i the middle Discrete variables: Mode = value which has the highest frequecy For grouped cotiuous variables: Formula is give later Applicable to followig variable types -ratio ad iterval level variables -also opiio survey aswers, with scale 1-5 -ratio,iterval ad ordial level variables -all discrete variable types -also applicable o grouped cotiuous variables 3

4 Example: I a school class of 0 studets followig biology grades were give: 4, 4, 5, 5, 5, 5, 6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 9, 9, 10 Calculate a) mea b) media c) mode a) 6.65 (Excel fuctio average) b) ME = (7+7)/ = 7 c) Md = 7 ( value with highest frequecy) 8. Measures of variatio Also these equivalet formulas ca be used 4

5 Example: calculate stadard deviatio of values 3, 5, 6, 5 ad 4 a) maually b) usig Excel fuctio STDEVP Visualisatio of data Pie chart Bar chart Lie diagram Histogram 5

6 Grouped data Rules: 1. The suitable umber of groups should be The umber of groups should ot be square root of sample size 3. The lower bouds of each group should be roud umbers 4. The upper bouds should ot overlap with ext lower boud. They should be less tha ext lower boud by measuremet accuracy Results of aalysis 1. Frequecy table ad histogram should be created. CDF fuctio should be formed Mea, media ad mode should be calculated Stadard deviatio ad quartiles should be calculated Formulas for measures of cetral tedecy 6

7 Stadard deviatio of grouped data Quartiles Q1 ad Q3 7

8 Solved example A compay has 130 workers. The ages are i the table below Classes Frequecy table. 5 classes are created Age group frequecy f Make complete aalysis of data: * mea, media, mode, stadard deviatio, quartiles. * Histogram, cdf(x) 8

3.10.013 a) Average age is calculated usig class ceters b) Media age CF ME L i f 130 51 ME 40 10 43. 1years 45 Media class is the class, where worker r 65 is located if the workers are sorted by age.

9 a) Average age is calculated usig class ceters b) Media age CF ME L i f ME years 45 Media class is the class, where worker r 65 is located if the workers are sorted by age. Media class is L = lower boud of the media class = 40 = total umber of workers = 130 CF = cumulative frequece of classes below media class = 19+3 = 51 f = frequecy of the media class = 45 i = width of the media class = 10 9

3.10.013 b) Mode of age Md f L f i 1 i f i 1 f L i 1 i 1 i 1 3 40 5 50 Md 44.

ext class. The weights are frequecies of the eighborig classes of the modal class.

10 b) Mode of age Md f L f i 1 i f i 1 f L i 1 i 1 i Md 44. 4years 3 5 Mode = weighted average of the lower boud of modal class ad lower boud of the ext class. The weights are frequecies of the eighborig classes of the modal class. s Stadard deviatio fixi ( fixi ) ( ) s 130 ( ) s 11. years

3.10.013 CDF(x) ad quartiles Quartiles are Q1 = 34 years Q3 = 51 years From CDF(x) graph we see that 1st quartile = 34 years ( = such age, that 5%

11 CDF(x) ad quartiles Quartiles are Q1 = 34 years Q3 = 51 years From CDF(x) graph we see that 1st quartile = 34 years ( = such age, that 5% of employees are below that) ad 3rd quartile (75% splittig poit) is 51 years. d quartile ( 50% splittig poit) is called Media, which was 43 years. 11

12 Cofidece itervals 1. Cofidece itervals of variables with ormal distributio 1

13 Defiitio: Cofidece itervals idicate the reliability of parameter estimate (usually estimate of mea) obtaied from samples. It gives a iterval ( mi, max ), where the true mea is located at a give probability. Most used probabilities are 95% ad 99%. They are called cofidece levels. Ofte we also use their complemets 5% ad 1%, which are called risk levels CASE1: Goal of a medical survey was to determie the average cholesterol level i blood of all 6000 studets of Rovaiemi The research group chose a sample of 00 studets. The mea value i this sample was 5.5 ad sample stadard deviatio was 0.35 Estimate the mea cholesterol value of all studets by calculatig cofidece iterval correspodig 95% cofidece level Assume that the distributio of x is ormal with average ad stadard deviatio. The the sample mea of x : (x 1 +x +. + x )/ is also ormally distributed with the same average ad stadard deviatio /. x ~ N(, ) x ~ N(, ) x x 1 x... x Sample mea ad variable itself have same distributio average, but stadard deviatio of sample mea depeds o sample size * The bigger sample, the smaller is the st.dev. of its mea

4.10.013 x x1 x... x The error margi, whe we estimate populatio mea with sample mea, is called cofidece iterval ( x t s, x t s ) s = sample stadard deviatio, = sample size, t = 1.

