Thomas Whitham Form Centre

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1 Thomas Whitham Form Cetre Statistics Module

2 . The Uiform Distributio (or rectagular distributio) U(a, b) a b Pdf f x a x b a b otherwise cdf E F x a b var, b a x a, b a, by symmetry x a a x b x b (proof by itegratio highly ulikely!) Measuremets made i cm are recorded correct to the earest cm. Let = true legth recorded legth. U(.5,.5) A strig AB of legth cm is cut at a radom poit. Let be the legth of the loger portio. U(5, )

3 U(, k) ad (i) the value of k (ii) P / x (iii) E, var, f x i the specified iterval. Fid 6 E ad E (i) k 6 6 k k 4 (ii) P / x - 4 (usig reduced domai) (iii) E var 6 E E 4 E E E 8 E. The Normal Distributio N(, ) E var The pdf is ot required so wo t be metioed here. The graph is symmetrical ad bell shaped. The theoretical distributio takes all values of from observatios lie withi of the mea. to but almost all It provides a model for data arisig out of aturally occurrig pheomea.

4 is trasformed by Z so that Z N(, ) with areas ad therefore probabilities coserved. The distributio fuctio of Z is tabulated for values of Z P Z z I a separate table, values of Z are tabulated for some values of PZ z N(,6) Fid (i).4 quartile of z P (ii) the upper (i) z x P.4 PZ. PZ Z (ii).75 q x z q q Z.674

5 4 Normal approximatio to the Biomial distributio As, Bi(, p) N(p, pq), hece we might be thikig of usig a ormal approximatio whe is large. Other cosideratios are as follows (i) (ii) (iii) A good approximatio is give for smaller values of whe p is close to.5. However the larger the value of, the better the approximatio for ay value of p. p should ot be too close to or uless. Both p ad q should be greater tha 5 to give a good approximatio. I usig a ormal approximatio (cotiuous) to a biomial distributio (discrete) a cotiuity correctio has to be applied. A coi is biased so that the probability that it comes dow heads is double the probability that it falls tails. The coi is tossed 8 times. What is the probability that there will be o more tha 45 tails? O ay oe throw P(tails) = = umber of tails i 8 throws Bi 8, p N 6,4 approximately P 45 P 45.5N PZ.9 PZ x z

6 5 Normal approximatio to the Poisso distributio As, P o N,, ad so we look for large for a good approximatio. The larger the value of, the better the approximatio. A guide is 5; do t forget the cotiuity correctio. The umber of accidets o a certai railway occurs o average at the rate of oe every two moths. Give that the accidets occur radomly fid the probability that there will be at least accidets i 5 years. A Poisso model for the umber of accidets will be appropriate = umber of accidets i 5 years P o N, approximately P P 9.5N PZ.9 PZ z.9 4. Two or more radom variables (i) Additive property of two idepedet Poisso radom variables Where P ad Y P + Y P o Extesios follow to three or more radom variables NB Y does ot follow a Poisso distributio o o

7 6 Durig a weekday heavy lorries pass a cesus poit P o a village high street idepedetly ad at radom times. I ay miute period, a average of westward travellig lorries ad eastward travellig lorries pass P. Fid the probability that a total of lorries pass P i a miute period. Fid also the probability that, give lorries pass P i a miute period at least oe was travellig westward. W = umber of lorries passig P per mi W P Similarly E P o P, ad T W E 5! T e e. 4 P 5 o o P W (ii) / T P W T P T P W you will fid!, E or W, E, or W P T Sums ad differeces, icludig multiples, E If ad Y are ay two radom variables, discrete or cotiuous E Y E EY Whe ad Y are idepedet var Y var vary Further, for ay two radom variables ad Y Note + oly

8 (iii) (iv) E a by ae bey var 7 a by a var b vary, ad whe ad Y are idepedet Extesios of the above follow to or more radom variables ote the idepedece requiremet for variace. Idepedet observatios from oe distributio Let,...,, be radomly observed values of the radom variable where,... E E... var var Products where ad Y are ay idepedet radom variables Y E EY E. E E E. E. E Y Y Y Not E Y Lear the method for variace, rather tha formula var Y E Y E E EY EY Y E EY. E Y E Extesios follow to three or more idepedet radom variables ad Y are idepedet radom variables where U,5 ad Y U,7 idepedet observatios of Fid (a) E, var, E Y, Y (b) E5 Y, var Y (c) var var, E, E Y where,, are

