OCR Statistics 2 Module Revisio Sheet The S2 exam is 1 hour 30 miutes log. You are allowed a graphics calculator. Before you go ito the exam make sureyou are fully aware of the cotets of theformula booklet you receive. Also be sure ot to paic; it is ot ucommo to get stuck o a questio (I ve bee there!). Just cotiue with what you ca do ad retur at the ed to the questio(s) you have foud hard. If you have time check all your work, especially the first questio you attempted...always a area proe to error. Cotiuous Radom Variables J.M.S. A cotiuous radom variable (crv) is usually described by meas of a probability desity fuctio (pdf) which is defied for all real x. It must satisfy f(x)dx = 1 ad f(x) 0 for all x. Probabilities are represeted by areas uder the pdf. For example the probability that X lies betwee a ad b is P(a < X < b) = b a f(x)dx. It is worth otig that for ay specific value of X, P(X = value) = 0 because the area of a sigle value is zero. The media is the value m such that m f(x)dx = 1 2. That is; the area uder the curve is cut i half at the value of the media. Similarly the lower quartile (Q 1 ) ad upper quartile (Q 3 ) are defied Q1 The expectatio of X is defied f(x)dx = 1 4 E(X) = ad Q3 xf(x)dx. f(x)dx = 3 4. Compare this to the discrete defiitio of xp(x = x). Always be o the lookout for symmetry i the distributio before carryig out a log itegral; it could save you a lot of time. You should therefore always sketch the distributio if you ca. The variace of X is defied Var(X) = x 2 f(x)dx µ 2. Agai, compare this to the discrete defiitio of x 2 P(X = x) µ 2. Do t forget to subtract µ 2 at the ed; someoe always does! The mai use for this chapter is to give you the basics you may eed for the ormal distributio. The ormal distributio is by far the most commo crv. www.mathshelper.co.uk 1 J.M.Stoe
The Normal Distributio The ormal distributio (also kow as the Gaussia distributio 1 ) is the most commo crv. It is foud ofte i ature; for example daffodil heights, huma IQs ad pig weights ca all be modelled by the ormal curve. A ormal distributio ca be summed up by two parameters; its mea (µ) ad its variace (σ 2 ). For a radom variable X we say X N(µ,σ 2 ). As with all crvs probabilities are give by areas; i.e. P(a < X < b) = b a f(x)dx. However the f(x) for a ormal distributio is complicated ad impossible to itegrate exactly. We therefore eed to use tables to help us. Sice there are a ifiite umber of N(µ,σ 2 ) distributios we use a special oe called the stadard ormal distributio. This is Z N(0,1 2 ). The tables give to you work out the areas to the left of a value. The otatio used is Φ(z) = z f(z)dz. So Φ(0.2) is the area to the left of 0.2 i the stadard ormal distributio. The tables do ot give Φ(egative value) so there are some tricks of the trade you must be comfortable with. These ad they are always helped by a sketch ad rememberig that the area uder the whole curve is oe. For example Φ(z) = 1 Φ( z) P(Z > z) = 1 Φ(z) Real ormal distributios are related to the stadard distributio by Z = X µ σ ( ). So if X N(30,16) ad we wat to aswer P(X > 24) we covert X = 24 to Z = (24 30)/4 = 1.5 ad aswer P(Z > 1.5) = P(Z < 1.5) = 0.9332. Aother example; If Y N(0,5 2 ) ad we wish to calculate P(90 < Y < 5). Covertig to P( 2 < Z < 1) usig. The fiish off with P( 2 < Z < 1) = Φ(1) Φ( 2) = Φ(1) (1 Φ(2)) = 0.8413 (1 0.9772) = 0.8185. You must also be able to do a reverse lookup from the table. Here you do t look up a area from a z value, but look up a z value from a area. For example fid a such that P(Z < a) = 0.65. Draw a sketch as to what this meas; to the left of some value a the area is 0.65. Therefore, reverse lookig up we discover a = 