14 x x1 x... x The error margi, whe we estimate populatio mea with sample mea, is called cofidece iterval ( x t s, x t s ) s = sample stadard deviatio, = sample size, t = 1.96, for 95% cofidece level t =.60, for 99% cofidece level x =sample mea CASE: Goal of a medical survey was to determie the average cholesterol level i blood of all 6000 studets of Rovaiemi The research group chose a sample of 00 studets. The mea value i this sample was 5.5 ad sample stadard deviatio was 0.35 Estimate populatio mea ad estimate also the magitude of samplig error. Aswer: mea cholesterol level of studets = 5.5 Cofidece iterval at 95% sigificace level ( *0.35/ 00, *0.35/ 00 ) = (5.0, 5.30) Web-calculator: 3

15 IN EXCEL FUNCTION CONFIDENCE CALCULATES CONFIDENCE MARGIN = Cofidece itervals of relative proportio 4

4.10.013 CASE: OPINION SURVEYS A ewspaper makes a opiio survey, where it iterviews a radomly selected sample of people askig if they support cadidate A or ot.

16 CASE: OPINION SURVEYS A ewspaper makes a opiio survey, where it iterviews a radomly selected sample of people askig if they support cadidate A or ot. The support percet of the sample is almost allways differet from the true support percet i the whole populatio. We call this differece samplig error. How does the samplig error deped o the sample size? How is the error margi of opiio surveys calculated? Cofidece level of relative proportio If relative proportio obtaied from the sample is p, the the cofidece iterval of relative proportio i the whole populatio is ( p t p(1 p), p t p(1 p) ) where t = 1.96, for 95% cofidece level t =.60, for 99% cofidece level CASE: ASSUME THAT A SAMPLE OF SIZE 1000 GAVE 35% SUPPORT FOR CANDIDATE A FOR PRESIDENT Estimate the true support by calculatig the cofidece iterval (95%) 0.35* 0.65 ( , * 0.65 ) 1000 (0.3,0.38) 5

4.10.013 Web calculator of cofidece margi of relative proportio http://www.surveysystem.com/sscalc.htm Iterpretatio: Cofidece iterval = ( 35.95, 35 +.

17 Web calculator of cofidece margi of relative proportio Iterpretatio: Cofidece iterval = ( 35.95, ) percetage = (3%, 38%) footote Coectio with biomial distributio You may otice that previous example of a cadidates support percet is idetical with a repeated experimet case: A experimet with success probability p = 0.35 is repeated times. Calculate expected value of succeeded experimets ad calculate stadard deviatio of succeeded meas. From the theory of biomial distributio we kow that expected value = *p stadard deviatio = ( p(1-p)) The relative proportio of succeeded experimets is *p / = p Stadard deviatio of relative proportio = ( p(1-p))/ = p ( 1 p) Thus cofidece Iterval is ( p t p(1 p), p t p(1 p) ) 6

4.10.013 CALCULATION OF SAMPLE SIZE FROM CONFIDENCE MARGIN Solvig from p t p( 1 p) gives p(1 p) t p What should be the sample size, if error of p= ± % is allowed ad we kow that last survey gave

18 CALCULATION OF SAMPLE SIZE FROM CONFIDENCE MARGIN Solvig from p t p( 1 p) gives p(1 p) t p What should be the sample size, if error of p= ± % is allowed ad we kow that last survey gave cadidate A 35% support 0.35* 0.65* If the researcher wats to press samplig error uder %, the sample size should be 00 Sample size calculator 7

19 Exercise1: The age of visitors i a art museum was researched by askig 600 radomly chose visitors about their age. Sample average was 9 years ad sample stadard deviatio 7.5 Estimate the average age of all visitors. Calculate 95% cofidece iterval. Exercise: BBC wats to fid out the opiio of British people o some political issue, where support percet was earlier close to 54. Calculate the sample size, if the wated error margi should be 3%. 8