9 8 (d) E Y ad var Y (e) E Y (a) E, var, E Y 5, var (b) E 5 Y 5E EY var 4 Y 4var 9varY 4 (c) var 9 var (d) Y E EY E, Y, E 7, E Y 79 var Y E. EY E EY (e) E E E. E. dy Y Y Y y 4 7. y dy (v) y 7 7 Liear combiatios of Normal radom variables, If N( ) ad Y N( ) the so log as ad Y are idepedet, a by Na b, a b,,...,, If N(, ) ad are idepedet observatios of... N, Extesios follow

10 9 A food maufacturer delivers tis of baked beas packed i cardboard boxes, each box cotaiig 4 tis. The mass F of a full ti of beas is ormally distributed with mea 5g ad stadard deviatio.5g. The mass B of a empty cardboard box is ormally distributed with mea 5g ad stadard deviatio.5g. Fid the probability that the total mass of a radomly chose box of 4 tis will exceed.4kg. F N 5,6.5 B N 59,.5 Let = mass of a box of 4 tis F F... F4 B E 4EF EB 9 var 4varF varb 5. 5 N 9,5.5 P 4 PZ.8 P z Furthermore, it is kow tha the mass E of a empty ti used by the maufacturer is ormally distributed with mea 55g ad stadard deviatio.5g. statig clearly ay assumptio regardig idepedece, fid the probability that the mass C of the cotets of a radomly chose ti of baked beas will exceed 45g. We have Mass of full ti F N 5,6.5

11 Now Mass of empty ti E N 55,.5 Mass of cotets C F C E NB it is easy to go wrog from here! EF EC EE 5 E C 55 E C 455 Also varf varc vare ad this is true sice the mass of a empty ti will be idepedet of the mass of the cotets varc. 5 varc NB writig C F E EC EF EE However varc varf vare is okay! is wrog! Why? It is because the weight of a full ti of beas F is ot idepedet of the weight of the empty ti E. 5. Samplig from N, Where N, i.e.,...,, are idepedet observatios from where, let deote the mea... It is a simple matter to show that E ad var

12 It follows that N, ad this is kow as the samplig distributio of the mea. The stadard deviatio of this samplig distributio is,4 P ad is otherwise kow as the stadard error of the mea A sample of size 6 is take from a ormal populatio i which N,4. What is the probability that the sample mea exceeds? N. N, 4 PZ. ad Y are idepedet radom variables where N,4.5 ad Y N,4. A sample of 9 observatios is take from ad a sample of 6 observatios is take from Y. Fid P Y z First N, ad Y N, The Y N, Now P Y P Y P Z.7.4 z.7

13 Approximately what sample size should be take from N,5 to esure that the sample mea is less tha.7 with 9% certaity? Let N,5 5 N, where is the sample size Require P Z.9.7 " z x " The Cetral Limit Theorem If,..., sample size = 6,, are idepedet radom variables ad Y the the distributio of Y will be approximately ormal for i large values of whatever the distributios of. If the i approx all have the same distributio i.8, the Y N,

14 . If the i are all ormal the Y will be ormal.. Where the i are radom ad idepedet observatios from N, N, approx It is this special case of the CLT which is so importat..4,.5. Fid the probability that 6 radomly observed values of will (i) Exceed i total 46 (ii) Have a mea less tha, (Assume that is a cotiuous radom variable) (i) T = N 446.4,8 approx 46 PZ.5 P T z 8.5 (ii) N.4,.65 P PZ z.65.6 Fid the probability that throws of a fair die will produce a mea score of 4 or more. Let = score o oe throw. r. 5 E {symmetry} P for r =,,, 4, 5, 6 6

15 var x px It follows that E. 5 ad var 5 5 so that N.5, 6 6 approx. However usig the distributio of to fid 4 cotiuity correctio of Cosider istead the distributio of E Y.5 5 var 5.5 ad that would be messy. 6 Y Y Y 5,87.5 Now 4 PY PY 9.5 N approx P would ivolve a P applyig a cotiuity correctio of.5 much easier to deal with the..(etc) 7. Hypothesis Tests Recall that the purpose of samplig is to discover characteristics of a populatio whe, for whatever reaso, ivestigatio of the etire populatio is ot feasible. A umerical characteristic of the populatio is called a parameter ad the correspodig umerical characteristic of a sample is called a statistic. A statistical hypothesis is a assertio cocerig a characteristic of a populatio. A hypothesis test is a procedure for establishig a set of rules, which lead to acceptace, or rejectio of a assertio. Specifically i this sectio we will coduct hypothesis tests o the values of from Bi, from P o

