Continuous Data that can take on any real number (time/length) based on sample data. Categorical data can only be named or categorised
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1 Questio 1. (Topics 1-3) A populatio cosists of all the members of a group about which you wat to draw a coclusio (Greek letters (μ, σ, Ν) are used) A sample is the portio of the populatio selected for aalysis (Roma letter (x, s, ) are used for sample data) A parameter is a umerical measure that describes a characteristic of a populatio A statistic is a umerical measure that describes a characteristic of a sample rage Class itervals: Width of iterval Arithmetic Mea: X X 1+X 2 + X Media (Positio): +1 Rage X o.of desired class groupigs 2 max X mi Z Score: Z X X Z Outliers > 3.0 or <-3.0 S Measures of Cetral Tedecy: Arithmetic Mea, Media, Mode Quartile (Positio): Q ( + 1), Q ( + 1), Q ( + 1) Iter-Quartile Rage: IQR Q 3 Q 1 Measures of Dispersio: Variace, Stadard Deviatio, Coefficiet of Variatio Covariace tells us oly the directio of associatio Sample coefficiet of correlatio r: r covar where s s x s x & s y S.Dev formula y Numerical Descriptive Measures Reordered data: 3, 4, 7, 9 Variace: firstly fid x 5.75 s 2 i1 (x x )2 Sample Variace 1 [(3 7) 2 + (4 7) 2 + (7 7) 2 + (9 7) 2 ] [( 4)2 + ( 3) 2 + (0) 2 + (2) 2 ] Stadard deviatio: s s Coefficiet of variatio: CV s x 100% % 61.7% 4 Questio 3 Cotiuous Probability Distributio Fid the followig probabilities 1. P(Z < 1.67) Read straight from the table. Note: P(Z<1.846) we ca oly look up z values to two decimal places so roud up to P(Z > 2.78)? 1 P(Z < 2.78)? P(0.15 < Z < 1.99)? P(Z < 1.99) P(Z < 0.15) Solve the followig iverse problems for the stadard ormal distributio P(Z > ) 0.01 Look up the Iverse Normal Table P(Z > ) 0.01 The Iverse table oly gives the Z values for upper-tail areas, but because the ormal distributio is symmetric about zero, we fid the upper-tail Z value, ad the lower-tail Z value that we eed is the same value but egative. Fid the two values of Z (symmetrically distributed aroud the mea) such that the followig statemets are true: P( < Z < ) 0.80 Each tail will have a area of 0.10, so lookig up the Iverse table to get the two Z values: Z LOWER Z UPPER P( < Z < ) 0.80 Samplig Distributio cot. Numerical data is measured o a atural umerical scale (age) Iferetial Statistics - Drawig coclusios about a populatio Cotiuous Data that ca take o ay real umber (time/legth) based o sample data Discrete - Coutable umber of resposes (caot have 0.5) Frequecy Distributios - summary table i which data are Categorical data ca oly be amed or categorised arraged ito umerically ordered classes or itervals Nomial o order, o respose is cosidered better (geder) Ordered array: sequece of data i rak order Ordial There is a order (very good, good, average) Time Series Data collect through time (Moths sales for May) Descriptive Statistics - Collect, Preset, Characterise data Cross Sectioal Collected for a poit i time (My height today) Sample of 4: (2, 3), (7, 9), (4, 5), (4, 6) x y x y (x x ) (y y ) (x x )(y y ) (x x )(y y ) (x x )(y y ) covariace (Directio) correlatio r covar 5.09 s x s y (Stregth) Cotiuous Probability Distributio cot. Betwee what two values of Z (symmetrically distributed aroud the mea) will 68.26% of all possible Z values be cotaied? Each tail has a area, α (i.e. ( )/2, so if we use the Cumulative Normal Distributio table ad look for the area of , we fid that P(Z < -1) Therefore the right tail where Z +1 has the same area. So the two values of Z that we are lookig for are -1 ad +1. i.e. P( -1 < Z < 1) as i the diagram. Usig Iverse Normal table, oly look up a area to two decimal places: 0.16 (i.e rouded to two decimal places) ad we would coclude that the two values of Z were Z ad Z i.e. P( < Z < ) 0.68 Samplig Distributio cot. I Iterpretig Correlatio Coefficiet r Iterpretatio r -1 PERFECT egative liear -1 < r -0.7 STRONG egative liear -0.7 < r -0.3 MODERATE egative liear -0.3 < r < 0 WEAK egative liear r 0 No relatioship 0 < r < 0.3 WEAK positive liear 0.3 r < 0.7 MODERATE positive liear 0.7 r < 1 STRONG positive liear 1 PERFECT positive liear Populatio mea μ Sample mea - X Populatio variace - 2 Sample Proportio p Stadard Deviatio S Variace S 2 Questio 4. Samplig Distributio Estimatio cot. / Cofidece Itervals. Estimatio Studet Name: Studet No: Is it for μ? No X 2 ( 1) 2 Yes Is kow? No t X μ 2 S Yes Quatitative Z X μ Qualitative Z p π π(1 π)
2 Questio 2 Simple Liear Regressio & Probability Probability & Discrete Probability Distributios Probability & Discrete Probability Distributios Biomial Distributio (Questio will provide, x ad % (portio) L Questio 5 Hypothesis testig Hypothesis Testig cot. Two populatio Proportio Example Two Sample (Rejectio regio use iverse ormal table) Pooled-Variace t Test Example Two Sample (Sigma Ukow, Variace Equal, Assume 30mi (Cetral Limit T) F Test Example Two Sample (F table for reject regios) (t0.05, 1998) df FL 1 Fu Fu F 0.025, 99, 71 F 0.025, 60, Fu* F 0.025, 71, 90 F 0.025, 60, Aalysis of Variace (ANOVA)
3 BSB123 Data Aalysis Semester Workshop 8 (Week 10) Estimatio Questio 1 The quality cotrol maager at a light bulb factory eeds to estimate that mea life of a large shipmet of light bulbs. The stadard deviatio is 100 hours. A radom sample of 64 light bulbs idicates a sample mea life of 350 hours. (a) Costruct a 95% cofidece iterval estimate of the populatio mea life of light bulbs i this shipmet. (b) Do you thik that the maufacturer has the right to state that the light bulbs last a average of 400 hours? Explai. The first approach is purely to say it s outside the cofidece iterval. The secod approach is to take that value of 400 covert it to a Z value, so you ca determie the probability that the statemet is correct.
4 (c) Must you assume that the populatio of light bulb life is ormally distributed? Explai. No because my sample size is >30. Therefore accordig to the CLT (cetral limit theorem) at the very least I will ed up with approximate ormal distributio I other words if we have 30 observatios or more, uder the CLT we have a Normal Questio 2 If X 75, S 24, 36, ad assumig that the populatio is ormally distributed, costruct a 95% cofidece iterval estimate of the populatio mea μ.
5 Questio 3 A study coducted by the Australia Stock Exchage foud that 46% of 2,405 Australia adults surveyed i 2006 held shares, either directly or idirectly through maaged fuds or self-maaged superauatio fuds (2006 Australia Share Owership Study, ASX). (a) Costruct a 95% cofidece iterval for the proportio of Australia adults who held shares i Whe dealig with populatios proportios we always use a Z. (b) Iterpret the iterval costructed i (a). As above. I am 95% cofidet that the true proportio of Australia adults who held shares i 2006 is betwee 44 ad 48% (c) To costruct a follow-up study to estimate the populatio proportio of adults who curretly hold shares to withi 0.01 with 95% cofidece, how may adults would you iterview?
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