20 Hypothesis tests Hypothesis tests are used whe somethig should be proved with statistical research Questio asked: Is the result statistically sigificat or is it just ormal statistical variatio 1. Oe sample t- test is used to prove that give, assumed mea value is too low or too high 1

10.10.013 Studet s t - test Whe sample sizes are relatively small, we caot use pdf ad cdf (probability desity ad cumulative distributio fuctio) of Normal distributio.

i the picture df = degrees of freedom = 1 = sample size - 1 EXAMPLE: A light bulb maufacturer has declared i the techical data that the average duratio of their light bulb is 1100 hours.

21 Studet s t - test Whe sample sizes are relatively small, we caot use pdf ad cdf (probability desity ad cumulative distributio fuctio) of Normal distributio. Istead Studet s distributio is used. As show i the picture, the shape of pdf is same as i Normal distributio, but at low sample size the peak is ot so high. i the picture df = degrees of freedom = 1 = sample size - 1 EXAMPLE: A light bulb maufacturer has declared i the techical data that the average duratio of their light bulb is 1100 hours. Cosumer associatio tests a sample of 50 bulbs, ad i the sample mea duratio is 1080 hours ad stadard deviatio is 55 hours. Does this prove that maufacturer has give false iformatio. Termiology used: H1 hypothesis: Mea duratio give by maufacturer is too high H0 hypothesis: There is o proof that H1 hypothesis is true Maufacturer: light bulb duratio = 1100 h

10.10.013 Method : Calculate test value t ad compare it to the critical values. If critical value is exceeded, H1 hypothesis is accepted. t x 1080 1100 t.

22 Method : Calculate test value t ad compare it to the critical values. If critical value is exceeded, H1 hypothesis is accepted. t x t. 57 s / 55/ 50 Studet s oe sided t-test critical values df Critical value 95% Critical value 99% 0 1, , , , , , , , , , ,65.33 Test parameter t > both 95% ad 99% critical values at sampe size ~50 Coclusio: Maufacturer has give false iformatio degrees of freedom df = 1 (sample size 1) ONLINE T - TEST CALCULATOR Test parameter t Critical value Test parameter t =,57 > critical value => hypothesis H1 is accepted. Maufacturer has give too large duratio value for bulbs. 3

23 Two sample t- test is used to compare two groups ad prove that the mea of group A is bigger or less tha mea of group B EXAMPLE: The same mathematics test was performed i two test groups: School A: sample size 60 studets, mea result 1.4 poits, stdev 3.5 School B: sample size 50 studets, mea result 0.1 poits, stdev.9 (both schools are big, more that 000 studets) H1 hypothesis: Studets of School A are better i mathematics H0 hypothesis: Test does ot prove ay differece. Result is ormal statistical fluctuatio 4

24 Method : Calculate test value t ad compare it to the critical values. If critical value is exceeded, H1 hypothesis is accepted. t x1 x t. 13 s 1 s df Critical value 95% Critical value 99% 0 1, , , , , , , , , , ,65.33 Test parameter t =.13 > 95% critical value Coclusio: H1 accepted at 95% cofidece level: School A is better i math at 95% cofidece (but ot at 99% cofidece level) degrees of freedom df = 1 + if st.deviatios are same or ot too differet Exercise 1: There is a text i the loaf of bread that its weight is 1000 g. I a idepedet research 80 loafs of bread were weighed. Average was 990 g ad stadard deviatio was 0 g. Does this proof that the maufacturer has cheated? t x s / / Test parameter 3.35 exceeds critical values => Maufacturer cheats 5

25 Exercise : A factory presets a ew medicie for high cholesterol. It was tested i two radomly chose test groups, both cosistig of elderly persos. The results were followig: Group A (50), got medicie, cholesterol mea 5.4, stdev 1.8 Group A (60) got placebo, cholesterol mea 5.7, stdev 1.9 Does this prove the effect of the medici? placebo = fake pills t x x s s > critical values Test proves that medicie lowers cholesteri. ( Completely differet questio is if the effect is maybe too small from the poit of view of the patiet 6