16 from N, 5 The p value approach (i) Defie the radom variable ad state its distributio (ii) State the Null hypothesis H ad the Alterative hypothesis H. State whether the test is oe tailed (lookig for defiite icrease or decrease) or two tailed (lookig for chage either way). (iii) State the distributio of the radom variable uder H i.e. give H is true. (iv) (v) Write dow the value, t say, of the test statistic give by the sample. For a oe sided test, the p value is equal to the probability that a value at least as extreme as t would occur if H were true. For a two sided test, the p value is twice that probability. Decisio guidelies are, (a) p value <. : Very strog evidece for rejectig H (b) p value <.5 : Strog evidece for rejectig H (c) p value >.5 : Isufficiet evidece for rejectig H NB For a p value >.5 we sometimes write, ad is see writte the decisio accept H. This is ot to say that H is true. A hypothesis test ca ot possibly determie the value of a parameter. It ca oly accept or reject a particular value with associated ucertaity.

17 6 I a horticultural experimet the survival rate for a certai type of plat was foud to be 4%. Whe the coditios of the experimet were chaged oly out of a sample of plat survived. Have the ew coditios lowered the survival rate? = Number of survivors out of Bi(=, ) H :.4 H :. 4 oe tailed test Uder H, Bi(=,.4) Test statistic is x p value = P /.4 =.84 > 5% Decisio Isufficiet evidece to reject H. The ew coditios have ot lowered the survival rate. Over a log period of time a machie has produced metal rods with a legth which has bee ormally distributed with mea 7mm ad stadard deviatio.mm. Periodically the machie is ispected. At the latest ispectio, to test whether the mea legth was still equal to 7mm, a radom sample of rods was take, ad their legths measured; the mea legth of the sample was 7.mm. What coclusio would be draw?

18 7 = legth of rods sample of N,.9 N,.9 H, N 7,.9 H : 7 H : 7 two tailed test Uder Test statistic is x 7. p value = P 7./ 7 = P Z. =.5 < 5% Decisio Strog evidece to reject mea legth of rods is ow bigger tha 7mm. H ad sice 7 x coclude that the The fire statio i Burham usually receives geuie callouts to fires at radom ad o average 64 per week. Durig the first week of the firema s strike there were 5 callouts. Are the good people of Burham takig more care? = weekly umber of geuie callouts P o H : 64 H : 64 oe tailed test Uder H, P 64 Approximately N 64,64 Test statistic is x 5 p value = Po o P 5/ 64 P 5. 5 = N = P Z. =.95 > 5% z 64.

19 8 Decisio Isufficiet evidece to reject H, so coclude that the mea umber of callouts has ot falle. Sigificace level The sigificace level of a test is the probability of rejectig H whe H is true, ad is usually give as a percetage. Commoly used sigificace levels are % ad 5% but others could well be used. I some tests, a predetermied sigificace level is set by the ivestigator, ad is a measure of what degree of risk she/he is prepared to accept i rejectig H whe H is true. I respect if % ad 5% levels of sigificace the followig statemets might be made. (i) (ii) Test at 5% level p-value >.5 H will ot be rejected. The test statistic could be described as beig ot sigificat. p-value <.5 H will be rejected. The test statistic could be described as beig sigificat. Test at % level p-value >. H will ot be rejected. The test statistic could be described as beig ot sigificat. p-value <. H will be rejected. The test statistic could be described as beig sigificat. Other predetermied levels of sigificace ca be, ad are used. If the sigificace level is % remember that P(rejectig H /H is true) = The p-value > H is ot rejected. p-value < H is rejected both at the % level of sigificace.

20 9 Tests ad decisios rules A test might br coducted o the basis of rejectig H if the test statistic is extreme. i.e. beyod a critical value (oe tail test) or outside critical values (two tail test). Suppose for example that ~N μ, σ with ukow. For samples of size take from, ~N μ, σ. (i) Uder H H : μ = μ H : μ > μ Oe tailed test at % level of sigificace % (ii) Uder H H : μ = μ Critical value reject H if x falls i this critical regio H : μ μ Two tailed test at % level of sigificace % reject H % reject H For each situatio the decisio rule is reject H if the test statistic falls ito the critical regio