0.385. Harder example; Fid b such that P(Z > b) = 0.9. Agai a sketch shows us that the area to the right of b must be 0.9, so b must be egative. Cosiderig the sketch carefully, we discover P(Z < b) = 0.9, so reverse look up tells us b = 1.282, so b = 1.282. Reverse look up is the combied with i questios like this. For X N(µ,5 2 ) it is kow P(X < 20) = 0.8; fid µ. Here you will fid it easier if you draw both a sketch for the X ad also for Z ad markig o the importat poits. The z value by reverse look up is foud to be 0.842. Therefore by we obtai, 0.842 = (20 µ)/5, so µ = 15.79. 1 I do wish we would call it the Gaussia distributio. Carl Friedrich Gauss. Arguably the greatest mathematicia ever. Germa... www.mathshelper.co.uk 2 J.M.Stoe
Harder example; Y (µ,σ 2 ) you kow P(Y < 20) = 0.25 ad P(Y > 30) = 0.4. You should obtai two equatios; 0.674 = 20 µ σ ad 0.253 = 30 µ σ µ = 27.27 ad σ =.79. The biomial distributio ca sometimes be approximated by the ormal distributio. If X B(,p)adp > 5adq > 5thewecauseV N(p,pq)asaapproximatio. Because we are goig from a discrete distributio to a cotiuous, a cotiuity correctio must be used. For example if X B(90, 1 3 ) we ca see p = 30 > 5 ad q = 60 > 5 so we ca use V N(30,20). Some examples of the coversios: The Poisso Distributio P(X = 29) P(28.5 < V < 29.5), P(X > 25) P(V > 25.5), P(5 X < 40) P(4 1 2 < V < 391 2 ). The Poisso distributio is a discrete radom variable (like the biomial or geometric distributio). It is defied P(X = x) = e λλx x!. X ca take the values 0,1,2,... ad the probabilities deped o oly oe parameter, λ. Therefore we fid x 0 1 2 3... P(X = x) e λλ0 0! e λλ1 1! e λλ2 2! e λλ3 3!... For a Poisso distributio E(X) = Var(X) = λ. We write X Po(λ) As for the biomial we use tables to help us ad they are give (for various differet λs) i the form P(X x). So if λ = 5 ad we wish to discover P(X < 8) we do P(X < 8) = P(X 7) = 0.8666. Also ote that if we wat P(X 4) we would use the fact that probabilities sum to oe, so P(X 4) = 1 P(X 3) = 1 0.2650 = 0.7350. The Poisso distributio ca be used as a approximatio to the biomial distributio provided > 50 ad p < 5. If these coditios are met ad X B(,p) we use W Po(p). [No cotiuity correctio required sice we are approximatig a discrete by a discrete.] For example with X B(60, 1 30 ) both coditios are met ad we use W Po(2). Therefore some example of some calculatios: P(X 3) P(W 3) = 0.8571 (from tables) P(3 < X 7) P(3 < W 7) = P(W 7) P(W 3) = 0.9989 0.8571 = 0.1418. The ormal distributio ca be used as a approximatio to the to the Poisso distributio if λ > 15. So if X Po(λ) we use Y N(λ,λ). However, here we are approximatig a discrete by a cotiuous, so a cotiuity correctio must be applied. www.mathshelper.co.uk 3 J.M.Stoe
For example if X Po(50) we ca use Y N(50,50) sice λ > 15. To calculate P(X = 49) we would calculate (usig Z = (X µ)/σ) Similarly Samplig P(X = 49) P(48.5 < Y < 49.5) = P( 0.212 < Z < 0.071) = P(0.071 < Z < 0.212) = Φ(0.212) Φ(0.071) = 0.5840 0.5283 = 0.0557. P(X < 55) P(Y < 54.5) ( = P Z < 54.5 50 ) 50 = P(Z < 0.6364) = 0.738. If a sample is take from a uderlyig populatio you ca view the mea of this sample as a radom variable i its ow right. This is a subtle poit ad you should dwell o it! If you ca t get to sleep sometime, you should lie awake thikig about it. (I had to.) If the uderlyig populatio has E(X) = µ ad Var(X) = σ 2, the the distributio of the mea of the sample, X, is E( X) = µ (the same as the uderlyig) ad Var( X) = σ2. This meas that the larger your sample, the less likely it is that the mea of this sample is a log way from the populatio mea. So if you are takig a sample, make it as big as you ca! If your sample is