26 Multivariable statistics Depedece of variables c test correlatio Liear Regressio y = kx + b Multiliear Regressio Y = a x 1 + b x + c No- liear regressio y = a e b x 1. Testig if two omial level variables are depedet χ test (Chi-Square test) χ ( o 1 e1 ) ( o e ) = e e

27 CASE1: I a research made to both Fiish ad Swedish families two variables: atioality ad owig a summer cottage were crosstabbed producig a followig table First we calculate colum ad row totals Fiish Swedish Σ Ow cottage No cottage Σ Brief look at the result seems to idicate that Owig a summer cottage is more commo amog Swedish tha Fiish people. I other words variable owig summer cottage seems to deped o atioality. Chi Square test Step1: Calculate theoretical frequecies e i, which are based o assumptio, that variables1 ad are ot depedet: I our example the relative proportios of atioalities are Fiish : 80/180 = , Swedish : 100/180 = 0,5555 Theoretical frequecies e i are calculated applyig the same relative proportios 80/180 ad 100/180 to cottage owers ad o-owers (that would be the case, if variables are ot depedet) Fiish Swedish Σ Ow cottage 80/180*84= /180*84= No cottage 80/180*96= /180*96= Σ

Step: Calculate test parameter Chi_Square with formula χ ( o 1 e1 ) ( o e ) = e 1 +... + o i = observed frequecies i the cells e i = expected frequecies (theoretical, based o idepedece assumptio) e = 0.

If test parameter > critical value, two variables are show to be depedet, otherwise there is o proof of depedece. Degree of freedom df = (-1)(m-1), where, m are umbers of classes of variables 1 ad.

28 Step: Calculate test parameter Chi_Square with formula χ ( o 1 e1 ) ( o e ) = e o i = observed frequecies i the cells e i = expected frequecies (theoretical, based o idepedece assumptio) e = Step3: Compare the test parameter with the value i the table of Chi Square critical values. If test parameter > critical value, two variables are show to be depedet, otherwise there is o proof of depedece. Degree of freedom df = (-1)(m-1), where, m are umbers of classes of variables 1 ad. Colum P=0.05 meas that the variable are depedet at probability 0,95 = 95% (error probability is thus 5%) I our example * Degrees of freedom df = (-1)(-1) = 1 Test value is much smaller tha critical value 3.84 Coclusio: there is o evidece, that rate of owig cottage would differ amog Fiish ad Swedish families. (olie calculator)

29 . Correlatio coefficiet = measure of liear depedece Properties of correlatio coefficiet Formula: Rage: -1 r 1

Sigificace of correlatio Iterpretatio of correlatio coefficiet depeds o sample size ad degrees of freedom df= - (sample size ) If r > critical value, we say that there is a sigificat positive or

30 Sigificace of correlatio Iterpretatio of correlatio coefficiet depeds o sample size ad degrees of freedom df= - (sample size ) If r > critical value, we say that there is a sigificat positive or egative (if r < 0) correlatio betwee variables. example Table cosists of math grades ad physics grades of a school class with 18 studets. Determie if there is a correlatio (liear depedece) betwee math ad physics grades. Calculatio gives r = 0.71 Iterpretatio: Degrees of freedom df = 18 = 16 Correspodig critical value = 0.48 It is exceeded => Coclusio There exists a sigificat positive correlatio betwee math grades ad physics grades

31 Excel fuctio CORREL Fuctio CORREL returs the correlatio coefficiet. Its argumets are datavalues of variables X ad Y

32 Liear regressio Liear regressio alias Least Square Sum Method 1

33 variables, tredlie y residual r i Observed pairs (x1,y1), (x1,y), are modeled usig liear model y = a x + b x * Residuals are differeces betwee observed y-values ad y- values calculated from model: r 1 = y 1 (ax 1 +b), r = y 1 (ax 1 +b), * They form a residual vector r = (r 1,r,,r ) * I LS method goal is to fid such a ad b, that the legth r ad its square (r 1 + r + r ) is at miimum. Miimum is foud at poit, where both partial derivatives of r are zero: Liear regressio with Excel 1 1. Write Data table. Create Graph of type XY - scatter 3. Click ay poit With right mouse butto ad choose add tredlie (liear) 4. Choose Optios - Show equatio - Show R -Square