21 A machie bags sugar with weights which follow a ormal distributio with mea kg ad stadard deviatio g. Periodically the machie is tested by takig a radom sample of 6 bags, weighig them ad computig the mea. If the sample mea lies betwee 99g ad g the machie is give a clea bill of health. Fid the probability of codemig the machie whe i fact it is workig properly? The decisio rule is to reject H if x > or x < 99 = weight of sugar i grammes 89 N N, for sample size 6 6, H : H : two tailed test at % level of sigificace Uder H, % 89 N, 6 % reject H 99 reject H P PZ.8 PZ z Sigificace level is approximately 7% usig this decisio rule ad

22 H / H true. 7 P rejectig The umber of emergecy calls per hour to a gas board over a weeked has a Poisso distributio with mea.7. It was suggested that the mea would be greater tha.7 over a very cold weeked. Over a very cold witer weeked the umber of emergecy calls was recorded. What decisio rule with a sigificace level as ear as possible to % would lead to the rejectio of the mea per hour beig.7 i favour of the hypothesis that the mea per hour is greater tha.7? = umber of emergecy calls i hours ~P λ Uder H ~P 7 H : λ =.7 H : λ >.7 Oe tailed test From tables P 9 =.85 P =.695 P =.95 P =.985 * P =.9467 P =.5 The decisio rule will be to reject H if ad the coclude that λ =.7

23 8. Iterval Estimatio If x is the mea of a radom sample of observatios from the radom variable where ~N μ, σ the through the distributio of will be N, x.96. a 95% (two sided, symmetrical) cofidece iterval for μ If is ot ormal, the so log as is large, the cetral limit theorem will tell us that will be approximately ormal The radom variable is ormally distributed with ukow mea cm ad stadard deviatio cm. a) A radom sample of 6 observatios of had values which summed to 8.4. Calculate a 95% cofidece iterval for the value of b) Fid the smallest possible sample size which would be take from for the width of the cofidece iterval for to be less tha cm. a) N,4;, N ; x % CI for = ,8.8 b) x x.96 4

24 4 Requires NB For Other % CI replace.96 with, for example.645 for 9% CI.6 for 98% CI 9. The differece of two meas The samplig distributio ca be writte as, ad will be ormal if ad are ormal, ad approximately ormal whe ad are large. Applicatio will be to probability problems, hypothesis tests ad iterval estimatio. A fruit grower uses two varieties of strawberries i the same field uder idetical coditios. Yields from both varieties are ormally distributed with variace 66 g but variety A produces o average 4g more per plat tha variety B. If 5 plats of each variety are take, what is the probability that the mea yield of variety A is 7g greater tha that of variety B? 66 ~ N, 5 = yield of A ~ N,66 66 ~ N, 5 = yield of B ~ N,66

25 ~ N, ~ N4,5.8 P PZ z 5.8. The fruit grower proposes to use a ew yield for his two varieties of strawberries, but is ot sure whether the soil coditios will affect the performace of each variety equally. He is prepared to accept that yield will ormally distributed ad the variace to be uaffected. He plated 5 of each variety i the ew field ad moitored yield. Variety A yielded o average 6g per plat ad variety B 96g. Usig a % level of sigificace would the grower coclude that ew soil coditios have affected comparative yield? ~ N,5.8 H : 4 H : 4 two tailed test Uder H ~ N4,5.8 Test statistic x x p-value = P / 4 P Z.74.8 %

26 5 Decisio will be to reject H at the % level of sigificace ad coclude that soil coditios have affected comparative yield. A compay has two factories producig a certai product. Daily outputs i toes are ormally distributed with factory havig a stadard deviatio f.5 toes ad factory Y havig a stadard deviatio of.6 toes. A check over days revealed mea productio figures of 5. toes ad.9 toes for ad Y respectively. Fid a symmetrical two sided 95% cofidece iterval for the differece of mea daily productio at the two factories. What decisio would be take i a 5% sigificace test H : Differece i meas is toes H : Differece i meas is ot toes Usig obvious otatio.5 ~ N x,.6 Y ~ N y, Y N,.6 ~ x y Sample value x y = 5..9 =.8 95% CI = = (.,.) Sice toes is ot i the 95% CI, H would be rejected at the 5% level of sigificace.

If, for instance, we were required to test whether the population mean μ could be equal to a certain value μ

If, for instance, we were required to test whether the population mean μ could be equal to a certain value μ STATISTICAL INFERENCE INTRODUCTION Statistical iferece is that brach of Statistics i which oe typically makes a statemet about a populatio based upo the results of a sample. I oesample testig, we essetially