sufficietly large (roughly > 30) the cetral limit theorem (CLT) states that the distributio of the sample mea is approximated by ) X N (µ, σ2 o matter what the uderlyig distributio is. If the uderlyig populatio is discrete you eed to iclude a 1 2 correctio factor whe usig the CLT. For example P( X > 3.4) for a discrete uderlyig with a sample size of 45 would mea you calculate P( X > 3.4+ 1 90 ). If the uderlyig populatio is a ormal distributio the o matter how large the sample is (e.g. just 4) we ca say ) X N (µ, σ2. If you have the whole populatio data available to you the to calculate the mea you use µ = ad to calculate the variace you use x σ 2 = x 2 x2 = x 2 x 2. www.mathshelper.co.uk 4 J.M.Stoe
However you do ot usually have all the data. It is more likely that you merely have a sample from the populatio. From this sample you may wat to estimate the populatio mea ad variace. As you would expect your best estimate of the populatio mea is the mea of the sample x. However the best estimate of the populatio variace is ot the variace of the sample. You must calculate s 2 where x s 2 2 x 2 = 1 = 1 ( x 2 x 2 ) = 1 ( x 2 x2 Sometextbooksuse ˆσ tomeas; theybothmea theubiasedestimatorofthepopulatio σ. So (Estimate of populatio variace) = (Sample variace). 1 You could be give raw data ({x 1,x 2,...x }) i which you just do a direct calculatio. Or summary data ( x 2, x ad ). Or you could be give the sample variace ad. From all of these you should be able to calculate s 2. It should be clear from the above sectio how to do this. Cotiuous Hypothesis Testig I ay hypothesis test you will be testig a ull hypothesis H 0 agaist a alterative hypothesis H 1. I S2, your H 0 will oly ever be oe of these three: H 0 : p =somethig H 0 : λ =somethig H 0 : µ =somethig Do t deviate from this ad you ca t go wrog. Notice that it does ot say H 0 = p =somethig. The book gives three approaches to cotiuous hypothesis testig, but they are all essetially the same. You always compare the probability of what you have see (uder H 0 ) ad aythig more extreme, ad compare this probability to the sigificace level. If it is less tha the sigificace level, the you reject H 0 ad if it is greater, the you accept H 0. Remember we coect the real (X) world to the stadard (Z) world usig Z = X µ σ. You ca do this by: 1. Calculatig the probability of the observed value ad aythig more extreme ad comparig to the sigificace level. 2. Fidig the critical Z-values for the test ad fidig the Z-value for the observed evet ad comparig. (e.g. critical Z-values of 1.96 ad 1.96; if observed Z is 1.90 we accept H 0 ; if observed is 2.11 the reject H 0.) 3. Fidig the critical values for X. For example critical values might be 17 ad 20. If X lies betwee them the accept H 0 ; else reject H 0. Example: P111 Que 8. Usig method 3 from above. Let X be the amout of magesium i a bottle. We are told X N(µ,0.18 2 ). We are takig a sample of size, so X N(µ, 0.182 ). Clearly H 0 : µ = 6.8 H 1 : µ 6.8. ). www.mathshelper.co.uk 5 J.M.Stoe
We proceed assumig H 0 is correct. Uder H 0, X N(6.8, 0.18 2 ). This is a 5% two-tailed test, so we eed 2 1 2 % at each ed of our ormal distributio. The critical Z values are (by reverse lookup) Z crit = ±1.960. To fid how these relate to X crit we covert thus Z crit = X crit µ σ 2 1.960 = X crit 6.8 0.18 2 ad 1.960 = X crit 6.8 0.18 2 These solve to X crit = 6.912 ad X crit = 6.688. The observed X is 6.92 which lies just outside the acceptace regio. We therefore reject H 0 ad coclude that the amout of magesium per bottle is probably differet to 6.8. [The book is i error i claimig that we coclude it is bigger tha 6.8.] Discrete Hypothesis Testig For ay test with