34 Iput of LINEST fuctio 1.. Pait target area : x 5 cells. Choose Isert Fuctio LINEST pait y s pait x s true Cost is left empty if you do ot wish to force costat b to 0 Stat = true meas that you wish to receive all statistical data Fiish with key combiatio CTRL- SHIFT-ENTER. This is used istead of OK or Eter for all multivalued Excel fuctios, where output area is a matrix. Output matrix of LINEST fuctio R-square F parameter Total square sum 1st lie: slope a ad costat b d lie: stadard errors of a ad b Stadard error of calculated y df : degrees of freedom Square of the residual vector Usually from output is take 1. Values of a ad b i model y = a x + b. Cofidece radiuses for a ad b are a = 1.96*STE(a) b = 1.96*STE(b) I the example: Y = a X + b a = ± 0.1 b = 8.08 ± 0.61 STE = stadard error 3

35 SOLUTION USING DIFFERENTIAL CALCULUS Calculatio methods 1) Form the square of residual vector ad set partial derivatives to zero. Solve the pair of equatios ) Excel fuctio LINEST returs slope a ad costat b ad also their stadard errors. 3) ONLINE calculators 4

36 Multiliear regressio z residual r i Example: 3 variables Observatios are triples : (x i, y i, z i ). Model is a plae z = ax + by + c y x Residuals r i = z i (ax i + by i + c) form vector r = (r1, r,, r ) Miimum is at poit where the three partial derivatives of r vaish. Multiliear regressio: Iput = observed data x1 x x3 y Output = model y = m 1 x 1 + m x + m 3 x 3 +b EXCEL fuctio LINEST returs also stadard errors for m i ad b

37 Multiliear regressio i practice 1) Excel fuctio LINEST Excel fuctio returs also parameters expressig goodess of model Stadard Error Goodess of the model is described by stadard deviatio of residuals r 1,r,, r, which is called Stadard Error R-Squared = coefficiet of determiatio R = correlatio coefficiet s aquare i case of variables: If R = 0.85, we say that model explais 85 % of the variatios of the depedet (explaied) variable. ( The rest 15% - is ot explaied by model) No-liear models, which ca be liearized Theoretical model: l = a T Task. Fid the parameter a based o measured values i the table. 6

38 l (meters) Calculate the squares T, ad perform liear regressio to Variables X = T ad Y = l. 1,80 1,60 1,40 y = 0,479x 1,0 1,00 0,80 0,60 0,40 0,0 0,00 0,00 1,00,00 3,00 4,00 5,00 6,00 7,00 T^ Result: I model l = a T Parameter a = Excel Add tredlie commad is used with Optio set itersectio = 0, which sets costat b to zero i y = a x + b No liear regressio Example: Fidig a. degree model y = ax + bx + c Priciple: Miimize square sum of residuals r =( y 1 ax 1 + bx 1 + c) + + (y 1 ax 1 + bx 1 + c) Miimum at poit, where partial derivatives vaish. r a r b r c 0 7

39 No-liear regressio i Excel I Excel you ca fid uder graphics - tredlie followig o-liear regressio models Model Formula Expoetial d defree polyimial y = A e b x y = a x + b x + c Logarithmic model y = A log(b x) 8

40 Formulas A. Grouped data mea, stdev, media, mode f i xi x f i = frequecy, x i = class ceter s fi xi ( fixi ) = sample size ( 1) CF ME L i L = lower limit of media class, =sample size f CF = cumulative frequecy of classes below media class, f = frequecy of media class i = width of media class f L f f L f i 1 i i 1 i 1 Md L i, L i+1 = lower limit of modal class i 1 i 1 f i-1, f i+1 = frequecies of classes eighborig to modal class B. Cofidece itervals CI of Normal distributio ( x t s, x t s ) s=sample stadard deviatio = sample size = sample mea CI of relative proportio ( p t p(1 p), p t p(1 p) ) p = relative proportio i the sample = sample size

1 Inferential Methods for Correlation and Regression Analysis

1 Inferential Methods for Correlation and Regression Analysis 1 Iferetial Methods for Correlatio ad Regressio Aalysis I the chapter o Correlatio ad Regressio Aalysis tools for describig bivariate cotiuous data were itroduced. The sample Pearso Correlatio Coefficiet