Math Third Midterm Exam November 17, 2010

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1 Math Treibergs σιι Third Midterm Exam Name: November 17, 1 1. Suppose that the mahie fillig mii-boxes of Fruitlad Raisis fills boxes so that the weight of the boxes has a populatio mea µ x = 14.1 grams ad a stadard deviatio σ x = 1.4 grams. Suppose that a radom sample of 49 boxes is seleted. What is the probability that the sample mea will be above 14. grams? There is a 99% hae that the sample mea will be above what value? What approximatio are you makig to aswer these questios ad why is it valid? Sie = 49 > 3, by the rule of thumb, we may assume that the sample mea is ormally distributed X N(µ x, σx ). Thus P ( X > 14.) = 1 P ( X 14.) ( = 1 P Z = X ) µ x σ x / / 49 =.5 = 1 Φ(.5) = = To fid x r so that with 99% probability the sample mea is above this umber, solve.99 = P ( X > x r ) = 1 P ( ) X xr ( = 1 P Z = X µ x σ x / x ) r µ x σ x / ( ) xr µ x = 1 Φ σ x / Remember that.1 = P (Z z.1 ) so that ( ) xr µ x.1 = Φ σ x / = Φ( z.1 ). Hee, so x r µ x σ x / = z 1 x r = µ x z 1 σ x = 14.1 (.36) = The approximatio used is that X N(µ x, σx ) beause of the Cetral Limit Theorem, whih aordig to the rule of thumb is aeptable for > 3. 1

2 . Let X {, 1,, 3} be the umber of ars ad Y {, 1, } be the umber of pikups i the Big Ed s parkig lot at oo. The joit pmf is give by p(x, y). What is the probability that X Y? What is the expeted umber of pikups E(Y )? Give that X = 1, determie the oditioal pmf of Y i.e., fid p Y X ( 1), p Y X (1 1), p Y X ( 1). Give that oe ar is i the lot, what is the probability that there is at least oe pikup? Are the umber of ars X ad the umber of pikups Y idepedet r.v. s? p(x, y) 1 3 x.1.1. y The evet that X Y osists of the pairs {(, ), (, 1), (, ), (1, 1), (1, ), (, )} whose probability is P = p(, ) + p(, 1) + p(, ) + p(1, 1) + p(1, ) + p(, ) = =.6. The margial probability for Y is gotte by summig rows, so p Y () = p Y (1) =.4 ad p Y () =.. Thus the expeted umber of pikups is E(Y ) = y p Y (y) = ()(.4) + (1)(.4) + ()(.4) =.8. y= The margial probability for X is gotte by summig olums, so p X () =.3, p X (1) =.4, p X () =.1 ad p X (3) =.. The oditioal probability is give by the formula Thus, p(x, y) p Y X (y x) = p X (x) p Y X (1 1) = so p Y X ( 1) = p(1, 1) p X (1) =.3.4 =.75, p Y X( 1) = p(1, ) p X (1) =.1.4 =.5, p(1, ) p X (1) =.4 =. P (Y 1 X = 1) = p Y X (1 1) + p Y X ( 1) =.75 + =.75. Fially, X ad Y are ot idepedet, beause, e.g., p X ()p Y () = (.3)(.4) =.1 = p(, ).

3 3. I a study of poisos i the eviromet, the followig measuremets were made of groudwater arsei i µg/l at a ertai ladfill. The sample mea ad stadard deviatio as foud by R are X = 4.4 ad s = Fid a two sided 9% ofidee iterval for µ, the populatio mea arsei level. What did you assume about the data i aswerig (a.)? Usig the probability plot geerated by R, ommet o the validity of your assumptios I this ase, there are = 19 observatios, whih is a small sample. Hee we use a CI based o the T -distributio. There are ν = 1 = 18 degrees of freedom. Thus for the 9% ofidee level, α =.1 ad the two sided ritial value is t α/,ν = t.5,18 = from Table A.5. The two-sided CI is thus s X ± t α/,ν = 4.4 ± (1.734) 4.91 = (.9, 6.19). 19 The data is assumed to be approximately ormally distributed. The poits lie up iely i the probability plot, so the assumptio that the data is approximately ormal is reasoable. (I fat, a ormal radom umber geerator was used to geerate this data. For small, this probability plot is as straight as it gets for ormal data.) 3

4 4. Suppose that reatio times are uiformly distributed i the iterval [, ]. Let X, Y be a radom sample of reatio times. Cosider two statistis ˆθ 1 ad ˆθ. Show that ˆθ 1 is a ubiased estimator for. Show that ˆθ is a ubiased estimator for. Fid the stadard error of ˆθ 1. [The stadard error for ˆθ is = σˆθ 3.] Whih is the better estimator for? Why? ˆθ 1 = X + Y, ˆθ = 3 max{x, Y }. For the uiform distributio o the iterval [, ], the pdf f(x) = 1/ if x ad f(x) = otherwise. The mea is µ = ad the expeted square ad variae is E(X ) = x dx R x f(x) dx = x dx =, = 3, σ = E(X ) E (X) = 3 4 = 1. X ad Y are idepedet variables take from this distributio. ˆθ 1 is ubiased sie E(ˆθ 1 ) = E(X + Y ) = E(X) + E(Y ) = + =. We ote that ˆθ = X if X Y ad ˆθ = Y if X Y. Thus, it is ubiased sie E(ˆθ ) = 3 The stadard error of ˆθ 1 is max{x, y} dy dx = 3 x σˆθ1 = V (X + Y ) = V (X) + V (Y ) = x dy dx x dy dx = 3 = = 6. The better estimator of two ubiased estimators has the smaller stadard error. Thus ˆθ is better tha ˆθ 1 beause = > σˆθ1 6 3 =. σˆθ 4

5 5. I a radom sample of Poatello telephoe ustomers who had trouble with their servie, 1 said that the problem was fixed the same day that it was reported. Estimate the true proportio p of the ustomers whose problem was fixed the same day that it was reported. What is the stadard error of the estimator you used? Fid a 9% lower ofidee boud for the true proportio p of phoe problems that are fixed o the same day. Let X be the umber of respodets who reported that their phoe problems were fixed o the same day. The estimate of p, the true proportio of phoe problems that are fixed o the same day is ˆp = X = 1 =.6. The estimated stadard error is σˆp = V (ˆp) = V ( ) X p q = ˆσˆp = ˆp ˆq (.6)(.4) = =.11. Sie ˆq = 8 is ot larger tha 1, we must use the urestrited oe sided CI for proportio. For α =.1, the ritial oe-sided z value is z α = z.1 = 1.8. Thus with 9% ofidee, a lower ofidee boud for the true proportio p of phoe problems that are fixed o the same day is give by p > ˆp + z α ˆp ˆq z α + z α z α =.6 + (1.8) 4 (1.8) (.6)(.4) 1 + (1.8) + (1.8) 16 =