discrete variables, it is usually best to fid the critical value(s) for the test you have set ad hece the critical regio. The critical value is the firstvalue at which you would reject the ull hypothesis. For example if testig X B(16,p) we may test (at the 5% level) H 0 : p = 5 6 H 1 : p < 5 6. We are lookig for the value at the lower ed of the distributio (remember the < acts as a arrow tellig us where to look i the distributio). We fid P(X 11) = 0.1134 ad P(X ) = 0.0378. Therefore the critical value is. Thus the critical regio is {0,1,2...9,}. So whe the result for the experimet is aouced, if it lies i the critical regio, we reject H 0, else accept H 0. Aother example: If testig X B(20,p) at the % level with H 0 : p = 1 6 H 1 : p 1 6. Here we have a two tailed test with 5% at either ed of the distributio. At the lower ed we fid P(X = 0) = 0.0261 ad P(X 1) = 0.1304 so the critical value is 0 at the lower ed. At the upper ed we fid P(X 5) = 0.8982 ad P(X 6) = 0.9629. Therefore P(X 6) = 1 P(X 5) = 1 0.8982 = 0.18 P(X 7) = 1 P(X 6) = 1 0.9629 = 0.0371 So at the upper ed we fid X = 7 to be the critical value. [Remember that at the upper ed, the critical value is always oe more tha the upper of the two values where the gap occurs; here the gap was betwee 5 ad 6 i the tables, so 7 is the critical value.] The critical regio is therefore {0,7,8...20}. There is a Poisso example i the Errors i hypothesis testig sectio. www.mathshelper.co.uk 6 J.M.Stoe
Errors I Hypothesis Testig A Type I error is made what a true ull hypothesis is rejected. A Type II error is made whe a false ull hypothesis is accepted. For cotiuous hypothesis tests, the P(Type I error) is just the sigificace level of the test. [This fact should be obvious; if ot thik about it harder!] For a Type II error, you must cosider somethig like the example o page 140/1 which is superbly explaied. From the origial test, you will have discovered the acceptace ad the rejectio regio(s). Whe you are told the real mea of the distributio ad asked to calculate the P(Type II error), you must use the ew, real mea ad the old stadard deviatio (with a ew ormal distributio; e.g. N(µ ew,σold 2 /)) ad work out the probability that the value lies withi the old acceptace regio. [Agai, the book is very good o this ad my explaatio is poor.] For discrete hypothesis tests, the P(Type I error) is ot merely the stated sigificace level of the test. The stated value (e.g. 5%) is merely the otioal value of the test. The true sigificace level of the test (ad, therefore, the P(Type I error)) is the probability of all the values i the rejectio regio, give the truth of the ull hypothesis. For example i a biomial hypothesis test we might have discovered the rejectio regio wasx 3adX 16. Iftheullhypothesiswas H 0 : p = 0.3, thethetruesigificace level of the test would be P(X 3 or X 16 p = 0.3). To calculate P(Type II error) you would, give the true value for p (or λ for Poisso), calculate the probability of the complemetary evet. So i the above example, if the true value of p was show to be 0.4, you would calculate P(3 < X < 16 p = 0.4). Worked example for Poisso: A hypothesis is carried out to test the followig: H 0 : λ = 7 H 1 : λ 7 α = % Two tailed test. Uder H 0, X Po(7). We discover the critical values are X = 2 ad X = 13. The critical regio is therefore X 2 ad X 13. Therefore P(Type I error) ad the true value of the test is therefore P(X 2 or X 13 λ = 7) = P(X 2)+P(X 13) = P(X 2)+1 P(X 12) = 0.0296+1 0.9730 = 0.0566 = 5.66%. Give that the true value of λ was show to be, the P(Type II error) would be P(2 < X < 13 λ = ) = P(X 12) P(X 2) = 0.7916 0.0028 = 0.7888 = 78.88%. www.mathshelper.co.uk 7 J.M.Stoe